## 1 Introduction

Sometimes theories are formulated with different sets of primitives and yet are definitionally equivalent. The non-logical primitives of a formalized language L are called its signature. There are many examples of theories (often involving formalized systems of arithmetic and set theory) formulated in very different signatures, which are nonetheless “equivalent”. To take a simple example, consider the theory $$T_1$$ of a reflexive relation:

\begin{aligned} T_1 : \qquad \forall x P(x,x), \end{aligned}

expressed with a binary relation symbol P. Suppose we introduce a new binary relation symbol, Q, and give an explicit definition of it—call this definition $$d_1$$—in terms of P as follows:

\begin{aligned} d_1: \qquad \forall x \forall y (Q(x,y) \leftrightarrow (P(x,y) \wedge x \ne y)). \end{aligned}

Call the extended theory $$T_1 + d_1$$. Then $$T_1 + d_1 \vdash \forall x \lnot Q(x,x)$$.

Let $$T_2$$ be the theory saying that Q is an irreflexive relation:

\begin{aligned} T_2 : \qquad \forall x \lnot Q(x,x). \end{aligned}

Consider following “inverse definition” of P in terms of Q, call it $$d_2$$:

\begin{aligned} d_2: \qquad \forall x \forall y(P(x,y) \leftrightarrow (Q(x,y) \vee x = y)). \end{aligned}

Call the extended theory $$T_2 + d_2$$. Then $$T_2 + d_2 \vdash \forall x P(x,x)$$. Moreover, we have:

\begin{aligned} T_1 + d_1 \vdash d_2&\text { and }&T_2 + d_2 \vdash d_1. \end{aligned}

These two theories, $$T_1$$ and $$T_2$$, are in fact equivalent with respect to these definitions. That is, $$T_1 + d_1$$ and $$T_2 + d_2$$ are logically equivalent:

\begin{aligned} \boxed {T_1 + d_1 \equiv T_2 + d_2.} \end{aligned}

We say that $$T_1$$ and $$T_2$$ have a “common definitional extension” and are therefore “definitionally equivalent” (see Definition 27 below).

As the reader may have guessed, this example derives from the standard mathematical method of passing between a non-strict preorder $$\le$$ and its corresponding strict preorder <: they may be defined in terms of each other. To put the above in the more standard notation:

\begin{aligned}&\{\forall x (x \le x), \forall x \forall y (x< y \leftrightarrow (x \le y \wedge x \ne y))\} \\&\qquad \qquad \qquad \qquad \qquad \qquad \equiv \{\forall x \lnot (x< x), \forall x \forall y(x \le y \leftrightarrow (x < y \vee x = y))\}. \end{aligned}

In effect, the explicit definitions $$d_1$$ and $$d_2$$ are “mutual inverses”. This notion will play a major role later.

Moving on to increasingly sophisticated examples, there are equivalent reformulations of Peano arithmetic $$\mathsf {PA}$$ with different primitives from the usual signature $$\{0,S,+,\times \}$$. For example, one might take exponentiation as the single basic notion, along with certain axioms for exponentiation, along with explicit definitions for $$0,S,+$$ and $$\times$$ (along with induction). The result is equivalent to $$\mathsf {PA}$$.

As is well-known, one can interpret a modification we shall call $$\mathsf {ZF}_0$$ of $$\mathsf {ZF}$$ set theory, obtained by removing the axiom of infinity, adding its negation, along with an additional axiom of “transitive containment”, into $$\mathsf {PA}$$. Moreover, the inverse of this interpretation interprets $$\mathsf {PA}$$ into $$\mathsf {ZF}_0$$.Footnote 1 So $$\mathsf {PA}$$ is definitionally equivalent to the theory of finite sets$$\mathsf {ZF}_0$$.

A similar relationship holds between the theory of formalized syntax and $$\mathsf {PA}$$. Suppose $$\textsf {S}_A$$ is the theory of concatenation for strings from alphabet A, with $$|A| \ge 2$$, and with the appropriate induction principle. Then $$\textsf {S}_A$$ is definitionally equivalent to $$\mathsf {PA}$$.Footnote 2

There are other examples—from mathematics, logic and philosophy of science.Footnote 3 To return to the broader point, the transformations between such equivalent formulations are rather like “basis transformations” in linear algebra and other parts of mathematics. In this paper and the follow-up, an analogous idea is investigated.Footnote 4

## 2 Syntax, Structures and Theories

Throughout, everything we consider is 1-sorted, relational and first-order.

