A Theory of Truthmaker Content I: Conjunction, Disjunction and Negation

Abstract

I develop a basic theory of content within the framework of truthmaker semantics and, in the second part, consider some of the applications to subject matter, common content, logical subtraction and ground.

Appendices

Appendix: Formal Appendix

Preliminaries

Recall that $\sqsubseteq$ is a partial order (po) on the set S if it is a reflexive, transitive and anti-symmetric relation on S. Given a po $\sqsubseteq$ on S, we shall make use of the following (mostly standard) definitions. Where s,t,u ∈ S and T ⊆ S:

• s is an upper bound of T if $t \sqsubseteq s$ for each t ∈ T;

• s is a least upper bound (lub) of T if s is an upper bound of T and $s \sqsubseteq s^{\prime }$ for any upper bound s ^′ of T;

• s is null if $s \sqsubseteq s^{\prime }$ for each s ^′∈ S and other wise is non-null;

• s is full if $s^{\prime }\sqsubseteq s$ for each s ^′∈ S;

• $s\sqsubset t$ (s is a proper part of t) if $s \sqsubseteq t$ but not $t \sqsubseteq s$;

• s overlaps t if for some non-null u, $u \sqsubseteq s$ and $u \sqsubseteq t$;

• s is disjoint from t if s does not overlap t.

An (unmodalized) state space S is a pair (S, $\sqsubseteq )$, where S (states) is a non-empty set and $\sqsubseteq$ is a binary relation on S subject to the following two conditions:

Partial Order (PO)

$\sqsubseteq$ is a po on S;

Completeness

Any subset of S has a least upper bound.

The least upper bound of T ⊆ S is unique (since if s and s ^′ are least upper bounds, then $s \sqsubseteq s^{\prime }$ and $s^{\prime }\sqsubseteq s$ and so, by anti-symmetry, s = s ^′). We denote it by $\bigsqcup T$ and call it the fusion of T (or of the members of T). When T = {t ₁,t ₂,...}, we shall often write $\bigsqcup T$ more perspicuously as t ₁ ⊔ t ₂ ⊔ ... . $\bigsqcup \emptyset$, which we denote by $\square$, is the bottom element of the space and $\bigsqcup S$, which we denote by ■, is the top element. Given that the least upper bound $\bigsqcup T$ always exists, it follows that the greatest lower bound $\bigsqcup T$ will always exist (defined by $\bigsqcup \{ s\!\!:~\mathrm {for~all}~t \in T, s \sqsubseteq t\})$.

A space S is said to be distributive if s ⊓ (t ₁ ⊔ t ₂ ⊔ ...) = (s ⊓ t ₁) ⊔ (s ⊓ t ₂) ⊔ ... for any s, t ₁, t ₂, ... ∈ S. In what follows, we shall assume that the space is distributive, although many of our results will not depend upon this assumption.

In a distributive space, the following principle will hold:

Overlap

If s overlaps t ₁ ⊔ t ₂ ⊔ ... then it overlaps some t _i.

For if s overlaps t ₁ ⊔ t ₂ ⊔ ..., then s ⊓ (t ₁ ⊔ t ₂ ⊔ ...) is non-null; so (s ⊓ t ₁) ⊔ (s ⊓ t ₂) ⊔ ... is non-null; so some s ⊓ t _i is non-null; and so s overlaps some t _i.

A modalized state space S - or M-space, for short - is an ordered triple (S, S ^♢, $\sqsubseteq )$, where (S, $\sqsubseteq )$ is an unmodalized state space and S ^♢ (possible states) is a non-empty subset of S subject to: Downward Closure t ∈ S ^♢ whenever s ∈ S ^♢ and $t \sqsubseteq s$.

We say that a state s is consistent or possible if s ∈ S ^♢ and inconsistent or impossible otherwise. Note that $\square$ is guaranteed to be a possible state since, given that S ^♢ contains a member s, its part $\square$ is also a member. A subset T of states is said to be compatible if their fusion belongs to S ^♢ and to be incompatible otherwise. Corresponding to each modalized stated space (S, S ^♢, $\sqsubseteq )$ is, of course, an unmodalized state space (S, $\sqsubseteq )$.

Of special interest are spaces that contain counterparts of possible worlds. Say that the state s of a modalized space S = (S,S ^♢, $\sqsubseteq )$ is a world-state if it is consistent and if any consistent state is either a part of s or incompatible with s; and say that the space S itself is a W-space if every consistent state of S is part of a world-state. It is often helpful to think of ourselves as working within a W-space, although many of our results will not depend upon this assumption.

A particular kind of W-space may be constructed from the sentential atoms p ₁, p ₂, ... of some language. Let us denote the negation ¬p of an atom p by $\bar {\mathrm {p}}$ and call the atoms p and their negations $\bar {\mathrm {p}}$ literals. Then the (modalized) canonical space S _c over the atoms {p₁,p ₂,...} is the triple (S, S ^♢, $\sqsubseteq )$, where:

• (i) S = {L : L is a set of literals };

• (ii) S ^♢ = {L ∈ S: L does not contain both a sentence letter p and its negation $\bar {\mathrm {p}} \}$; and

• (iii) $\sqsubseteq ~\!\!\!\!\,= \{(K, L)$: K ⊆ L ⊆ S}.

In this case, we may write p, for example, in place of {p} or pq in place of {p,q}. It is readily verified that S _c is a W-space whose world-states are all sets of literals containing exactly one of p or p ^′ for each atom p.

Regular Propositions

Given a state space, we take a (unilateral) proposition to be a set of states (intuitively, the set of states that verify the proposition). We denote unilateral propositions by the letters ‘ P’,‘ Q’,‘ R’ and the like. Such a proposition P ⊆ S is said to be:

• trivial if $\square \in P$

• verifiable if P is non-empty

• unverifiable or vacuous if P is empty

• closed (under fusion) if $\bigsqcup Q \in P$ for any non-empty subset Q of P

• convex if p ∈ P whenever q, r ∈ P and $q \sqsubseteq p \sqsubseteq r$,

• regular if it is closed and convex, and

• semi-regular if it is convex.

We shall use p and the like for an arbitrary member of P and p for $\bigsqcup P$. Given that P is non-empty and closed, p ∈ P and is the maximal verifier of P. We shall later identify p with the subject-matter of P.

There is a certain sense in which regular propositions occupy a span of logical space. Given R ⊆ S and t ∈ S, we take their span [ R, t] to be {s: for some r ∈ R, $r \sqsubseteq s \sqsubseteq t\}$. We call R the lower limit or base of the span and t its upper limit. We now have the following simple criteria for when a proposition is regular:

Lemma 1

The following conditions on the proposition P are equivalent:

• (i) P is regular

• (ii) P is convex and either P is empty or p ∈ P

• (iii) P = [ P , p]

• (iv) P is of the form[ R , t]for R ⊆ S and t ∈ S .

Proof

• (i) ⇒ (ii). Suppose P is regular. Then P is convex by definition; and, given that P is non-empty, p ∈ P by closure.

• (ii) ⇒ (iii). Suppose (ii). If P is empty then clearly P = [ P,p]. If p ∈ P,then P ⊆[ P,p] and [ P,p] ⊆ P by P convex.

• (iii) ⇒ (iv). Set R = P and t = p.

• (iv) ⇒ (i). Suppose P = [ R, t].Let p be the non-empty fusion p ₁ ⊔ p ₂ ⊔...for p ₁,p ₂,... ∈ P.Then for each i, there is an r _i ∈ R for which $r_{\mathrm {i}} \sqsubseteq p_{\mathrm {i}} \sqsubseteq t$.Hence $r_{\mathrm {i}} \sqsubseteq p \sqsubseteq t$and p ∈ P.Now suppose $p \sqsubseteq q \sqsubseteq r$with p, r ∈ P.Then $p^{\prime }\sqsubseteq p$for some p ^′∈ R and $r \sqsubseteq t$.Hence $p^{\prime }\sqsubseteq q \sqsubseteq t$and so q ∈ P.

Note that a regular proposition P may be identical to many different spans [ R, t]. The upper limit t of the span must always be p (for P non-empty). The largest lower limit R ⊆ P is P itself, though there will in general be many lower limits and not necessarily any smallest lower limit.

The above lemma provides us with a limited form of monotonicity for regular propositions - given a verifier p, any extension $q \sqsupseteq p$ will also be a verifier as long as it lies within the subject-matter p of the proposition. Thus the current view of propositions provides us with a kind of compromise between insistence on exact relevance without any form of monotonicity and adherence to an unrestricted form of monotonicity.

Given a proposition P, we use P ^∗ for the smallest proposition Q ⊇ P to be closed under fusion, P _∗ for the smallest proposition Q ⊇ P to be convex, and $P_{{\kern 1.5pt}\ast }^{\ast }$ for the smallest regular proposition Q to contain P.

We set:

• $T_{\square } = \{\square \}$

• F _■ = {■}

• $F_{\square } = \emptyset$, and

• T _■ = S.

We call $T_{\square }$, T _■, $F_{\square }$ and F _■ the extremal propositions and readily verify that they are regular.

We turn to the bilateral case. A bilateral proposition P is an ordered pair (P, P ^′) where P and P ^′ are unilateral propositions. Intuitively, P is the set of verifiers of the proposition and P ^′ its set of falsifiers. We shall often use P for the first component of P, its positive content, and P ^′ for the second component, its negative content (and similarly when subscripts or other letters are in place). Given P = (P, P ^′), we use $\boldsymbol {p}= \bigsqcup P$ for the positive subject-matter of P and p $^{\prime } = \bigsqcup P^{\prime }$ for the negative subject-matter of P.

The bilateral proposition P = (P, P ^′) is said to be convex (closed, regular, semi-regular) when both its positive content P and its negative content P ^′ are convex (closed, regular, semi-regular); and the bilateral proposition P = (P, P ^′) is said to be trivial (vacuous) if either P or P ^′ is trivial (vacuous).

We define bilateral analogues of the extremal propositions in the natural way:

• $\boldsymbol {T}_{\square } = (T_{\square }$, $F_{\square })$

• T _■ = (T _■, F _■)

• $\boldsymbol {F}_{\square } = (F_{\square }$, $T_{\square })$

• F _■ = (F _■, T _■).

Containment and Entailment

Given verifiable propositions P and Q from a state space, we say that Q contains P, or that P is a conjunctive part of Q, - in symbols, P≤_c Q - if (i) for all q ∈ Q there is a p ∈ P for which $q \sqsupseteq p$ and (ii) for all p ∈ P there is a q ∈ Q for which $p \sqsubseteq q$; and we say that P entails Q, or that P is a disjunctive part of Q, - in symbols, P≤_d Q - if P ⊆ Q. In what follows, we shall sometimes think of containment, in a mereological manner, as a relation from greater to lesser propositions and of entailment, in a logical manner, as a relation from stronger to weaker propositions and we shall sometimes drop the subscript from ≤_c, though never from ≤_d. Note that containment is only taken to be defined on verifiable propositions.

We readily verify the mereological behavior of the extremal propositions:

Lemma 2

For any proposition P ,

• (i) T _■≥_d P

• (ii) $F_{\square }\le _{\mathrm {d}} P$

• (iii) $F_{\blacksquare }\ngeq _{\mathrm {d}} T_{\square }$ and F _■≤_d T _■ in any non-singleton space S

• (iv) $T_{\square } \le _{\mathrm {c}} P$ for $P \ne F_{\square }$

• (v) F _■≥_c P for $P \ne F_{\square }$ .

Proof

(i) & (ii) are immediate from the definitions.

(iii) Since F _■and$T_{\square }$are non-empty and disjoint.

(iv) For each member p of P,$p \sqsupseteq \square$;and if P is non-empty and hence contains a member p,then $\square \sqsubseteq p$.

(v) For each member p of P,$p \sqsubseteq \blacksquare$;and if P is non-empty and hence contains a member p,then $\blacksquare \sqsupseteq p$.

Clauses (i) and (ii) show that T _■ and $F_{\square }$ are the top and bottom elements with respect to the disjunctive ordering ≤_d, clause (iii) shows that $T_{\square }$ and F _■ are incomparable with respect to the disjunctive ordering, and clauses (iv) and (v) show that $T_{\square }$ and F _■ are the bottom and top elements with respect to the conjunctive ordering ≤_c. It is important to note that ≤_d is defined on all propositions P while ≤_c is only defined on the propositions $P \ne F_{\square }$ (and so we do not have the usual bilattice structure).

A proposition P is said to be definite (or non-disjunctive) if it is of the form P = {p} for some state p and is otherwise said to be indefinite (or disjunctive). Definite propositions correspond to world-propositions within the possible worlds semantics and will later be of importance in the account of exclusionary negation. The criterion for containment simplifies when one of the propositions is definite:

Lemma 3

If both P and Q are verifiable and one is definite, then P ≥ Q iff for all p ∈ P and q ∈ Q , $p \sqsupseteq q$ .

Proof

Assume that P = {p ^′}and that Q contains the verifier q ^′(the other case is similar). Suppose P ≥ Q.Take any q ∈ Q.Then $p^{\prime }\sqsupseteq q$by clause (ii) in the definition of containment and so$p \sqsupseteq q$for any p ∈ P and q ∈ Q.Now suppose $p \sqsupseteq q$for any p ∈ P and q ∈ Q.Then $p^{\prime }\sqsupseteq q^{\prime }$and $q \sqsubseteq p^{\prime }$for each q ∈ Q;and so P ≥ Q.

Regular closure normally makes no difference to whether one proposition is contained in another:

Lemma 4

For verifiable propositions P and Q :

• (i) P ≥ Q implies $P_{\ast }^{\ast } \ge Q_{\ast }^{\ast }$ , and

• (ii) $P^{\ast }_{\ast } \ge Q_{\ast }^{\ast }$ implies P ≥ Q for closed P and Q .

Proof

• (i) Suppose P ≥ Q.Take $p^{\prime }\in P_{{\kern 1.2pt}\ast }^{\ast } =$[ P,p]. Then for some p ∈ P,$p \sqsubseteq p^{\prime }\sqsubseteq$ p. Since P ≥ Q,$p \sqsupseteq q$for some q ∈ Q and so $p^{\prime }\sqsupseteq q \in Q_{\ast }^{\ast }$.Now take $q^{\prime }\in Q^{\ast }_{\ast }$.Then $q^{\prime }\sqsubseteq$ q $= \bigsqcup Q = \bigsqcup Q^{\ast }_{\ast }$.But for each q _i ∈ Q there is a p _i ∈ P for which $q_{\mathrm {i}} \sqsubseteq p_{\mathrm {i}}$;and so $q^{\prime }\sqsubseteq \textbf {{q}}= q_{1}\, \sqcup \, q_{2}\, \sqcup \ldots \sqsubseteq p_{1}\, \sqcup \,p_{2}\, \sqcup \ldots \in P^{\ast }_{{\kern 1.3pt}\ast }$.

• (ii) Suppose $P^{\ast }_{{\kern 1.3pt}\ast } \ge Q^{\ast }_{\ast }$.Take $p^{\prime }\in P\subseteq P^{\ast }_{{\kern 1.3pt}\ast }$.Then $p^{\prime }\sqsupseteq q$for some $q \in Q_{\ast }^{\ast } =$[ Q,q], where $q^{\prime }\sqsubseteq q$for some q ^′∈ Q.But then $p^{\prime }\sqsupseteq q^{\prime }\in Q$.Now take $q^{\prime }\in Q \subseteq Q^{\ast }_{\ast }$.Then $q^{\prime }\sqsubseteq p$for some $p \in P_{{\kern 1.3pt}\ast }^{\ast } =$[ P,p], where $p \sqsubseteq$ p. But then $q^{\prime }\sqsubseteq$ p ∈ P,given that P is closed.

The relation ≥_d is anti-symmetric on all propositions and the relation ≥_c is anti-symmetric on all convex propositions:

Lemma 5

For arbitrary propositions P and Q,P≥_d Q and Q≥_d P implies P = Q ; and for convex propositions P and Q , P≥_c Q and Q≥_c P implies P = Q .

Proof

The result for ≥_dis immediate. For the case of ≥_c,suppose P≥_c Q and Q≥_c P and take p ∈ P.Since Q≥_c P,$p \sqsubseteq q$for some q ∈ Q.Since P≥_c Q,$p \sqsupseteq q^{\prime }$for some q ^′∈ Q.Thus $q^{\prime }\sqsubseteq p \sqsubseteq q$and so, by Q convex, p ∈ Q.The other direction is proved similarly.

The result for ≥_c does not hold for arbitrary propositions for, setting P = {p, pqr } and Q = {p, pq, pqr } within the canonical space, P ≥ Q and Q ≥ P. Part of the reason for insisting on convexity is to avoid having distinct propositions which cannot be distinguished in terms of containment.

We should note that, for regular propositions, the second clause in the definition of P ≥ Q may be simplified to q $\sqsubseteq$ p. For if q $\sqsubseteq p$ for some p ∈ P then q $\sqsubseteq$ $p \sqsubseteq \boldsymbol {p}\in P$ and if q $\sqsubseteq$ p then, for each q ∈ Q, $q \sqsubseteq$ q $\sqsubseteq \boldsymbol {p}\in P$. Thus the second clause amounts to the subject-matter of Q being part of the subject-matter of P; and so we can see containment as lying within the tradition of analytic implication initiated by Parry [21].

We turn to the bilateral case. Say that the bilateral proposition P = (P, P ^′) is verifiable if P is verifiable and falsifiable if P ^′ is verifiable. We may then extend the notions of containment and entailment to bilateral propositions P = (P, P ^′) and Q = (Q, Q ^′) as follows:

for verifiable P and Q, P≤_c Q - Q contains P - if P≤_c Q and P ^′≤_d Q ^′, and

for falsifiable P and Q, P≤_d Q - P entails Q - if P≤_d Q and P ^′≤_c Q ^′.

Note the restriction of ≤_c to verifiable propositions and the restriction of ≤_d to falsifiable propositions. If we wish to talk in the same breath about disjunctive and conjunctive part, then we should require the propositions under consideration to be non-vacuous (a property which will be preserved under negation and non-empty conjunction).

In analogy to lemma 5, we have:

Lemma 6

For convex bilateral propositions P and Q , P≥_d Q and Q ≥_d P implies P = Q and P≥_c Q and Q ≥_c P implies P = Q .

We also have the following analogue of lemma 2:

Lemma 7

For any bilateral proposition P = (P , P ^′),

• (i) T _■≥_d P for $P^{\prime }\ne F_{\square }$

• (ii) $\boldsymbol {F}_{\square } \le _{\mathrm {d}}$ P for $P^{\prime }\ne F_{\square }$

• (iii) $\boldsymbol {F}_{\blacksquare }\ngeq _{\mathrm {d}}$ T $_{\square }$ and $\boldsymbol {F}_{\blacksquare }\ngeq _{\mathrm {d}} \boldsymbol {T}_{\square }$ in any non-singleton space

• (iv) $\boldsymbol {T}_{\square }\le _{\mathrm {c}}$ P for $P \ne F_{\square }$

• (v) F _■≥_c P for $P \ne F_{\square }$

Proof

(i) $\boldsymbol {T}_{\square } = (T_{\square }$,$F_{\square })$and so we need to show that (a) $T_{\square } \ge _{\mathrm {d}} P$and that (b) $F_{\square } \ge _{\mathrm {c}} P$for $P \ne F_{\square }$.But (a) follows from lemma 2(i) and (b) from lemma 2(v). The other results are provedsimilarly.

Conjunction

Given the verifiable propositions P ₁, P ₂, ..., we let their conjunction P = P ₁ ∧ P ₂ ∧ ... be {p ₁ ⊔ p ₂ ⊔ ... : p ₁ ∈ P ₁, p ₂ ∈ P ₂, ... }. We do not take the conjunction P ₁ ∧ P ₂ ∧ ... to be defined when one or more of the propositions P ₁, P ₂, ... is not verifiable; and use of such expressions as ‘ P ₁ ∧ P ₂ ∧ ...’ will always presuppose that the context is one in which the conjunction is in fact defined. Note that, when there are no propositions P ₁, P ₂, ..., the resulting conjunction P will be the proposition $T_{\square }$.

Conjunction is indifferent to the order of the conjuncts and also to their repetition in the case of regular propositions:

Lemma 8

• (i) If the multi-sets[ P ₁ , P ₂ , ...] and[ Q ₁ , Q ₂ , ...] of propositions are the same then P ₁ ∧ P ₂ ∧ ... = Q ₁ ∧ Q ₂ ∧ ... ;

• (ii) P = P ∧ P ∧... for any non-empty repeating sequence P , P , ... of the closed proposition P ;

• (iii) If the sets {P ₁ , P ₂ , ... }and {Q ₁ , Q ₂ , ... }of regular propositions are the same then P ₁ ∧ P ₂ ∧ ... = Q ₁ ∧ Q ₂ ∧...

Proof

• (i) Evident from the definition of P = P ₁ ∧ P ₂ ∧... and Q ₁ ∧ Q ₂ ∧....

• (ii) Suppose p ∈ P.Then p ∈ P, p ∈ P,... and so p ∈ P ∧ P ∧.... Now suppose p ∈ P ∧ P ∧.... Then for p ₁ ∈ P, p ₂ ∈ P,..., p = p ₁ ⊔ p ₂ ⊔...; and so, by P closed, p ∈ P.

• (iii) Any multi-set [ P ₁, P ₂,...] of propositions can be put in the form[ P ₁, P ₁,..., P ₂, P ₂,..., ...], where the P ₁, P ₂,... are pairwise distinct. By (ii), each P _i ∧ P _i ∧... = P _iand so P ₁ ∧ P ₁ ∧... ∧ P ₂ ∧ P ₂ ∧... .... = P ₁ ∧ P ₂ ∧.... Suppose now that the sets {P ₁, P ₁,..., P ₂, P ₂,..., ... }and {Q ₁, Q ₁,..., Q ₂, Q ₂,..., ... }of regular propositions are the same. Then P ₁ ∧ P ₁ ∧... ∧ P ₂ ∧ P ₂ ∧... .... = P ₁ ∧ P ₂ ∧..., and Q ₁ ∧ Q ₁ ∧... ∧ Q ₂ ∧ Q ₂ ∧... ... = Q ₁ ∧ Q ₂ ∧..., and P ₁ ∧ P ₂ ∧... = Q ₁ ∧ Q ₂ ∧... by (i), given that [ P ₁, P ₂,...] = [ Q ₁, Q ₂,...]; and so P ₁ ∧ P ₁ ∧... ∧ P ₂ ∧ P ₂ ∧..., .... = Q ₁ ∧ Q ₁ ∧... ∧ Q ₂ ∧ Q ₂ ∧..., ... ..

The closure of a proposition can be obtained through iterated conjunction:

Lemma 9

For P a verifiable proposition of cardinality c :

• (i) P ^∗ = P ∧ P ∧... for c or more P , P , ...;

• (ii) P is closed under fusion iff P ∧ P ∧... = P for any conjunction P ∧ P ∧.....

Proof

• (i) Clearly, P ∧ P ∧... ⊆ P ^∗.Now suppose p = p ₁ ⊔ p ₂ ⊔... for p ₁, p ₂,... ∈ P.Without loss of generality, we can suppose that there are at most c elements p ₁, p ₂,..., since repetitions make no difference to the identity of p.But then p ∈ P ∧ P ∧....

• (ii) Suppose P is closed. Then P ∧ P ∧... = P by lemma 8(ii). Now suppose P ∧ P ∧... = P for any identical conjunction P ∧ P ∧.... Take any p ₁, p ₂,... ∈ P.Then p ₁ ⊔ p ₂ ⊔... ∈ P ∧ P ∧... = P;and so P is closed.

Containment may be defined in terms of conjunction:

Lemma 10

For regular verifiable propositions P and Q :

$$\begin{array}{@{}rcl@{}} P \ge_{\mathrm{ c}} Q &\text{iff}& P \wedge Q = P\\ &\text{iff}& \mathrm{for~some~verifiable~proposition} ~R, R \wedge Q = P. \end{array} $$

Proof

Suppose P≥_c Q.Take p ⊔ q ∈ P ∧ Q for p ∈ P and q ∈ Q.Since P≥_c Q,$p^{\prime }\sqsupseteq q$for some p ^′∈ P.Since P is closed, p ⊔ p ^′∈ P.So $p \sqsubseteq p \sqcup q \sqsubseteq p \sqcup p^{\prime }$with p, p ⊔ p ^′∈ P;and, since P is convex, p ⊔ q ∈ P.Now take p ∈ P.Since P≥_c Q,$p \sqsupseteq q$for some q ∈ Q.But then p ⊔ q = p ∈ P ∧ Q.

Suppose P ∧ Q = P.Then evidently, R ∧ Q = P for some verifiable proposition R.

Suppose R ∧ Q = P for some verifiable proposition R.Take p ∈ P.Then p ∈ R ∧ Q;and so p is of the form r ⊔ q for r ∈ R and q ∈ Q.But then $q \sqsubseteq p$,establishing the first clause of P≥_c Q.Now take q ∈ Q.Pick an r ∈ R.Then q ⊔ r ∈ Q ∧ R = P;and so $q \sqsubseteq q \sqcup r \in P$,establishing the second clause of P≥_c Q.