### Definition 1

Let $$P = \{P_i\}_{i \in I}$$ be a set and let $$a: P \rightarrow \mathbb {N}$$. The pair $$S = (P, a)$$ is called a one-sorted signature, and a is called the arity function for S. The multiset $$t = (a(P_i) \mid P_i \in P)$$ is called the similarity type of the signature S. If $$a(P_i) = 0$$, then $$P_i$$ is called a sentence letter (or a propositional atom). The alphabet of S is P.

### Definition 2

$$(P^c,a^c)$$ is a copy of (Pa) iff the similarity types of (Pa) and $$(P^c,a^c)$$ are the same. $$(P^c,a^c)$$ is a disjoint copy of (Pa) when, in addition, $$P \cap P^c = \varnothing$$.

### Definition 3

$$L_P$$ is the first-order language over the signature $$S = (P,a)$$, where each symbol $$P_i$$ is a primitive relation symbol of arity $$a(P_i)$$. We will sometimes call $$L_P$$ “the P-language”.

### Definition 4

Let

\begin{aligned} \Phi = \{\phi _{1}, \dots \} = \{\phi _i\}_{i \in I_{\Phi }} \end{aligned}

be a set of $$L_P$$-formulas, which will be called defining formulas. Given the $$\phi _i$$, a corresponding set

\begin{aligned} Q = \{Q_i\}_{i \in I_{\Phi }} \end{aligned}

of new relation symbols $$Q_i$$ is introduced, such that the arity of each $$Q_i$$ matches the arity of its corresponding $$\phi _i$$. The new language $$L_Q$$ will sometimes be called “the Q-language”. The combined language is then called $$L_{P,Q}$$.

A theory T in L is a set of L-sentences. When we require deductive closure, we say so. We use a deductive system such that, if $$\Delta \vdash \alpha$$, then $$\Delta \vdash \forall x \alpha$$, so long as x doesn’t appear free in any formulas in $$\Delta$$.Footnote 5$$T \vdash \alpha$$ means: there exists a derivation of $$\alpha$$ from the axioms/rules of T. An L-structure A will always interpret all variables. So, we can always write $$A \models \alpha$$, even where $$\alpha$$ has free variables, since A will assign a value $$x^A$$ to each variable $$x \in \text {FV}(\alpha )$$.Footnote 6 The Completeness Theorem holds in the usual form: $$\Delta \vdash \alpha \text { iff, for any } A \models \Delta , A \models \alpha$$. A theory T in $$L_P$$ is said to be deductively closed iff, for all $$\alpha \in \text {Sent}(L_P)$$, if $$T \vdash \alpha$$, then $$\alpha \in T$$. The deductive closure of T (written $$\text {DedCl}(T)$$) is $$\{\alpha \in \text {Sent}(L_P) \mid T \vdash \alpha \}$$.

So far as I can tell, nothing in this paper uses either proof theoretic methods or model theoretic methods beyond what is taught at intermediate logic.Footnote 7 We do introduce specific new terminology for the following notions:

 Definition systems $$d_{\Phi }$$. Definitional expansions/extensions $$A \mapsto A + d_{\Phi }$$ and $$T \mapsto T + d_{\Phi }$$. Definitional image operator $$A \mapsto D_{\Phi }A$$ and $$T \mapsto D_{\Phi }T$$. Definitional equivalence $$A \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} B$$ and $$T_1 \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} T_2$$.

### Definition 5

A structureA for the language $$L_P$$ specifies a non-empty domain, $$\text {dom}(A)$$; and interprets each variable x of $$L_P$$ as an element $$x^{A} \in \text {dom}(A)$$; and interprets each n-ary relation symbol $$P_i$$ as a n-ary relation $$(P_i)^{A} \subseteq (\text {dom}(A))^n$$.

### Definition 6

Given signature P, let $$P^c$$ be a disjoint copy of P. Let A be an $$L_P$$-structure. Then the disjoint copy$$A^c$$ of A in $$L_{P^c}$$ is defined by setting $$\text {dom}(A^c) = \text {dom}(A)$$ and, for each $$P_i$$, setting $$(P^c_i)^{A^c} = (P_i)^A$$. Let T be an $$L_P$$-theory. Then the disjoint copy$$T^c$$ of T in $$L_{P^c}$$ is defined by replacing every occurrence of $$P_i$$, in any sentence in T by the new symbol $$P^c_i$$.

We may next give inductive definitions of the denotation function $$t \mapsto t^{A}$$ (specifying what any term t in $$L_P$$ refers to in A) and the satisfaction relation $$\models$$ between A and $$L_P$$-formulas.

### Definition 7

If $$A_1$$ and $$A_2$$ are $$L_{P}$$-structures, then an isomorphism

\begin{aligned} f: A_1 \rightarrow A_2 \end{aligned}

is a bijection from $$\text {dom}(A_1)$$ to $$\text {dom}(A_2)$$ satisfying the preservation condition that $$f[(P_i)^{A_1}] = (P_i)^{A_2}$$, for each relation symbol $$P_i$$ in the signature P. This is written

\begin{aligned} A_1 \overset{f}{\cong }A_2 \end{aligned}

(or simply $$A_1 \cong A_2$$ if the isomorphism is left implicit).

### Definition 8

If A is an $$L_P$$-structure and $$T, T_1,T_2$$ are sets of $$L_P$$-sentences:

 (1) $$\text {Th}_{L_P}(A)$$ := the set of $$L_P$$-sentences true in A. (2) $$A \models T$$ := for all $$\alpha \in T$$, $$A \models \alpha$$. (3) $$\text {Mod}_{L_P}(T)$$ := the class of $$L_P$$-structures $$A \models T$$ (4) $$T_1 \equiv T_2$$ := $$\text {Mod}(T_1) = \text {Mod}(T_2)$$. (5) $$T_1 \vdash T_2$$ := for all $$\alpha \in T_2$$, $$T_1 \vdash \alpha$$.

$$\text {Th}_{L_P}(A)$$ is called $$\text {Th}(A)$$ if it’s clear from context what language is involved. Likewise, $$\text {Mod}_{L_P}(T)$$ is called $$\text {Mod}(T)$$ if it’s clear from context what language is involved.

If $$T_1$$ and $$T_2$$ are deductively closed theories, then $$T_1 \vdash T_2 \text { iff } T_2 \subseteq T_1$$. The Completeness Theorem tells us that $$T_1 \equiv T_2$$ iff $$T_1 \vdash T_2$$ and $$T_2 \vdash T_1$$.

### Definition 9

Let A be an $$L_P$$-structure and $$A^{+}$$ be an $$L_{P,Q}$$-structure. Then $$A^{+}$$ is an expansion of A iff for all $$P_i$$, $$(P_i)^{A^{+}} = (P_i)^A$$. This is equivalent to saying that A is a reduct of $$A^{+}$$. If $$A^{+}$$ is an $$L_{P,Q}$$-structure, its reduct to P is denoted $$A^{+} \upharpoonright _{P}$$ (an $$L_P$$-structure) and its reduct to Q is denoted $$A^{+} \upharpoonright _{Q}$$ (an $$L_Q$$-structure) and we have: $$(P_i)^{A^{+} \upharpoonright _{P}} = (P_i)^{A^+}$$ and $$(Q_i)^{A^{+} \upharpoonright _{Q}} = (Q_i)^{A^+}$$.

The central property of expansions is that the truth value of a formula in the smaller language $$L_P$$ remains invariant as we pass from an $$L_P$$-structure to an expanded structure for $$L_{P,Q}$$: if $$\alpha \in L_P$$ and an $$L_{P,Q}$$-structure $$A^{+}$$ is an expansion of an $$L_P$$-structure A, then $$A^{+} \models \alpha \text { iff } A \models \alpha$$.