It is important to note that the first equivalence above need not hold for arbitrary propositions, since P≥_c P even though it is not in general true that P ∧ P = P. If P = {p, q } within the canonical space, for example, then P ∧ P = {p, q, pq }. Also, the second equivalence need not hold if we drop the requirement that the propositions P and Q should be verifiable, since $P \wedge F_{\square } = F_{\square }$ even though it is not in general true that $P \ge _{\mathrm {c}} F_{\square }$.

For the case of arbitrary propositions, we might define a relation $P\,{\succcurlyeq }_{\mathrm {c}}\,Q$ by: P = Q ∧ R for some proposition R. It is then readily shown that ${\succcurlyeq }_{\mathrm {c}}$ is reflexive and transitive. It may also shown to be antisymmetric. For suppose $P\,{\succcurlyeq } _{\mathrm {c}} \,Q$ and ${Q\succcurlyeq } _{\mathrm {c }}P$ - that is, P = Q ∧ R for some R and Q = P ∧ R ^′ for some R ^′. Take p ∈ P. Then for some q ∈ Q and r ∈ R, p = q ⊔ r and so $q \sqsubseteq p$. But then for some p ^′∈ P and r ^′∈ R ^′, q = p ^′⊔ r ^′. So $r^{\prime }\sqsubseteq p$. But then p ⊔ r ^′ = p ∈ P ∧ R ^′ = Q. The relation ${\succcurlyeq }_{\mathrm {c}}$ will agree with $\succcurlyeq _{\mathrm {c}}$ on verifiable regular propositions but not on arbitrary propositions.

Regular closure distributes through conjunction:

Lemma 11

For regular verifiable propositions P ₁ , P ₂ , ...:

$$(P_{1\ast}^{{\kern2.3pt}\ast} \wedge P_{2\ast}^{{\kern2.3pt}\ast} \wedge ...) = (P_{1} \wedge P_{2} \wedge ...)_{\ast}^{\ast}. $$

Proof

Suppose s ∈(P ₁ ∧ P ₂ ∧...)$_{\ast }^{\ast } \;=$[(P ₁ ∧ P ₂ ∧...), p ₁ ⊔p ₂ ⊔...]. So p ₁ ⊔ p ₂ ⊔... $\sqsubseteq s \sqsubseteq \boldsymbol {p}_{1} \sqcup \boldsymbol {p}_{2} \,\sqcup$... for p ₁ ∈ P ₁, p ₂ ∈ P ₂,.... Now $p_{\mathrm {i}} \sqsubseteq s\, \sqcap$ p $_{\mathrm {i}} \sqsubseteq \boldsymbol {p}_{\mathrm {i}}$ and so s ⊓p _i∈P _i.But s = s ⊓(p ₁ ⊔p ₂ ⊔...) = (s ⊓ p ₁) ⊔(s ⊓p ₂) ⊔... by Distribution and so s = (s ⊓p ₁) ⊔(s ⊓p ₂) ⊔... ∈($P_{1\ast }^{{\kern 2.3pt}\ast } \wedge P_{2\ast }^{{\kern 2.3pt}\ast } \,\wedge$...).

For the other direction, suppose s ∈($P_{1\ast }^{{\kern 2.3pt}\ast } \wedge P_{2\ast }^{{\kern 2.3pt}\ast } \,\wedge$...). Then s is of the form s ₁ ⊔ s ₂ ⊔..., where, for each i, there is a p _i ∈ P _ifor which $p_{\mathrm {i}}\sqsubseteq s_{\mathrm {i}} \sqsubseteq$p _i.But then p ₁ ⊔ p ₂ ⊔... $\sqsubseteq s_{1} \sqcup s_{2}\; \sqcup$... $\sqsubseteq \boldsymbol {p}_{1} \sqcup \boldsymbol {p}_{2} \,\sqcup$... and so s ∈(P ₁ ∧ P ₂ ∧...)$_{\ast }^{\ast }$.

Note the use of Distribution in establishing $(P_{1} \wedge P_{2} \,\wedge ...)_{\ast }^{\ast } \subseteq$ ($P_{1\ast }^{{\kern 2.3pt}\ast } \wedge P_{2\ast }^{{\kern 2.3pt}\ast } \,\wedge$ ...). Without this assumption, we would need to define the conjunction of the propositions P ₁, P ₂, ... within a regular domain to be (P ₁ ∧ P ₂ ∧ ...)$_{\ast }^{\ast }$ rather than (P ₁ ∧ P ₂ ∧ ...).

We can now establish the following ‘external’ or ‘algebraic’ characterization of conjunction for regular propositions:

Theorem 12

Suppose that P ₁ , P ₂ , ... are regular verifiable propositions and that P = P ₁ ∧ P ₂ ∧.... Then:

• (i) $\boldsymbol {p}= \bigsqcup \boldsymbol {p}_{\mathrm {i}}$ ; and

• (ii) P is the least regular verifiable proposition to contain each of P ₁ , P ₂ , ....

Proof

• (i) $\boldsymbol {p}= \bigsqcup \{p_{1} \sqcup p_{2} \,\sqcup$... : p ₁ ∈ P ₁, p ₂ ∈ P ₂,... } = p ₁ ⊔p ₂ ⊔..., given that each p _iis the largest member of P _i.

• (ii) Suppose that P ₁, P ₂,... are regular verifiable propositions. We establish the following facts in turn:

• (1) P is verifiable. For each P _icontains a state p _iand so p ₁ ⊔ p ₂ ⊔... ∈ P.

• (2) P isregular.

  $$\begin{array}{@{}rcl@{}} P &=& P_{1} \wedge P_{2} \wedge ...\\ &=& P_{1\ast}^{{\kern2.3pt}\ast} \wedge P_{2\ast}^{{\kern2.3pt}\ast} \wedge ... \mathrm{since~each~of}~P_{1},~P_{2}, ... \mathrm{is~regular}\\ &=& (P_{1} \wedge P_{2} \wedge ...)_{\ast}^{\ast}~\mathrm{by~the~previous~lemma} \end{array} $$

  and so P is regular.

• (3) P containseach P _iby lemma 10.

• (4) P is the least propositionto contain each of P ₁, P ₂, .... Suppose theregular proposition Q contains each P _i. Itfollows that P ≤ Q. Forsuppose, first, that p ∈ P.Then p is ofthe form p ₁ ⊔ p ₂ ⊔...,with each p _i ∈ P _i.For each i, $p_{\mathrm {i}} \sqsubseteq q_{\mathrm {i}}$for some q _i ∈ Q and so p ₁ ⊔ p ₂ ⊔... $\sqsubseteq q_{1} \sqcup q_{2} \,\sqcup$... ∈ Q. Now suppose q ∈ Q. Then foreach i, $q \sqsupseteq p_{\mathrm {i}}$for some p _i ∈ P _iand so $q \sqsupseteq p_{1} \sqcup p_{2} \,\sqcup$... ∈ P.

The result that P = P ₁ ∧ P ₂ ∧ ... is the least proposition to contain each of P ₁, P ₂, .... does not hold for semi-regular propositions since it is not generally true that P ∧ P = P. Indeed, there may not even exist a least upper bound of two propositions P and Q. For set P = {p,r} and Q = {q,s} within the canonical space. Then it may be verified that {pq,rs} and {ps,rq} are both minimal upper bounds of P and Q; and so P and Q cannot have a least upper bound. Thus the standard lattice-theoretic approach to conjunction is not possible within this more general framework and it is not altogether clear what, if anything, should be put in its place. We might perhaps use a containment relation ≥_C that allows any number of propositions to occur on its right, as in Q≥_c P ₁, P ₂, ..., where the intended meaning of Q≥_c P ₁, P ₂, ... is Q≥_c P ₁ ∧ P ₂ ∧ ... .The conjunction P ₁ ∧ P ₂ ∧ ... would then be the least upper bound with respect to this relation - that is, we would have P ₁ ∧ P ₂ ∧ ... ≥_c P ₁, P ₂, ... and Q≥_c P ₁ ∧ P ₂ ∧ ... whenever Q≥_c P ₁, P ₂, ....

Disjunction

We may give different definitions of disjunction, depending upon whether a full, semi-regular or regular domain of propositions is in question. For arbitrary propositions P ₁, P ₂, ..., their disjunction P = P ₁ ∨ P ₂ ∨ ... is simply (P ₁ ∪ P ₂ ∪ ...); for semi-regular propositions P ₁, P ₂, ..., their disjunction P = P ₁ ∨ P ₂ ∨ ... is the convex closure (P ₁ ∪ P ₂ ∪ ...) _∗ of (P ₁ ∪ P ₂ ∪ ...); and for regular propositions P ₁, P ₂, ..., their disjunction P = (P ₁ ∨ P ₂ ∨ ...) is the regular closure (P ₁ ∪ P ₂ ∪ ...)$_{\ast }^{\ast }$ of (P ₁ ∪ P ₂ ∪ ...). We cannot, in the last two cases, take the disjunction P to be (P ₁ ∪ P ₂ ∪ ...) since (P ₁ ∪ P ₂ ∪ ...) may be neither convex nor closed.

In analogy to lemma 10, we have the following result (whose proof is straightforward):

Lemma 13

For any propositions P and Q ,

$$\begin{array}{@{}rcl@{}} P \le_{\mathrm{d}} Q &\text{iff}& P = Q \vee P.\\ &\text{iff}&~\mathrm{for~some~proposition}~R, P = Q \vee R. \end{array} $$

We also have the following external characterization of disjunction:

Theorem 14

Suppose that P ₁ , P ₂ , ... are regular propositions and that P = P ₁ ∨ P ₂ ∨.... Then:

• (i) $\boldsymbol {p} = \bigsqcup \boldsymbol {p}_{\mathrm {i}}$ ; and

• (ii) P is the strongest regular proposition to be entailed by each of P ₁ , P ₂ , ...

Proof

• (i) $\boldsymbol {p} = \bigsqcup (P_{1} \cup P_{2} \cup ... )_{\ast }^{\ast } = \bigsqcup (P_{1} \cup P_{2} \cup ... ) = \bigsqcup \boldsymbol {p}_{\mathrm {i}}$.

• (ii) (1) P is regular by construction. (2) P _i ⊆ P.For given p _i ∈ P _i,$p_{\mathrm {i}} \sqsubseteq p_{\mathrm {i}} \sqsubseteq \boldsymbol {p}_{\mathrm {i}} \sqsubseteq \boldsymbol {p}_{1} \sqcup \boldsymbol {p}_{2} \,\sqcup$... and so p _i ∈ P.(3) If each P _ientails Q,for Q regular, then P entails Q.For (P ₁ ∪ P ₂ ∪...) ⊆ Q,given that each P _ientails Q;and so $P = (P_{1} \cup P_{2} \cup ...)_{\ast }^{\ast } \subseteq Q_{\ast }^{\ast } = Q$,given that Q is regular.

A similar result holds for semi-regular propositions and is proved in a similar way. Thus disjunction in the general case, in contrast to conjunction, does have a straightforward lattice-theoretic characterization.

Let $\mathcal {P}$ be the set of all propositions and $\mathcal {P}^{\prime }$ the set of all verifiable propositions. We may then readily establish the following characterization of the extremal propositions in terms of conjunction and disjunction:

Lemma 15

$$\begin{array}{@{}rcl@{}} T_{\square} &=&\bigwedge \emptyset\\ T_{{\blacksquare}}&=&\bigvee\mathcal{P}\\ F_{\square}&=&\bigvee \emptyset \\ F_{{\blacksquare}}&=&\bigwedge \mathcal{P}^{\prime}. \end{array} $$

We have the normal distribution principles for conjunction and disjunction within a regular domain:

Theorem 16

For regular verifiable propositions P , Q and R :

• (i) P ∧ (Q ∨ R) = ( P ∧ Q) ∨ (P ∧ R)

• (ii) P ∨( Q ∧ R) = ( P ∨ Q) ∧ ( P ∨ R)

Proof

• (i) Suppose s ∈ P ∧(Q ∨ R).Then s is of the form p ⊔ t for p ∈ P and t ∈[ Q ∪ R,q ⊔r]. Without loss of generality, assume $q \sqsubseteq t \sqsubseteq$ q ⊔r for some q ∈ Q.Since p,$q \sqsubseteq s$,$p \sqcup q \sqsubseteq s$and, since $t \sqsubseteq$ q ⊔r, $t \sqsubseteq$p ⊔q ⊔r. Thus $p \sqcup q \sqsubseteq s \sqsubseteq \boldsymbol {p}\,\sqcup$ q ⊔r and, given that p ⊔ q ∈ P ∧ Q, s ∈(P ∧ Q) ∨(P ∧ R).

  For the other direction, it is clear that(P ∧ Q) ⊆ P ∧ (Q ∨ R)and that (P ∧ R) ⊆ P ∧ (Q ∨ R);and so (P ∧ Q) ∪(P ∧ R) ⊆ P ∧ (Q ∨ R).Hence $(P \wedge Q) \vee (P \wedge R) = [(P \wedge Q) \,\cup \, (P \wedge R)]_{\ast }^{\ast } \subseteq [P \wedge (Q \vee R)]_{\ast }^{\ast } = [P \wedge (Q \vee R)]$,given that P ∧ (Q ∨ R)is regular.