### Definition 10

A theory $$T^{+}$$ is an extension of T iff T is a subset of $$T^{+}$$. Let signatures PQ, and corresponding languages $$L_P, L_Q$$ and $$L_{P,Q}$$ be given. An extension $$T^{+}$$ in $$L_{P,Q}$$ of T in $$L_P$$ is called a conservative extension of T with respect to $$L_P$$-formulas iff, for any $$L_P$$-formula $$\alpha$$,

\begin{aligned} T^{+} \vdash \alpha \Rightarrow T \vdash \alpha \end{aligned}

## 3 Definition

### Definition 11

Given the set $$\Phi = \{\phi _i\}_{i \in I}$$ of $$L_P$$-formulas, we introduce a disjoint set $$Q = \{Q_i\}_{i \in I}$$ of new relation symbols, with $$\text {card} \ Q = \text {card} \ \Phi$$, and with the arity of $$Q_i$$ matching the arity of $$\phi _i$$, and let $$n_i$$ be $$a(\phi _i)$$. The definition system over $$\Phi$$, which we write as,

\begin{aligned} d_{\Phi } \end{aligned}

is the set of explicit definitions,

\begin{aligned} \forall x_1 \dots x_{n_i}(Q_i(x_1, \dots , x_{n_i}) \leftrightarrow \phi _{i}) \end{aligned}

where $$\{x_1,\dots x_{n_i}\} = \text {FV}(\phi _i)$$. These define the new symbols $$Q_i$$ in terms of the $$L_P$$-formulas $$\phi _i$$. We shall sometimes write $$\forall \overline{x}(Q_i( \overline{x}) \leftrightarrow \phi _{i})$$ instead of $$\forall x_1 \dots x_n(Q_i(x_1, \dots , x_n)\leftrightarrow \phi _{i})$$.Footnote 8

### Definition 12

If A is an $$L_P$$-structure and $$\phi (x_1,\dots ,x_n)$$ is an $$L_P$$-formula, then

\begin{aligned} \phi (A): = \{(a_1,\dots ,a_n) \in (\text {dom}(A))^n \mid A \models \phi [a_1, \dots , a_n]\} \end{aligned}

is the relation that $$\phi$$ defines in A.

### Definition 13

An $$L_{P,Q}$$-structure $$A^{+}$$ is a definitional expansion of an $$L_P$$-structure A with respect to the definition system $$d_{\Phi }$$ just if $$A^{+}$$ is an expansion of A interpreting each $$Q_i$$, and $$A^{+} \models d_{\Phi }$$. This ensures that $$(Q_i)^{A^{+}} = \phi _i(A^{+})$$, for each $$Q_i$$.

Given any $$L_P$$-structure A, it is clear that there is a unique definitional expansion $$A^{+} \models d_{\Phi }$$. We introduce the following notation for this expansion:

### Definition 14

$$A + d_{\Phi }$$ := the unique definitional expansion $$A^{+} \models d_{\Phi }$$.

We are going to treat the definitional expansion map

\begin{aligned} A \mapsto A + d_{\Phi } \end{aligned}

as a unary operator $$+ d_{\Phi }$$ (indexed by $$\Phi$$), taking us from $$L_P$$-structures to $$L_{P,Q}$$-structures. It is clear that it is well-defined (i.e., unique, given $$\Phi$$). It also satisfies the following useful “right cancellation” law (this amounts, in essence, to taking a reduct):

### Lemma 1

If $$A_1 + d_{\Phi } \cong A_2 + d_{\Phi }$$ then $$A_1 \cong A_2$$.

### Definition 15

A relation symbol $$P_i$$ in the signature P is explicitly definable in T just if

\begin{aligned} T \vdash P_i(\overline{x}) \leftrightarrow \theta \end{aligned}

for some formula $$\theta$$ in the language of the subsignature $$P {\setminus } P_i$$. We say that $$\theta$$ is a defining formula for $$P_i$$ in T.

### Definition 16

A relation symbol $$P_i$$ in the signature P is implicitly definable in T in $$L_P$$ just in case, given any pair of models $$A, B \models T$$, with $$\text {dom}(A)=\text {dom}(B)$$ and which assign the same extension to all $$P_j$$except$$P_i$$, we have $$(P_i)^A = (P_i)^B$$.

Beth’s Theorem states that a relation symbol $$P_i$$ is implicitly definable in T iff $$P_i$$ is explicitly definable in T.Footnote 9

### Definition 17

Given a definition system $$d_{\Phi }$$, the definitional extension of T wrt $$\Phi$$ is $$T + d_{\Phi }$$. We say that $$T^{+}$$ in $$L_{P,Q}$$ is a definitional extension of T in $$L_P$$ just if

\begin{aligned} T^{+} \equiv T + d_{\Phi }, \end{aligned}

for some definition system $$d_{\Phi }$$, where $$\Phi$$ is some set of $$L_P$$-formulas.