• (ii) P ⊆(P ∨ Q)and P ⊆(P ∨ R);and so P ⊆(P ∨ Q) ∧(P ∨ R).Also, (Q ∧ R) ⊆(P ∨ Q) ∧(P ∨ R);and so P ∪(Q ∧ R) ⊆(P ∨ Q) ∧(P ∨ R).But then $P \,\vee \, (Q \,\wedge \, R) = [P \cup (Q \wedge R)]_{\ast }^{\ast } \subseteq [(P \vee Q) \,\wedge \, (P \,\vee \, R)]_{\ast }^{\ast } = (P \,\vee \, Q)\, \wedge \, (P\, \vee \, R)$,given that (P ∨ Q) ∧(P ∨ R)is regular.

For the other direction, take s ∈(P ∪ Q) ∧(P ∪ R). There arefour cases. (a) s ∈ P.But then s ∈ P ∨(Q ∧ R). (b) s is of theform p ⊔ r for p ∈ P and r ∈ R.But $p \sqsubseteq p \sqcup r \sqsubseteq \boldsymbol {p}\;\sqcup$ q ⊔r andso s = p ⊔ r ∈ P ∨(Q ∧ R). (c) s is of theform q ⊔ p for q ∈ Q and p ∈ P. Similar tocase (b). (d) s is of the form q ⊔ r for q ∈ Q and r ∈ R. Butthen s ∈ (Q ∧ R) ⊆ P ∨ (Q ∧ R).So in each case:

$$(*) (P \cup Q) \wedge (P \cup R) \subseteq P \vee (Q \wedge R). $$

Hence:

$$\begin{array}{@{}rcl@{}} (P \vee Q) \wedge (P \vee R) &=& (P \cup Q)_{\ast}^{\ast} \wedge (P \cup R)_{\ast}^{\ast}\\[-2pt] &=&[(P \cup Q) \wedge (P \cup R)]_{\ast}^{\ast}~\mathrm{by~lemma}~11\\[-2pt] &\subseteq& [P \vee (Q \wedge R)]_{\ast}^{\ast}~\text{by}~(*)~\text{above}\\[-2pt] &=& P \vee (Q \wedge R) \end{array} $$

The infinitary form of these results may be proved in the same way. The first distribution principle holds for arbitrary propositions, but the second distribution principle does not even hold for semi-regular propositions. For in the canonical space, let P = {p},Q = {q} and R = {r}. Then P ∨ (Q ∧ R) = {p,q r} while (P ∨ Q) ∧ (P ∨ R) = {p,p r,q p,q r}.

Conjunction and Disjunction - the Bilateral Case

Given the bilateral propositions P ₁ = (P ₁, $P_{1}^{\prime }$), P ₂ = (P ₂, $P_{2}^{\prime }$), ..., we let:

$$\begin{array}{@{}rcl@{}} \boldsymbol{P}_{1} \wedge \boldsymbol{P}_{2} \wedge ... &=& (P_{1} \wedge P_{2} \wedge ..., P^{\prime}_{1} \vee P^{\prime}_{2} \vee ...)~\mathrm{for~verifiable} \boldsymbol{P}_{1}, \boldsymbol{P}_{2}, ...;~\text{and}\\[-2pt] \boldsymbol{P}_{1} \vee \boldsymbol{P}_{2} \vee ... &=& (P_{1} \vee P_{2} \vee ..., P^{\prime}_{1} \wedge P^{\prime}_{2} \wedge ..)~\mathrm{for~falsifiable} \boldsymbol{P}_{1}, \boldsymbol{P}_{2}, .... \end{array} $$

We have some analogues of the previous results for unilateral propositions:

Lemma 17

• (i) For regular verifiable propositions P = (P,P ^′)and Q = (Q,Q ^′):

  $$\boldsymbol{P} \ge_{\mathrm{c}} \boldsymbol{Q}~\text{iff}~\boldsymbol{P}~\wedge \boldsymbol{Q} = \boldsymbol{P}~\text{iff}~\boldsymbol{R} \wedge \boldsymbol{Q} = \boldsymbol{P}~\mathrm{for~some~verifiable~proposition}~\boldsymbol{R}. $$

• (ii) For regular falsifiable propositions P = (P,P ^′)and Q = (Q,Q ^′):

  $$\boldsymbol{P} \ge_{\mathrm{d}} \mathrm{Q}~\text{iff}~\boldsymbol{P} \vee \boldsymbol{Q} =\boldsymbol{P}~\text{iff}~\mathrm{R}~\vee~\boldsymbol{Q} = \boldsymbol{P}~\mathrm{for~some~falsifiable~proposition}~\boldsymbol{R} $$

Proof

• (i) Suppose P≥_c Q.Then P≥_c Q and P ^′≥_d Q ^′.By lemmas 10 and 13, P ∧ Q = P and P ^′∨ Q ^′ = P ^′.But then P ∧ Q = (P ∧ Q, P ^′∨ Q ^′) = (P, P ^′) = P.

  Clearly, P ∧Q = P implies R ∧Q = P for some verifiable proposition R.

  Now suppose R ∧Q = P for some verifiable proposition R = (R,R ^′).Then (R ∧ Q,R ^′∨ Q ^′) = (P,P ^′)and so (a) R ∧ Q = P and (b) R ^′ ∨ Q ^′ = P ^′.Again by lemmas 10 and 13, P≥_c Q and P ^′≥_d Q ^′.But then P≥_c Q.

• (ii) is proved similarly.

Theorem 18

Suppose that P ₁,P ₂ , ... are regular verifiable propositions and that P = P ₁ ∧P ₂ ∧.... Then:

• (i) $\boldsymbol {p} = \bigsqcup \boldsymbol {p}_{\mathrm {i}}$ and $\boldsymbol {p}'= \bigsqcup \boldsymbol {p}_{i}^{\prime };$

• (ii) P is the least regular proposition to contain each of P ₁,P ₂ , ....

Proof

P is of the form (P ₁ ∧ P ₂ ∧..., $P^{\prime }_{1} \vee P^{\prime }_{2} \vee ...)$.From theorem 13(ii), P ₁ ∧ P ₂ ∧... is the least proposition to contain each of P ₁, P ₂,... and, from theorem 14(ii), $P^{\prime }_{1} \,\vee P^{\prime }_{2}\,\vee ...$is the greatest regular proposition to be entailed by each of$P^{\prime }_{1}, P^{\prime }_{2}, ... .$

Theorem 19

Suppose that P ₁ , P ₂ , ... are regular falsifiable propositions and that P = P ₁ ∨P ₂ ∨.... Then:

• (i) $\boldsymbol {p}= \bigsqcup \;$ p _i and p $^{\prime } = \bigsqcup \boldsymbol {p}_{i}^{\prime }$ ;

• (ii) P is the strongest regular proposition to be entailed by each of P ₁ , P ₂ , ....

Proof

Similar to the proof of the previous theorem.

Theorem 20

For regular non-vacuous propositions P , Q and R :

• (i) P ∧ (Q ∨ R) = (P ∧ Q) ∨ (P ∧ R)

• (ii) P ∨ (Q ∧ R) = (P ∨ Q) ∧(P ∨ R).

Proof

We may prove (i) by using theorem 16(i) on the ‘left’ and theorem 16(ii) on the ‘right’;and similarly, in reverse, for (ii).

Note that neither Distribution principle will hold for arbitrary or for semi-regular propositions, since each Distribution principle for bilateral propositions involves both distribution principles for unilateral propositions on the left and on the right.

Let $\mathcal {P}$ be the set of all falsifiable propositions (P, P ^′) and $\boldsymbol {\mathcal {P}}^{\prime }$ the set of all verifiable propositions (P, P ^′). Then:

Lemma 21

$$\begin{array}{@{}rcl@{}} \boldsymbol{T}_{\square} &=& \bigwedge \emptyset\\ \boldsymbol{T}_{{\blacksquare}} &=& \bigvee \mathcal{P}\\ \boldsymbol{F}_{\square}&=& \bigvee \emptyset\\ \boldsymbol{F}_{{\blacksquare}}&=&\bigwedge \mathcal{P}^{\prime}. \end{array} $$

These results presuppose that we are working within the full space of bilateral propositions P = (P, P ^′). But they, or analogues of them, also hold under various natural constraints on the domain.

Negation

The treatment of negation is somewhat less straightforward. In the case of unilateral propositions, we may define their negation in terms of a relation of exclusion. Say that a relation ⊥⊆ S × S is exclusionary if it satisfies the following three conditions:

ᅟ:

Upward Exclusion if q⊥p and $p \sqsubseteq p^{+}$ then q⊥p ⁺;

ᅟ:

Downward Exclusion if q⊥p ₁ ⊔ p ₂ ⊔ ... , then q ∈{r : for some i, r⊥p _i};

ᅟ:

Null Exclusion never $\square \bot s$ or $s \bot \square$ and, for each $s \ne \square$, there is a t for which s⊥t and a t for which t⊥s.

We also say that s excludes t when s⊥t holds. Upward Exclusion tells us that an excluder of a state will always exclude an extension of the state; Downward Exclusion tells us that the excluder of the fusion of some states may be derived from the excluders of the component states; and Null Exclusion tells us that a state will exclude or be excluded just in case it is not the null state.

We have formulated Downward Exclusion in a somewhat weak form. In many cases, we can insist upon the following stronger condition:

if q⊥p ₁ ⊔ p ₂ ⊔ ..., then q is the fusion of some q _k1, q _k2, ..., where each q _kj excludes some p _i.

Note that Downward Exclusion implies part of Null Exclusion, viz., that no state q excludes $\square$, for if q excludes $\square$, which is the empty fusion p ₁ ⊔ p ₂ ⊔ ..., then $q \in \{r\!\!:~\mathrm {for~some~ i},~r \bot p_{\mathrm {i}}\}_{\ast }^{\ast } = \emptyset _{\ast }^{\ast } = \emptyset$. For many applications, the second part of Null Exclusion is not required.

Given an exclusionary relation ⊥, we say that Q excludes P if, for each q ∈ Q, there is a p ∈ P for which q⊥p and, for each p ∈ P, there is a q ∈ Q for which q⊥p (thus exclusion between propositions is analogous to containment but with the exclusion relation ⊥ in place of the containment relation $\sqsupseteq )$, and we say that the state q excludes P − q⊥P - if, for some Q, Q excludes P and q = ⊔Q.

Note that F _■ = ∅ is the one and only proposition to exclude F _■ and hence $\square = \bigsqcup F_{\square }$ is the one and only state to exclude $F_{\square }$. Also, no state q can exclude a trivial proposition P since q would then have to be a fusion of states one of which excluded $\square$.

We take the exclusive negation − P of P to be {q : q excludes P} and the exclusionary negation ∼ P of P to be the regular closure $(-P)_{\ast }^{\ast }$ of − P.

Lemma 22

When P is non-trivial then ∼ P is verifiable and when P is verifiable then ∼ P is non-trivial.

Proof

Suppose P = {p ₁,p ₂,...}is non-trivial. Since each $p_{\mathrm {i}} \ne \square$,it follows by Null Exclusion that some q _i⊥p _i;and so q = q ₁ ⊔ q ₂ ⊔... ∈∼ P and ∼ P is verifiable.

Now suppose P is verifiable and hence contains a verifier s.Then $\square \notin \, \sim \!\!P$since, by Null Exclusion, never $\square \bot s$;and hence ∼ P is non-trivial.

The following two results will later be useful:

Lemma 23

• (i) $(-(P{{~}_{\ast }^{\ast }}))_{\ast }^{\ast } = (-P){{~}_{\ast }^{\ast }}$ for any regular unilateral proposition P

• (ii) ∼ (P ₁ ∧ P ₂ ∧ ...) = (∼ P ₁ ∨∼ P ₂ ∨ ...)for definite propositions P ₁,P ₂,....

Proof

• (i) Suppose $q \in -(P^{~}{{~}_{\ast }^{\ast }})$(to show $q \in (-P){{~}_{\ast }^{\ast }})$.Then q is of the form q ₁ ⊔ q ₂ ⊔ ...,where $P^{~}{{~}_{\ast }^{\ast }} =\{p_{1}, p_{2}, ...\}$and each q _i⊥p _i.Now each member of P = {p _k1, p _k2,...}is a member of $P{{~}_{\ast }^{\ast }}$and hence will be excluded by a q _ki.Let p ^′ = q _k1 ⊔ q _k2 ⊔.... Then p ^′∈−P and p $^{\prime } \sqsubseteq q$.Since each q _iexcludes $p_{\mathrm {i}} \in P^{~}{{~}_{\ast }^{\ast }}$,with $p_{\mathrm {i}} \sqsubseteq \boldsymbol {p}= p_{\mathrm {k1}} \sqcup p_{\mathrm {k2}} \sqcup$..., it follows by Upward Exclusion that each q _i⊥p; and so, by Downward Exclusion, $q_{\mathrm {i}} \in \{q\!\!: q ~\mathrm {excludes~some}~p_{\text {kj}}\}_{\ast }^{\ast }$.Thus each $q_{\mathrm {i}} \sqsubseteq \bigsqcup _{\ast }^{\ast }\!\!-\!\!P \in (-P){{~}_{\ast }^{\ast }}$and so $q = q_{1} \sqcup q_{2} ~\sqcup$... $\in (-P){{~}_{\ast }^{\ast }}$.Since $-(P^{~}{{~}_{\ast }^{\ast }}) \subseteq (-P){{~}_{\ast }^{\ast }}, (-(P^{~}{{~}_{\ast }^{\ast }})) _{\ast }^{\ast }\subseteq (-P)_{\ast }^{\ast }{~}_{\ast }^{\ast }= (-P)^{~}{{~}_{\ast }^{\ast }}$.

  Now suppose q ∈−P(to show $q \in -(P{{~}_{\ast }^{\ast }}))$.Then q is of the form q ₁ ⊔ q ₂ ⊔..., where P = {p ₁,p ₂,...}and each q _i⊥p _i.Now for each $p \in P{{~}_{\ast }^{\ast }}$,there is a p _i ∈ P for which $p_{\mathrm {i}} \sqsubseteq p$.Since q _i⊥p _i, q _i⊥p by Upward Exclusion. But it is then clear that$q \bot P _{\ast }^{\ast }$and so $q \in -(P{{~}_{\ast }^{\ast }})$.Since $-\!\!P \subseteq -(P{{~}_{\ast }^{\ast }}), (-P){{~}_{\ast }^{\ast }} \subseteq (-(P{{~}_{\ast }^{\ast }})){{~}_{\ast }^{\ast }}$.

• (ii) Each P _iwill be of the form {p _i}for some state p _i.Suppose q ∈ ∼(P ₁ ∧ P ₂ ∧...) = ∼ {p ₁ ⊔ p ₂ ⊔... }.Then q⊥p ₁ ⊔ p ₂ ⊔.... By Downward Exclusion, q ∈{r: r excludes some $p_{\mathrm {i}}\}^{~}{{~}_{\ast }^{\ast }}$.But {r: r excludes some p _i}⊆∼{p ₁} ∪∼ {p ₂} ∪...; so {r: r excludes some $p_{\mathrm {i}}\}^{~}{{~}_{\ast }^{\ast }} \subseteq$[ ∼ {p ₁} ∪∼ {p ₂} ∪...]${{~}_{\ast }^{\ast }}$;and so q ∈∼ {p ₁} ∨∼ {p ₂} ∨... .

  Now suppose q ∈∼ {p ₁}∨∼ {p ₂} ∨.... Then for some p _iand some q ^′⊥p _i,$q^{\prime }\sqsubseteq q \sqsubseteq q_{1} \sqcup q_{2} ~\sqcup$..., where each q _jexcludes some p _j.By Upward Exclusion, each q _jexcludes p ₁ ⊔ p ₂ ⊔...; and, since q ^′⊥p _i,it follows, again by Upward Exclusion, that q ^′⊥p ₁ ⊔ p ₂ ⊔.... But then by the convexity of ∼[ {p ₁}∧{p ₂} ∧...], q ∈∼[ {p ₁}∧{p ₂} ∧...].

We can establish the De Morgan Laws for exclusionary negation:

Lemma 24

For regular propositions P ₁ , P ₂ , ...,

• (i) ∼(P ₁ ∨ P ₂ ∨...) = ∼ P ₁ ∧∼ P ₂ ∧..., P ₁ , P ₂ , ... non-trivial

• (ii) ∼(P ₁ ∧ P ₂ ∧...) = ∼ P ₁ ∨∼ P ₂ ∨..., P ₁ , P ₂ , ... verifiable;

Proof

• (i) Note that if P ₁, P ₂, ... arenon-trivial then ∼ P ₁, ∼ P ₂,... are verifiable. We then have:

  $$\begin{array}{@{}rcl@{}} (\sim \!\!P_{1}\, \wedge \sim \!\!P_{2} \wedge ...) &=& (-P_{1})_{\ast}^{\ast} \wedge (-P_{2}){{~}_{\ast}^{\ast}} \wedge ...~\mathrm{by~definition~of}~\sim\\ &=& (-P_{1} \wedge -P_{2} \wedge ...){{~}_{\ast}^{\ast}}~\mathrm{by~lemma}~11\\ &=& (-(P_{1} \cup P_{2} \cup ...)){{~}_{\ast}^{\ast}}~\mathrm{by~definition~of}~\wedge~\text{and}~-\\ &=& (-((P_{1} \cup P_{2} \cup ...){{~}_{\ast}^{\ast}})){{~}_{\ast}^{\ast}}~\mathrm{by~lemma}~23\\ &=& (-(P_{1} \vee P_{2} \vee ...)){{~}_{\ast}^{\ast}}~\mathrm{by~definition~of} \vee\\ &=& \sim (P_{1} \vee P_{2} \vee ...)~\mathrm{by~definition~of} \sim \end{array} $$

• (ii) Suppose first that each of P ₁, P ₂, ... is trivial. Then P ₁ ∧ P ₂ ∧... is trivial; andso $\sim \!(P_{1} ~\wedge P_{2~} \wedge ...) = F_{\square } = F_{\square } ~\vee F_{\square } ~\vee ... ={~}\sim \!P_{1~} \vee \sim \!\!P_{2} ~\vee$.... So we maysuppose that one of P ₁, P ₂, ... is trivial.Now $(P_{1} \wedge P_{2} \wedge ...) ={}\bigvee \{ \{p_{1}\ \sqcup \,p_{2}\ \sqcup ...\}: p_{1} \in P_{1}, p_{2} \in P_{2}, ...\}$. So: $\sim \!(P_{1} \wedge P_{2} \wedge ...) ={} \sim {} \bigvee \{ \{p_{1} \sqcup p_{2} \sqcup ...\}: p_{1} \in P_{1}, p_{2} \in P_{2}, ...\}\\ \hspace *{5.8pc}= \bigwedge \{\sim \{p_{1} ~\sqcup p_{2} ~\sqcup {\kern 1pt} ...\}: p_{1} \in P_{1}, p_{2} \in P_{2}, ...\}$by part (i) given that each {p ₁ ⊔ p ₂ ⊔ ...}is non-trivial

  $$\begin{array}{@{}rcl@{}} \hspace*{0pc} &=& \bigwedge \{\sim \!\{p_{1}\}\vee \sim \!\{p_{2}\}\vee ...: p_{1} \in P_{1}, p_{2} \in P_{2}, ...\} \text{ by lemma 23(ii)}\\ &=& \bigvee_{\mathrm{i}} \bigwedge \{\sim \{p\}: p \in P_{\mathrm{i}}\}\!~\mathrm{by\!~General\!~Distribution}\!~(\text{Theorem} 16)\\ &=& \bigvee_{\mathrm{i}}\sim\bigvee \{\{p\}: p \in P_{\mathrm{i}}\}~\text{by}~\text{(i)}\\ &=& \sim P_{1} \vee \sim \!P_{2} \vee .... \end{array} $$

There is no general guarantee that ∼∼ P = P. To adapt a previous example, suppose that there are three states r (red), b (blue), g (green) with any two of them excluding the third. Let RED be the proposition {r}. Then ∼RED = {b, g, b ⊔ g} while ∼∼RED = {r, r ⊔ g, r ⊔ b, r ⊔ g ⊔ b}. However, the Law of Double Negation will hold if it holds for all definite propositions:

Theorem 25

If ∼∼ P = P for each definite unilateral proposition P then ∼∼ P = P for every unilateral proposition whatever.

Proof

This is clear if $P = F_{\square }$.So let us suppose P is verifiable. Now $P = \bigvee \{\{p\}$: p ∈ P}. So:

$$\begin{array}{@{}rcl@{}} \sim\sim \!\!P &=& \sim \sim \bigvee \{\{p\}: p \in P\}\\ &=& \sim \bigwedge \{\sim \!\{ p\}: p \in P\}~\mathrm{by~lemma}~24(\mathrm{i})\\ &=& \bigvee \{\sim\sim\{p\}: p \in P\}~\mathrm{by~lemma}~24\text{(ii)}\\ &=& \bigvee \{\{p\}: p \in P\}~\mathrm{by~assumption}\\ &=& P. \end{array} $$

Negation - the Bilateral Case

The (classical) negation of a bilateral proposition is simply obtained by reversing its verifiers and its falsifiers. Thus when P = (P, P ^′), ¬ P = (P ^′, P).

From the previous definitions, we immediately obtain:

Lemma 26

For non-vacuous propositions P and Q ,

• (i) P≤_c Q iff¬P ≤_d¬Q , and

• (ii) P≤_d Q iff¬P ≤_c¬Q .

We see from this result that containment and entailment, which seem very different from the unilateral perspective, can be seen to be two sides of the same coin when viewed from the bilateral perspective.

It is readily shown that Double Negation, De Morgan and Distribution hold in the bilateral case:

Theorem 27

(i) ¬¬P = P

• (ii) ¬(P ₁ ∧P ₂ ∧...) = (¬ P ₁ ∨¬ P ₂ ∨...), P ₁, P ₂,... verifiable,

• (iii) ¬(P ₁ ∨P ₂ ∨...) = (¬ P ₁ ∧¬ P ₂ ∧...), P ₁, P ₂,... falsifiable.

Proof

• (i) For P = (P, P ^′),¬¬P = ¬¬(P, P ^′) = ¬(P ^′, P) = (P, P ^′) = P.

• (ii) For P _i = (P _i ,$P^{\prime }_{i}$):

  $$\begin{array}{@{}rcl@{}} \lnot (\boldsymbol{P}_{1} \wedge \boldsymbol{P}_{2} \wedge ...) &=& \lnot (P_{1} \wedge P_{2} \wedge ..., P^{\prime}_{1} \vee P^{\prime}_{2} \vee ...)\\ &=& (P^{\prime}_{1} \vee P^{\prime}_{2} \vee ..., P_{1} \wedge P_{2} \wedge ...)~\mathrm{by~definition~of}~\lnot\\ &=& (P^{\prime}_{1}, P_{1}) \vee (P^{\prime}_{2}, P_{2}) \vee ...~\mathrm{by~definition~of} \vee\\ &=& \lnot (P_{1}, P^{\prime}_{1}) \vee \lnot (P_{2}, P^{\prime}_{2}) \vee ...~\mathrm{by~definition~of}~\lnot\\ &=& (\lnot \boldsymbol{P}_{1} \vee \lnot \boldsymbol{P}_{2} \vee ...) \end{array} $$

• (iii) Similarly.

Suppose given a set $\mathcal {Q}$ of bilateral propositions. We then let $\mathcal {Q}^{\mathrm {n}}$ be the closure $\mathcal {Q}\cup \, \{$ ¬P: P $\in \mathcal {Q}\}$ of $\mathcal {Q}$ under negation; and similarly, we let $\mathcal {Q}^{\mathrm {c}}$ be the closure of $\mathcal {Q}$ under conjunction and $\mathcal {Q}^{\mathrm {d}}$ the closure of $\mathcal {Q}$ under disjunction. We also let $\mathcal {Q}^{\mathrm {b}}$ be the closure of $\mathcal {Q}$ under negation, disjunction and conjunction, and let $\mathcal {Q}^{\mathrm {ncd } }=$ (($\mathcal {Q}^{\mathrm {n}})^{\mathrm {c}})^{\mathrm {d}}$; and similarly for $\mathcal {Q}^{\text {cd}}$ and the like. Note that $\mathcal {Q}^{\text {ncd}}$, in contrast to $\mathcal {Q}^{\mathrm {b } }$, requires that the negations, conjunctions and disjunctions be formed in a certain order. We say, in case $\mathcal {P}=\mathcal {Q}^{\mathrm {b}}$, that $\mathcal {P}$ is the Boolean closure of $\mathcal {Q}$; and we say that $\mathcal {P}$ is a Boolean domain if $\mathcal {P}=\mathcal {P}^{\mathrm {b}}$, i.e. if $\mathcal {P}$ is closed under the Boolean operations. It is important to bear in mind that, for the purpose of these definitions, conjunction is only defined on verifiable propositions and disjunction only defined on falsifiable propositions.