So, each new relation symbol $$Q_i$$ is explicitly defined in $$T + d_{\Phi }$$ and the defining formula for $$Q_i( \overline{x})$$ is simply $$\phi _i$$. Analogous to what we did with structures, we are going to treat the definitional extension map

\begin{aligned} T \mapsto T + d_{\Phi } \end{aligned}

for theories as a unary operator $$+ d_{\Phi }$$ (indexed by $$\Phi$$), taking us from $$L_P$$-theories to $$L_{P,Q}$$-theories. Again, it is well-defined (i.e., unique, given $$\Phi$$) and satisfies an analogous “right cancellation” law for $$L_P$$-theories:

### Lemma 2

The following are straightforward consequences of the definitions:

1. (1)

If $$T_1 + d_{\Phi } \equiv T_2 + d_{\Phi }$$ then $$T_1 \equiv T_2$$.

2. (2)

$$A \models T$$ iff $$A + d_{\Phi } \models T + d_{\Phi }$$.

3. (3)

If $$B \models T + d_{\Phi }$$, then $$B = A + d_{\Phi }$$, for some $$A \models T$$.

Before moving on to translations, we give three standard lemmas about definitional and conservative extensions (the converse of Lemma 3 is far from being true):

### Lemma 3

If $$T^{+}$$ in $$L_{P,Q}$$ is a definitional extension of T in $$L_P$$, then $$T^{+}$$ is a conservative extension of T wrt $$L_P$$-formulas.

### Lemma 4

$$T + d_{\Phi }$$ is a conservative extension of T for $$L_P$$-formulas.

### Lemma 5

Let T in $$L_P$$ and $$T^{+}$$ in $$L_{P,Q}$$ be such that $$T \subseteq T^{+}$$. Suppose that, for any model $$A \models T$$, there is an expansion $$A^{+} \models T^{+}$$. Then $$T^{+}$$ is a conservative extension of T for $$L_P$$-formulas.

## 4 Translation

### Definition 18

Let a definition system $$d_{\Phi }$$ be given. Define the translation, induced by $$\Phi$$

\begin{aligned} \tau _{\Phi }: L_Q \rightarrow L_P \end{aligned}

as follows. For symbols $$Q_i$$, variables $$x,y, \overline{x}$$, and for $$L_Q$$-formulas $$\alpha , \alpha _1, \alpha _2$$:

 (1) $$\tau _{\Phi }(Q_i(\overline{x}))$$ $$:=$$ $$(\phi _i)^{\prime }$$ (2) $$\tau _{\Phi }(x=y)$$ $$:=$$ $$(x=y)$$ (3) $$\tau _{\Phi }(\lnot \alpha )$$ $$:=$$ $$\lnot \tau _{\Phi }(\alpha )$$ (4) $$\tau _{\Phi }(\alpha _1 \# \alpha _2)$$ $$:=$$ $$\tau _{\Phi }(\alpha _1) \# \tau _{\Phi }(\alpha _2)$$ (5) $$\tau _{\Phi }(\mathbf {q} x \alpha )$$ $$:=$$ $$\mathbf {q} x \tau _{\Phi }(\alpha )$$.

$$\#$$ is any binary connective, $$\mathbf {q}$$ is a quantifier and $$(\phi _i)^{\prime }$$ is the result of ensuring that the free variables appearing $$\phi _i$$ are relabelled, to match those of $$Q_i( \overline{x})$$. We call $$\tau _{\Phi }$$ the translation induced by $$\Phi$$. It maps from the new language $$L_Q$$ back to the original language $$L_P$$.Footnote 10

### Lemma 6

For any $$\alpha , \beta \in L_Q$$, if $$\alpha \vdash \beta$$ then $$\tau _{\Phi }(\alpha ) \vdash \tau _{\Phi }(\beta )$$.

### Proof

We will prove this using a lemma below. Suppose $$\tau _{\Phi }(\alpha ) \nvdash \tau _{\Phi }(\beta )$$. This gives us a model $$A \models \tau _{\Phi }(\alpha )$$ and $$A \not \models \tau _{\Phi }(\beta )$$. By Lemma 15(1) below, $$D_{\Phi }A \models \alpha$$ and $$D_{\Phi } A \not \models \beta$$. So, $$\alpha \nvdash \beta$$. $$\square$$

Lemma 6 is a general property of translations, but its converse is not true in general.