We have the following abstract version of the standard disjunctive and conjunctive normal form theorems in classical propositional logic:

Theorem 28

Suppose $\mathcal {P}$ is the Boolean closure of $\mathcal {Q}$ . Then:

• (i) $\mathcal {P}=\mathcal {Q}^{\text {ncd}}$ , and

• (ii) $\mathcal {P}=\mathcal {Q}^{\text {ndc}}$

Proof

By a straightforward induction on the formation of the propositions in$\mathcal {P}$using the De Morgan, Distribution and Double Negation Laws.

We have the following natural requirements on trivial and vacuous propositions:

• Triviality Any trivial proposition is vacuous;

• Vacuity Any vacuous proposition is trivial;

• Strong Triviality Any trivial proposition is identical to $\boldsymbol {T}_{\square }$ or $\boldsymbol {F}_{\square }$.

These requirements are preserved under Boolean combination.

Theorem 29

If the propositional subdomain $\mathcal {Q}$ satisfies Triviality (or Vacuity or Strong Triviality) then so does its Boolean closure $\mathcal {P}=\mathcal {Q}^{\mathrm {b}}$ .

Proof

We prove the result by induction on the formation of a proposition P in$\mathcal {Q}^{\mathrm {b}}$.If P is of the form ¬Q then the result is trivial, since it is evident from the symmetry of therequirements that if Q satisfies the requirement then so does ¬Q. So, in each case, we needonly show that the requirements are preserved under conjunction. We go through each casein turn.

Triviality. Take P of the form P ₁ ∧P ₂ ∧... for verifiable P ₁ = (P ₁,$P^{\prime }_{1}$),$\boldsymbol {P}^{\prime }_{2} =$(P ₂,$P^{\prime }_{2}$),... (recall that conjunction is only defined on verifiable propositions), so that P = (P, P ^′) = (P ₁ ∧ P ₂ ∧..., $P^{\prime }_{1} \vee P^{\prime }_{2}\,\vee$...); and suppose that P is trivial. There are two cases:

• (a) P is trivial. But then each P ₁, P ₂,... is trivial. By IH, each $P^{\prime }_{\mathrm {i}}$is vacuous; so $P^{\prime }= P^{\prime }_{1} \vee P^{\prime }_{2}\,\vee$... and hence P is vacuous.

• (b) P ^′is trivial. But then a $P^{\prime }_{\mathrm {i}}$is trivial. By IH, P _iis vacuous; and so the conjunction is illegitimate.

Vacuity Take P of the form P ₁ ∧P ₂ ∧... for verifiable P ₁ = (P ₁,$P^{\prime }_{1}$), P ₂ = (P ₂,$P_{2}^{\prime }$), ...;and suppose P is vacuous. There are two cases:

• (a) P is vacuous. But then some P _iis vacuous and the conjunction is illegitimate.

• (b) P ^′is vacuous. But then each $P^{\prime }_{\mathrm {i}}$is vacuous. By IH, each P _iis trivial; and so P = P ₁ ∧ P ₂ ∧... and hence P is trivial.

Strong Triviality Take P of the form P ₁ ∧P ₂ ∧... for verifiable P ₁ = (P ₁,$P^{\prime }_{1}$), P ₂ = (P ₂,$P^{\prime }_{2}$), ...;and suppose P is trivial. There are two cases:

• (a) P is trivial. But then each P ₁, P ₂,... is trivial. By IH, each P _iis identical to $\boldsymbol {T}_{\square }$or $\boldsymbol {F}_{\square }$;and hence each $\boldsymbol {P}_{\mathrm {i}} = \boldsymbol {T}_{\square }$and $\boldsymbol {P}=\boldsymbol {T}_{\square }$.

• (b) P ^′is trivial. But then a $P^{\prime }_{\mathrm {i}}$is trivial. By IH, P _iis identical to $\boldsymbol {T}_{\square }$or $\boldsymbol {F}_{\square }$and hence, given that P is trivial, P _iis identical to $\boldsymbol {F}_{\square }$;and so the conjunction is illegitimate.

Given a unilateral proposition P, let < P > be the bilateral proposition (P, ∼ P). We might call a proposition of the form < P > standard. We have the following results for standard propositions:

Lemma 30

For regular propositions P , P ₁ , P ₂ , ...:

• (i) < P ₁ ∧ P ₂ ∧... > = < P ₁ > ∧ < P ₂ > ∧..., for P ₁ , P ₂ , ... verifiable;

• (ii) < P ₁ ∨ P ₂ ∨... > = < P ₁ > ∨ < P ₂ > ∨..., for P ₁ , P ₂ , ... non-trivial;

• (iii) $<\hspace *{-3.5pt}P\hspace *{-3.5pt} >\, =\; \bigvee \{ <\hspace *{-3.5pt} p\hspace *{-3.5pt} >$ : p ∈ P}, for P non-trivial.

Proof

• (i)
  $$\begin{array}{@{}rcl@{}} <P_{1} \wedge P_{2} \wedge ...>&=& (P_{1} \wedge P_{2} \wedge ..., \sim \!(P_{1} \wedge P_{2} \wedge ...))\\ &=& (P_{1} \wedge P_{2} \wedge ..., \sim\!\! P_{1} \vee \sim\! P_{2} \vee ...)~\mathrm{by~lemma}~24(\text{ii})\\ &=& (P_{1}, \sim\!\! P_{1}) \wedge (P_{2}, \,\sim\!P_{2}) \wedge ...\\ &=& <P_{1}>\wedge <P_{2}>\wedge ... \end{array} $$
• (ii)
  $$\begin{array}{@{}rcl@{}} <P_{1} \vee P_{2} \vee ...> &=& (P_{1} \vee P_{2} \vee ..., \sim\!(P_{1} \vee P_{2} \vee ...))\\ &=& (P_{1} \vee P_{2} \vee ..., \sim\!\! P_{1} \wedge \sim \!\!P_{2} \wedge ...)~\mathrm{by~lemma}~24(\mathrm{i})\\ &=& (P_{1}, \sim\!\! P_{1}) \wedge (P_{2}, \sim \!\!P_{2}) \wedge ...\\ &=& <P_{1}>\wedge <P_{2}>\wedge... \end{array} $$

• (iii) For non-trivial P = {p ₁,p ₂,...},

  $$\begin{array}{@{}rcl@{}} <P> &=& <\{p_{1}\} \vee \{ p_{2}\} \vee ...>\\ &=& <\{p_{1}\}\>\vee <\{p_{2}\}> \vee ... \mathrm{by~part}~(\text{ii})\\ &=& \bigvee \{<p>: p \in P\}. \end{array} $$

We turn to the connection between the operations ∼ and ¬ of exclusionary and classical negation. With any exclusionary relation ⊥, we may associate two subdomains of propositions:

ᅟ:

the narrow subdomain $\mathcal {Q}_{\mathrm {\bot } }= \{ < \{s\} >: s \in S\}$ ; and

ᅟ:

the broad subdomain $\mathcal {Q}^{+}_{\mathrm {\bot }} = \{<P>: P$ a regular verifiable proposition ⊆ S}

and two corresponding Boolean domains:

ᅟ:

the narrow domain $\mathcal {P}_{\mathrm {\bot }} = (\mathcal {Q}_{\mathrm {\bot }})^{\mathrm {b}}$; and

ᅟ:

the broad domain $\mathcal {P}^{+}_{\perp }= (\mathcal {Q}^{+}_{\perp })^{\mathrm {b}}$.

Since each definite proposition {s} is regular and verifiable, $\mathcal {Q}_{\perp } \subseteq \mathcal {Q}_{\perp }^{+}$ and hence $\mathcal {P}_{\perp } \subseteq \mathcal {P}_{\perp }^{+}$. For non-trivial P, $<P>\,=\, \bigvee \{< \!\!\!p\!\! > : p \in P\}$ by lemma 30(iii) and hence $<P>\, \in \mathcal {P}_{\mathrm {\bot }}$. Thus $\mathcal {P}_{\mathrm {\bot }}^{+}$ can also be taken to be the Boolean closure $(\mathcal {Q}^{\prime }_{\!\!\bot })^{\mathrm {b}}$ of ${\mathcal {Q}}^{\prime }_{\perp }={\mathcal {Q}}_{\mathrm {\bot }} \cup \{ < \!P > : P~\text {trivial}~\}$; and so the difference from $\mathcal {P}_{\mathrm {\bot } }$ can be seen to consist in the fact that the propositions in $\mathcal {P}_{\perp }^{+}$ can be formed with the help of additional trivial propositions of the form (P, $F_{\square })$.

Corollary 31

The broad domain $\mathcal {P}_{{\perp }}^{+}$ (and hence the narrow domain $\mathcal {P}_{\mathrm {\bot } })$ conform to the Triviality and Vacuity requirements; and the narrow domain $\mathcal {P}_{\mathrm {\bot } }$ also conforms to the Strong Triviality requirement.

Proof

Given theorem 29, we need only show, for the first part, that$\mathcal {Q}_{{\perp }}^{+}$conforms to the Triviality and Vacuity requirements.

Suppose first that$\boldsymbol {P}\!= (P, \sim \!\!P) \in \mathcal {Q}_{\mathrm {\bot }}^{+}$is trivial. There are two cases. (a) P is trivial. But then by lemma 22, ∼ P is vacuous and hence P is vacuous. (b) ∼ P is trivial. But then by lemma 22, P and hence P is vacuous.

Suppose next that P = (P, ∼ P) ∈Q $_{\mathrm {\bot }}^{+}$is vacuous. There are two cases. (a) P is vacuous. But then by lemma 22, ∼ P and hence P is trivial. (b) ∼ P is vacuous. But then by lemma 22, P and hence P is trivial.

We now show that Q _⊥conforms to Strong Triviality. Suppose that P = (P,∼ P) ∈Q _⊥is trivial. There are two cases. (a) P is trivial. But then P must be of the form $\{\square \}$,$\sim \!\! P = F_{\square }$and so $\boldsymbol {P}= \boldsymbol {T}_{\square }$.(b) ∼ P is trivial. But by Null Exclusion, this is impossible given that P is definite.

Corollary 32

For any exclusionary relation $\bot , \text {\textbf {F}}_{\square }, \text {\textbf {T}}_{\square }\in \mathcal {P}_{\mathrm {\bot } }$ and hence $\text {\textbf {F}}_{\square }, \text {\textbf {T}}_{\square } \in \mathcal {P}_{\mathrm {\bot }}^{+}$ but $\text {\textbf {F}}_{{\blacksquare }}, \text {\textbf {T}}_{{\blacksquare }} \notin \mathcal {P}_{\mathrm {\bot }}^{+}$ and hence $\text {\textbf {F}}_{{\blacksquare }}, \text {\textbf {T}}_{{\blacksquare }} \notin \mathcal {P}_{\mathrm {\bot }}$ .

Proof

$\boldsymbol {T}_{\square } = (T_{\square }, F_{\square }) = (\{\square \}, \sim \!\!\{\square \}) \in \mathcal {Q}_{\mathrm {\bot } } \subseteq \mathcal {P}_{\mathrm {\bot }}$;and so $\text {\textbf {F}}_{\square } =$¬$\text {\textbf {T}}_{\square } \in \mathcal {P}_{\mathrm {\bot } }$.For the second part, we need only note that F _■and T _■are trivial yet non-vacuous propositions.

Although $\mathcal {P}_{\mathrm {\bot }}$ and $\mathcal {P}{^{+}_{\bot }}$ do not contain F _■ or T _■, they contain counterparts of F _■ and T _■. For let $T^{\prime }_{\blacksquare }= T_{{\blacksquare }} - \{\square \}, \text {\textbf {F}}_{\blacksquare }^{\prime }= (F_{\square }, T_{\blacksquare }^{\prime })$ and $\text {\textbf {T}}_{\blacksquare }^{\prime }= (T_{{\blacksquare }}, F_{\blacksquare }^{\prime })$ . Then within ${\mathcal {P}}_{\mathrm {\bot }}$ and $\mathcal {P}_{{\perp }}^{+}$, F $_{\blacksquare }^{\prime }$ will be the conjunction of all verifiable propositions and T $_{\blacksquare }^{\prime }$ the disjunction of all falsifiable propositions.

The next result relates exclusion to falsification and provides conditions under which all falsifiers will be excluders or all excluders will be falsifiers within a narrow domain:

Theorem 33

Suppose that P ^′⊆∼∼ P (resp. P ⊇∼∼ P)holds for all the definite propositions in $\mathcal {Q}_{\mathrm {\bot } }$ . Then P ^′⊆ ∼ P (resp. P ^′⊇ ∼ P)holds for all propositions P = ( P , P ^′ ) in $\mathcal {P}_{\mathrm {\bot } }$ .

Proof

$\mathcal {P}$is of the form $\mathcal {Q}^{\text {ncd}}$for $\mathcal {Q}$a definite domain $\mathcal {Q}_{\mathrm {\bot } }$;and so we may prove the results by induction on the generation of P via$\mathcal {Q}^{\text {ncd}}$.We focus on the case P ^′⊆ ∼ P,since the proof for P ^′⊇ ∼ P is exactly the same but with ⊇replacing ⊆.