Corresponding to a translation $$\tau _{\Phi } : L_Q \rightarrow L_P$$ is its “lift” $$\tau ^{+}_{\Phi } : L_{P,Q} \rightarrow L_P$$ from the combined language $$L_{P,Q}$$ down to $$L_P$$:

### Definition 19

Let $$\Phi$$ be given, along with definition system $$d_{\Phi }$$. Define the lifted translation$$\tau ^{+}_{\Phi }$$ induced by $$\Phi$$

\begin{aligned} \tau ^{+}_{\Phi }: L_{P,Q} \rightarrow L_P \end{aligned}

as follows. For symbols $$Q_i$$, $$P_i$$, variables $$x,y, \overline{x}$$:

 (1) $$\tau ^{+}_{\Phi }(Q_i(\overline{x}))$$ $$:=$$ $$(\phi _i)^{\prime }$$ (2) $$\tau ^{+}_{\Phi }(x=y)$$ $$:=$$ $$(x=y)$$ (3) $$\tau ^{+}_{\Phi }(P_i(\overline{x}))$$ $$:=$$ $$P_i( \overline{x})$$

Along with the requirement that $$\tau ^{+}_{\Phi }$$ commutes with the logical operators on the full language $$L_{P,Q}$$.

Thus, the translation $$\tau _{\Phi }$$ is the restriction to $$L_Q$$ of its lift, $$\tau ^{+}_{\Phi }$$.

Note that because the translations we are interested in always act as the identity on equations, it is always the case that $$\vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ if $$\alpha$$ is an equation. Thus, in inductive proofs establishing biconditionals of the form $$\alpha \leftrightarrow \tau _{\Phi }(\alpha )$$, we only need to check the condition holds for atomic formulas which are not identity formulas.

### Lemma 7

We have:

 (1) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ for any $$\alpha \in L_Q$$. (2) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau ^{+}_{\Phi }(\alpha )$$ for any $$\alpha \in L_{P,Q}$$. (3) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ for any $$\alpha \in L_Q$$. (4) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau ^{+}_{\Phi }(\alpha )$$ for any $$\alpha \in L_{P,Q}$$.

### Proof

For (2), we reason by induction. Let $$\alpha$$ be an atomic $$L_{P,Q}$$-formula. As noted above, for equations $$(x=y)$$, the translation $$\tau _{\Phi }(x=y)$$ is trivially $$(x=y)$$: so $$\alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ always holds for equations. Suppose $$\alpha$$ is an atomic formula of the form $$Q_i( \overline{x})$$. Then its translation $$\tau ^{+}_{\Phi }(\alpha )$$ is $$\phi _i$$. Since we have $$A + d_{\Phi } \models \forall \overline{x}(Q_i( \overline{x}) \leftrightarrow \phi _i)$$, we have: $$A + d_{\Phi } \models Q_i( \overline{x}) \leftrightarrow \tau _{\Phi }(Q_i( \overline{x}))$$. Instead let $$\alpha$$ be $$P_i( \overline{x})$$. Then its translation $$\tau ^{+}_{\Phi }(\alpha )$$ is simply $$\alpha$$. So, $$A+ d_{\Phi } \models \alpha \leftrightarrow \tau ^{+}_{\Phi }(\alpha )$$. The other cases are shown by induction on the construction of $$\alpha$$.

For (1), the result follows from (2), by restricting to $$L_Q$$-formulas (since $$\tau ^{+}_{\Phi }(\alpha ) = \tau _{\Phi }(\alpha )$$ for $$\alpha \in L_Q$$).

For (4), reasoning by induction, let $$\alpha$$ be an atomic $$L_P,Q$$-formula. If $$\alpha$$ is atomic, then the condition is trivial. Suppose $$\alpha$$ has the form $$P_i( \overline{x})$$. Again, he condition is trivial, since $$\tau ^{+}_{\Phi }(P_i( \overline{x}))$$ is $$P_i( \overline{x})$$. Instead, suppose $$\alpha$$ has the form $$Q_i( \overline{x})$$. Then its translation $$\tau ^{+}_{\Phi }(\alpha )$$ is $$\phi _i$$. $$Q_i( \overline{x}) \leftrightarrow \phi _i$$ is a theorem of $$T + d_{\Phi }$$, by construction. The other cases are shown by induction on the construction of $$\alpha$$.