Suppose P = (P, P ^′)$\in \mathcal {Q}$.The result is immediate since P ^′ = ∼ P.

Suppose P is of the form ¬Q for Q = ({p}, Q ^′)$\in \mathcal {Q}$.Thus P = (Q ^′, {p}), where P = Q ^′, P ^′ = {p}and Q ^′ = ∼ P ^′.By the supposition, P ^′⊆ ∼∼ P ^′ = ∼ Q ^′ = ∼ P.

Suppose P is of the form Q ∧R for verifiable Q = (Q, Q ^′),R = (R, R ^′)and P = (P, P ^′) = (Q ∧ R, Q ^′∨ R ^′)(the general case is similar). By IH, Q ^′⊆ ∼ Q and R ^′⊆ ∼ R;and so P ^′ = Q ^′∨ R ^′⊆∼ Q ∨∼ R =∼ (Q ∧ R) = ∼ P,by De Morgan for ∼.

Suppose P is of the form Q ∨R. Similar to the previous case.

Corollary 34

Suppose that P = ∼∼ P holds for all definite propositions P . Then, for any proposition P = ( P , P ^′ ) in $\mathcal {P}_{\mathrm {\bot } }$ , P ^′ = ∼ P and hence each proposition P = ( P , P ^′ ) in $\mathcal {P}_{\mathrm {\bot } }$ is of the form < P > .

The assumption that P ⊆ ∼∼ P is quite reasonable and hence it is quite reasonable to suppose that any falsifier of a bilateral proposition P = (P, P ^′), i.e. any member of P ^′, is also an excluder, i.e. a member of ∼ P. This means that, given a narrow domain of propositions $\mathcal {P}_{\mathrm {\bot } }$, we can take the (maximal) underlying exclusionary relation ⊥ from which it is generated to be {(s,t): for some proposition P = ({s}, P ^′), t ∈ P ^′}, thereby determining ⊥ from $\mathcal {P}_{\mathrm {\bot } }$.

The converse assumption that P ⊇ ∼∼ P, by contrast, is not at all reasonable, as is evident from the red/green/blue example above; and so we cannot, in general, assume that the falsifiers of a given bilateral proposition will be a function of its verifiers.

Classical Truth-conditions

We turn to the modal aspects of the theory of Boolean propositions. We should now take ourselves to be working within a modalized state space. Within such a space, there are two important conditions that may be imposed on the relationship between positive and negative content. Say that a bilateral proposition P = (P, P ^′) is exclusive if no member of P is compatible with a member of P ^′, that P is exhaustive if any consistent state is compatible with a member of P or with a member of P ^′, and that P is classical if it is both exclusive and exhaustive.

Within a W-space, these conditions take a more familiar form. Given an arbitrary bilateral proposition P = (P, P ^′) and state s, say s|| > P if $s \sqsupseteq p$ for some member p of P and that s < ||P if $s \sqsupseteq p^{\prime }$ for some member p ^′ of P ^′. In the particular case in which s is a world- state w, we say P is true (false) at w if w|| > P (w < ||P).

Theorem 35

Within a W-space:

• (i) P is exclusive iff there is no world-state at which P is glutty, i.e. both true and false;

• (ii) P is exhaustive iff there is no world-state at which P is gappy, i.e. neither true nor false:

• (iii) P is classical iff at any world-state P is bivalent, i.e. either true or false but not both.

Proof

• (i) Suppose P is not exclusive. Then some state p ∈ P is compatible with a state p ^′∈ P ^′.So p ⊔ p ^′is consistent and hence is contained in a world-state w.But then w contains both p and p ^′and P is both true and false at w.Now suppose P is both true and false at the world-state w,so that P contains both a p ∈ P and a p ^′∈ P ^′.Then since p and p ^′are both parts of w,they are compatible, contrary to the exclusivity of P.

• (ii) Suppose P is exhaustive and take a world-state w.Since w is consistent, it is compatible with a p ∈ P or with a p ^′∈ P ^′.Hence w contains such a p or a p ^′and P is either true or false. Now suppose P is either trueor false at each world-state, so that each world-state contains a p ∈ P or a p ^′∈ P ^′.Take any consistent state q.Then q is part of a world-state w;and since w is compatible with a p or a p ^′,then so is q.

• (iii) From (i) and (ii).

Theorem 36

For arbitrary propositions P = (P , P ^′), P ₁ = (P ₁ , $P_{1}^{\prime }$), P ₂ = (P ₂ , $P_{2}^{\prime }$), ... and any state s :

• (i)(a) s|| > ¬P iff s < || P

• (b) s < ||¬P iff s|| > P

• (ii)(a) s|| > ( P ₁ ∧P ₂ ∧...) iff s|| > P ₁and s|| > P ₂ and ...

• (b) s < ||( P ₁ ∧P ₂ ∧...) iff s < ||P ₁or s < ||P ₂ and ...

• (iii)(a) s|| > ( P ₁ ∨P ₂ ∨...) iff s|| > P ₁or s|| > P ₂ or ....

• (b) s < ||( P ₁ ∨P ₂ ∨...) iff s < || P ₁and s < ||P ₂and ....

Proof

• (i)(a) For P = (P, P ^′), s|| > ¬P iff $s \sqsupseteq$a member of P ^′given that ¬ P = (P ^′, P),and this holds iff s < ||P.

• (i)(b) Similar.

• (ii)(a) s|| > (P ₁ ∧P ₂ ∧...) iff$s \sqsupseteq$a memberof (P ₁ ∧ P ₂ ∧...)

  • iff $s \sqsupseteq p_{1} \sqcup p_{2}~ \sqcup$for some p ₁ ∈ P ₁, p ₂ ∈ P ₂,...

  • iff s|| >P ₁and s|| >P ₂and ....

• (ii)(b) s < ||(P ₁ ∧P ₂ ∧...) iff$s \sqsupseteq$a memberof ($P_{1}^{\prime }\vee P_{2}^{\prime }~\vee$...)

  • iff $s \sqsupseteq$ a member of $P_{1}^{\prime }\cup P_{2}^{\prime }~\cup$...

  • iff s < ||P ₁or s < ||P ₂and ...

• (iii)(a)-(b) Similar to (ii) or from (i) and (ii) by De Morgan.

Combining the previous two results, we can derive the classical truth-conditions for Boolean compounds of bivalent propositions:

Corollary 37

For bivalent propositions P, P ₁,P ₂,... and any world-state w :

• (i) w| = ¬P iff not w| = P

• (ii) w| = ( P ₁ ∧P ₂ ∧...) iff w| = P ₁and w| = P ₂and ...

• (iii) w| = ( P ₁ ∨P ₂ ∨...) iff w| = P ₁or w| = P ₂or ....

We can also show that the properties of being exclusive or being exhaustive are inherited under Boolean compounding:

Theorem 38

• (i)(a) If P is exclusive then so is¬P

• (b) If P is exhaustive then so is¬P

• (ii)(a) If P ₁ , P ₂ , ... are exclusive then so is ( P ₁ ∧P ₂ ∧...)

• (b) If P ₁ , P ₂ , ... are exhaustive then so is ( P ₁ ∧P ₂ ∧...) - within a W-space

• (iii)(a) If P ₁ , P ₂ , ... are exclusive then so is ( P ₁ ∨P ₂ ∨...)

• (b) If P ₁ , P ₂ , ... are exhaustive then so is ( P ₁ ∨P ₂ ∨...) - within a W-space.

Proof

• (i)(a)-(b). By the symmetry of the properties of being exclusive and being exhaustive.

• (ii) Take P ₁ = (P ₁,$P_{1}^{\prime }$), P ₂ = (P ₂,$P_{2}^{\prime }$)..., so that P = (P, P ^′) = ((P ₁ ∧ P ₂ ∧...), ($P_{1}^{\prime }\vee P_{2}^{\prime }\vee$...)). To establish (a), suppose that P ₁, P ₂,... are exclusive and suppose that some member p of (P ₁ ∧ P ₂ ∧...) is compatible with a member p ^′of ($P_{1}^{\prime }\vee P_{2}^{\prime }\vee$...). Then $p^{\prime \prime }\sqsubseteq p^{\prime }$for some p ^″∈ P and, given that p is of the form p ₁ ⊔ p ₂ ⊔... for p ₁ ∈ P ₁, p ₂ ∈ P ₂,..., p _i ∈ P _iis compatible with $p^{\prime \prime }\in P_{\mathrm {i}}^{\prime }$.

  To establish (b), suppose that P ₁, P ₂,... are exhaustive and take an arbitrary possible state s.Then some world-state $w \sqsupseteq s.$There are two cases. (1) w is compatible with a member of each P _iand hence $w \sqsupseteq p_{\mathrm {i}} \in P_{\mathrm {i}}$for each i = 1, 2, .... But then $w \sqsupseteq p = p_{1} \sqcup p_{2}~\sqcup$... ∈ P andso s is compatible with p.(2) w is not compatible with any member of some P _i.But then w and hence s is compatible with a memberof $P_{\mathrm {i}}^{\prime }$.

• (iii) Proved similarly to (ii) or on the basis of (ii) by De Morgan.

The results under (ii)(b) and (iii)(b) can be proved within an arbitrary state space for finitary conjunctions and disjunctions but not for infinitary conjunctions or disjunctions (see Fine [10] for further consideration of this case).

Say that a domain Q of bilateral propositions is bivalent (classical) if each proposition in Q is bivalent (classical). From the theorem, we immediately obtain:

Corollary 39

If Q is bivalent (within a W-space) then so is Q ^b .

We consider the special case in which Q is a basis for the broad domain $\mathcal {P}_{\mathrm {\bot }}^{+}$. Say that the exclusionary relation ⊥ is classical if:

• (i) if s⊥ t then s is incompatible with t;

• (ii) if s and t are consistent and s is incompatible with t then, for some $s^{\prime }\!\sqsubseteq \!s$, s ^′⊥ t; and

• (iii) if t is inconsistent then every consistent state is compatible with some t ^′ for which t ^′⊥ t.

Corollary 40

If ⊥is classical then so is $\mathcal {P}_{\mathrm {\bot }}^{+}$ .

Proof

By the previous corollary, it suffices to show that$\mathcal {Q}_{\perp }^{\prime } = \mathcal {Q}_{\perp }\cup \{ <\!\!P\!\! >$: P trivial }is classical, given that ⊥is classical. It is evident that trivial < P > satisfies exclusivity, since either P or ∼ P is empty, and it is evident that it satisfies Exhaustivity, since either P or ∼ P contains $\square$,which is compatible with any state. So it remains to show that$\mathcal {Q}_{\mathrm {\bot }}$is classical. Take a proposition ({p}, ∼{p}) in $\mathcal {Q}_{\mathrm {\bot }}$and some arbitrary member q of ∼ {p}.Then $q \sqsupseteq q^{\prime }$for some q ^′which excludes p.But then, by condition (i) above, q ^′and hence q is incompatible with p.For Exhaustivity, suppose s is consistent yet incompatible with p.If p is inconsistent, then it follows from (iii) that s iscompatible with a p ^′⊥p.So suppose p is consistent. Then by (ii), for some $s^{\prime }\sqsubseteq s$, s ^′⊥p.But then s is compatible with s ^′∈ ∼ {p}.

Semantics and Logic

Our focus has been on developing an abstract theory of truthmaker content, but it is worth pointing out how the abstract theory might relate to the interpretation of a formal language (for further discussion, consult Fine [7]). We suppose given an infinitude of sentence letters p ₁, p ₂, .... Formulas are then formed from these sentence letters using the Boolean operators ∧, ∨, and ¬ in the usual way (the double use of the constants ∧, ∨, and ¬ as operators and as operations should cause no confusion). A (state) model M is an ordered triple (S, $\sqsubseteq$, |⋅|), where (S, $\sqsubseteq )$ is a state space and |⋅| (valuation) is a function taking each sentence letter p into a bilateral proposition, i.e. a pair (V,F) of subsets of S. When |p| = (V,F), we let |p|⁺ = V and $|p|^{\mathrm {-}} = F$.

Given a model M $=(S, \sqsubseteq , |\cdot |)$, we define what it is for an arbitrary formula A to be verified by a state s (s||- A) or falsified by the state s (s- || A):

• (i) ⁺ s ||- p if $s \in |$p |⁺;

• (i) ⁻ s- || p if $s \in |$p |⁻;

• (ii) ⁺ s ||- ¬B if s - || B;

• (ii) ⁻ s- || ¬B if s ||- B;

• (iii) ⁺ s ||- B ∧ C if for some t and u, t ||- B, u ||- C and s = t ⊔ u;

• (iii) ⁻ s- || B ∧ C if s - || B or s - || C;

• (iv) ⁺ s ||- B ∨ C if s ||- B or s ||- C;

• (iv) ⁻ s- || B ∨ C if for some t and u, t - || B, u - || C and s = t ⊔ u.

Alternatively, we can think in terms of extending the map |p| from sentence letters to bilateral propositions to all formulas whatever. Thus in place of clauses (ii)-(iv)(a)(b), we have:

• (ii) |¬B| = ¬|B|

• (iii) |B ∧ C| = |B|∧|C|

• (iv) |B ∨ C| = |B|∨|C|,

using the previous definitions of propositional negation, conjunction and disjunction.

Let us temporarily use [A] ambiguously for |A|, |A|^∗, |A|_∗ and $|\text {A}|_{\ast }^{\ast }$, thereby reflecting the various closure conditions that might be imposed upon a proposition. We may then distinguish the following notions of verification (relative to a given model M):

ᅟ:

Exact Verification s|| - A if s ∈ [A]

ᅟ:

Inexact Verification s|| > A if for some $s^{\prime }\sqsubseteq s$, s ^′ ||- A

ᅟ:

Loose Verification s⊧ A if any state compatible with s is compatible with some t ∈ [A].

These notions of verification will be ambiguous given the ambiguous meaning of [A], but it turns out that the ambiguity only matters in the case of exact verification; that is, inexact and loose verification will coincide whatever the underlying notion of verification given by [A]. Note that the classical notion of loose verification must be defined by reference to a modalized state space or model, in contrast to the notions of exact and inexact verification; and, indeed, it is a general characteristic feature of the classical approach that it takes heed of the distinction between consistent and inconsistent states.

Given these different notions of verification, we may distinguish corresponding notions of consequence (either relative to a given model M or allowing M to vary arbitrarily):

ᅟ:

Exact Consequence A ⊧_e C if s||- A implies s|| C

ᅟ:

Inexact Consequence A | = _i C if s|| > A implies s|| > C

ᅟ:

Analytic Consequence (Containment) A > C if [A] ≥_c [C]

ᅟ:

Loose (or Classical) Consequence A | = _l C if s| = A implies s| = C.

There will, of course, be corresponding notions of equivalence, according to which A and C are exactly (inexactly, analytically, classically) equivalent if each is exactly (inexactly, analytically, loosely) a consequence of the other.

The logic of each of these notions of consequence (or equivalence) can be determined. This is not my task here, but we may note that loose consequence corresponds to the familiar notion of classical consequence, containment for regular non-vacuous propositions [A] and [C] to Angell’s logic of analytic entailment [1], inexact consequence to the logic of first degree entailment, and exact consequence to a logic that has been explored in Correia [5] and in unpublished work of the author and Mark Jago.

Minimal Verifiers

Our own account of partial content is hyper-intensional; intensionally equivalent - even truth-functionally equivalent - propositions may not contain or be contained in the same propositions. Yablo and Gemes have developed intensional accounts of partial content, in which intensionally equivalent propositions will be indistinguishable with respect to their containment relationships. In the present section, we discuss Yablo’s approach in terms of ‘minimal’ verifiers; and, in the next section, we discuss Gemes’ approach in terms of what I shall call ‘compact’ verifiers. In each case, we shall show how to accommodate the approach within the more abstract setting of our own framework and show how it may be extended, though with some artificiality, to cases in which the required minimal or compact verifiers do not exist.

Recall that s| = P is loose verification (we must therefore be working within a modalized space). We say that s-minimally verifies the proposition P - in symbols, s| = ^m P - if s| = P and for no $s^{\prime }\sqsubset s$ does s ^′| = P. The proposition P itself is said to be minimal if (i) p| = ^m P for any p ∈ P and (ii) for any p| = P there is a p ^′∈ P for which $p^{\prime }\sqsubseteq p$. It follows that, when P is minimal, then every minimal verifier p ^′ of P will belong to P, since there will be a p ∈ P for which $p \sqsubseteq p^{\prime }$ and hence for which p = p ^′. Thus a minimal proposition will be identical to its set of minimal verifiers.

Let us say that two propositions P and Q are loosely (or intensionally) equivalent if their loose verifiers are the same (this is equivalent, within a W-space, to their being true in the same possible worlds). Then it is readily shown that any two intensionally equivalent minimal propositions P and Q will be identical. For suppose p ∈ P. Then it will loosely verify P and so will loosely verify Q. If p is a minimal verifier of Q, then p ∈ Q and we are done. If p is not a minimal verifier of Q, then p ^′∈ Q for some $p^{\prime }\sqsubset p$. But then p ^′ is a loose verifier of Q and hence of P and so, for some p ^″∈ P, $p^{\prime \prime }\sqsubseteq p^{\prime }\sqsubset p$, and p is not a minimal verifier of P after all.

Partial content (and some other notions) can be defined for minimal propositions in the usual way. Thus given minimal propositions P and Q, we can say that P ≥ Q if every verifier of P contains a verifier of Q and every verifier of Q is contained in a verifier of P. The definition may then be extended to propositions P ^′ and Q ^′ intensionally equivalent to minimal propositions P and Q by taking P ^′≥ Q ^′ to hold just in case P ≥ Q holds. However, it is not generally true that a proposition P is intensionally equivalent to a minimal proposition P ^′; and so one must find some other way of extending the definition to arbitrary propositions.

One way to do this is in terms of ‘approximations’ to minimal propositions. A proposition Q is said to be minimalist if (i) q ∈ Q, $q^{\prime }\sqsubseteq q$ and q ^′ | = Q implies q ^′∈ Q and (ii) for any $q|\!\!=Q, q \sqsupseteq q^{\prime }$ for some q ^′∈ Q. (i) is a downward closure condition and (ii) is an adequacy condition to the effect that every loose verifier must lie above a verifier in the proposition. Q is said to be a minimalist basis for P if Q is a minimalist proposition intensionally equivalent to P. Note that a minimalist basis Q for a proposition P will always exist since we can let Q be the set of all loose verifiers of P; and note also that any minimal proposition P will be a minimalist basis for any intensionally equivalent proposition.

We may now extend the relation of partial content to all propositions as follows (Y stands for Yablo):

P≥^Y Q if for any minimalist basis P ^′ for P there is a minimalist basis Q ^′ for Q for which P ^′≥ Q ^′.

Thus no matter how closely we approximate to P via a minimalist basis we can approximate sufficiently closely to Q to ensure that the one proposition contains the other.

We would like to show that this definition agrees with the standard definition of partial content in its application to minimal propositions. We first show:

Lemma 33

Suppose that P is a minimal proposition and that P ^′ is a minimalist basis for P . Then P ^′≥ P .

Proof

(a) Suppose first that p ^′∈ P ^′.Since P ^′is intensionally equivalent to P, p ^′ | = P and, since P is minimal, $p^{\prime }\sqsupseteq p$for some p ∈ P.(b) Now suppose p ∈ P.Since P ^′is intensionally equivalent to P, p | = P ^′and, since P ^′is a minimalist basis for P,$p \sqsupseteq p^{\prime }$for some p ^′∈ P ^′.If p = p ^′,we are done. So suppose p≠p ^′.Then p ^′⊧P and p is not a minimal verifier of P(in fact, this shows P ⊆ P ^′).

We now establish:

Theorem 34

For minimal propositions P and Q , P ≥ Q iff P≥^Y Q .

Proof

Suppose P ≥ Q for minimal propositions P and Q.Take any minimalist basis P ^′for P.Then P ^′≥ P by the lemma. We let the minimalist basis Q ^′for Q be Q itself. Given that P ≥ Q, P ^′≥ P ≥ Q ^′;and so P≥^Y Q.Now suppose P≥^Y Q.Given that P is minimal, it is a minimalist basis for P.So for some minimalist basis Q ^′for Q, P ≥ Q ^′.But Q ^′≥ Q by the lemma; and so P ≥ Q.

Compact Verifiers

Gemes develops his account of partial content in terms of what one might call compact verification. Given a sentential formula A, a sentence letter will be relevant to A if ‘toggling’ its truth-value within some assignment of truth-values (i.e. changing its true-value from T to F) can make a difference to the truth-value of A. A compact verifier A is then an assignment of truth-values to the relevant sentence-letters of A under which it is true; and given the notion of a compact verifier, partial content can be defined in the usual way.

For our purposes, such a definition of partial content is far too linguistic; we need a definition that will apply purely at the level of content, without any reference to language. There are perhaps various ways to do this, but one fairly natural way is based on the idea that the assignments of truth-values to the relevant sentence letters will provide a minimally sufficient basis for settling the truth-value of the given formula.

To translate this idea to a more abstract setting, we make the following definitions (which are also of some independent interest). Given some subject-matter s, we say that w is an s -world if w is a consistent part of s and any part of s is either a part of w or inconsistent with w. An s-world is a kind of mini-world with respect to the subject-matter s. When s is the full state ■, then an s-world is a full world, or what we previously called a world-state.

The subject-matter s is said to be comprehensive if any consistent state is compatible with some s-world. In case s is the full state ■, it will be comprehensive just in case the space is what we called a W-space, in which every consistent state is part of a world-state. The subject-matter s is said to comprehend the proposition P if it is comprehensive and every s-world loosely verifies or loosely falsifies P; and s is said to be compact for P if it comprehends P and if s $\sqsubseteq$ t for any subject-matter t that comprehends P.Wenowsaythat s compactly verifies theproposition P-insymbols, s| = ^c P- if s| = P and s is an s-world for some compact subject-matter s for P.

Partial content can now be defined for compact propositions in the usual way. Thus given two compact propositions P and Q, we may say that P ≥ Q if every compact verifier of P contains a compact verifier of Q and every compact verifier of Q is contained in a compact verifier of P; and the definition may be extended to propositions P ^′ and Q ^′ intensionally equivalent to the compact propositions P and Q by taking P ^′≥ Q ^′ to hold just in case P ≥ Q holds. However, it is not in general true that a proposition P is intensionally equivalent to a compact proposition P ^′; and so, again, one must find some way of extending the definition to these other propositions.

We may do something analogous to what we already did in the case of minimal propositions. Say that the proposition Q is comprehensive if it is a set of s-worlds for some comprehensive subject-matter s and that Q is a comprehensive basis for P if Q is a comprehensive proposition intensionally equivalent to P. Note that, in a W-space, a comprehensive basis Q for a given proposition P will always exist since we can let Q be the set of world-states in which P is true.

We may now extend the relation of partial content to all propositions as follows (G for Gemes):

P≥^G Q if for any comprehensive basis P ^′ for P there is a comprehensive basis Q ^′ for Q for which P ^′≥ Q ^′.

Thus no matter how closely we approximate to P via a comprehensive basis we can approximate sufficiently closely to Q to ensure that the one proposition contains the other.

We may now establish analogues of lemma 30 and theorem 31:

Lemma 35

Suppose that P is a compact proposition and P ^′ a comprehensive basis for P . Then P ^′≥ P .

Proof

Given that P ^′is a comprehensive basis for P,it will consist of a set of s ^′-worldsfor some subject-matter s ^′comprehensive for P ^′and hence for P and, given that P is a compact proposition, it will consist of a setof s-worlds for some subject-matter s compact for P;and so s $\sqsubseteq$ s ^′.(a) Suppose first that w ^′∈ P ^′.Since w ^′is consistent, it is compatible with some s-world w and, since s $\sqsubseteq$ s ^′and w ^′is an s ^′-world,$w \sqsubseteq w^{\prime }$.But w either loosely verifies or loosely falsifies P and, since w ^′loosely verifies P,it is clear that w loosely verifies P and hence that w ∈ P.(b) Now suppose w ∈ P.Then w is compatible with some s ^′-world w ^′and so $w \sqsubseteq w^{\prime }$.But since w ^′either loosely verifies or loosely falsifies P and, since w loosely verifies P,it is clear that w loosely verifies P ^′and hence that w ∈ P ^′.

Theorem 36

For compact propositions P and Q , P ≥ Q iff P≥^G Q .

Proof

Suppose P ≥ Q for compact propositions P and Q.Take any comprehensive basis P ^′for P.Then P ^′≥ P by the lemma. We let the comprehensive basis Q ^′for Q be Q itself. Given that P ≥ Q, P ^′≥ P ≥ Q ^′;and so P≥^G Q.Now suppose P≥^G Q.Given that P is compact, it is a comprehensive basis for P.So for some comprehensive basis Q ^′for Q, P ≥ Q ^′.But Q ^′≥ Q by the lemma; and so P ≥ Q.

These results (theorems 34 and 36) do not, of course, establish the adequacy of the proposed definition of P≥^Y Q for cases in which P and Q are not both minimal or the adequacy of the proposed definition of P≥^G Q for cases in which P and Q are not both compact. I suspect that there is no reasonable alternative to our treatment of these cases, but further work would be required to show that this was indeed so.