For (3), the result follows from (4), by restricting to $$L_Q$$-formulas. $$\square$$

### Definition 20

Let $$\tau _{\Phi } : L_Q \rightarrow L_P$$ be the translation induced by $$d_{\Phi }$$. If $$T_2$$ is a theory in $$L_Q$$, then the image of $$T_2$$ under $$\tau _{\Phi }$$ is the set of $$L_P$$-sentences:

\begin{aligned} \tau _{\Phi }[T_2] := \{\tau _{\Phi }(\alpha ) \in L_P \mid \alpha \in T_2\} \end{aligned}

If $$T_1$$ is a theory in $$L_P$$, then the pre-image of $$T_1$$ under $$\tau _{\Phi }$$ is the set of $$L_Q$$-sentences:

\begin{aligned} (\tau _{\Phi })^{-1}[T_1] := \{\beta \in L_Q \mid \tau _{\Phi }(\beta ) \in T_1\} \end{aligned}

Similarly, if $$\Theta = \{\theta _i\}_{i \in I_P}$$ is a set of $$L_Q$$-formulas and $$d_{\Theta }$$ is the corresponding definition system over $$\Theta$$ (for the primitives $$P_i$$ of $$L_P$$), we can define a translation

\begin{aligned} \tau _{\Theta }: L_P \rightarrow L_Q \end{aligned}

by requiring that $$\tau _{\Theta }$$ commute with the logical operators and, for atomic $$L_P$$-formulas:

 (i) $$\tau _{\Theta }(P_i( \overline{x})) :=$$ $$(\theta _i)^{\prime }$$ (ii) $$\tau _{\Theta }(x=y) :=$$ $$(x=y)$$

Likewise, we can also define the lifted translation $$\tau ^{+}_{\Theta }: L_{P,Q} \rightarrow L_Q$$.

### Definition 21

Let $$\tau : L_Q \rightarrow L_P$$ be a translation. Let $$T_1$$ be a theory in $$L_P$$ and $$T_2$$ be a theory in $$L_Q$$. Then we say:

1. (1)

$$\tau$$interprets$$T_2$$ into $$T_1$$ iff $$T_1 \vdash \tau [T_2]$$.

2. (2)

$$\tau$$faithfully interprets$$T_2$$ into $$T_1$$ iff $$T_1 \equiv \tau [T_2]$$.

One may compare the condition $$T_1 \equiv \tau [T_2]$$ with Visser’s definition of faithful interpretability:

We write $$K : U \triangleleft _{\text { faith}} V$$ for: K is a faithful interpretation of U in V. This means that: for all U-sentences A, we have: $$U \vdash A$$ iff $$V \vdash A^{\tau _{K}}$$. (Visser [19], p. 6).

Thus $$\tau _{\Phi } : T_2 \triangleleft _{\text { faith}} T_1$$ holds iff, for all $$\alpha \in L_{T_2}$$, we have: $$T_2 \vdash \alpha$$ iff $$T_1 \vdash \tau _{\Phi }(\alpha )$$. Thus, $$\tau _{\Phi } : T_2 \triangleleft _{\text { faith}} T_1$$ iff $$T_1 \equiv \tau _{\Phi }[T_2]$$. This establishes:

### Lemma 8

$$\tau _{\Phi }$$ faithfully interprets $$T_2$$ into $$T_1$$ iff, for all $$\alpha \in L_Q$$, we have: $$T_2 \vdash \alpha$$ iff $$T_1 \vdash \tau _{\Phi }(\alpha )$$.

### Definition 22

Let $$\tau _{\Phi }: L_Q \rightarrow L_P$$ and $$\tau _{\Theta } : L_P \rightarrow L_Q$$ be translations induced by $$d_{\Phi }$$ and $$d_{\Theta }$$. Let $$T_1$$ be an $$L_{P}$$ theory. Let $$T_2$$ be an $$L_Q$$ theory. Then $$\tau _{\Theta }$$ is an right inverse of $$\tau _{\Phi }$$ in $$T_1$$ iff, for any $$\alpha \in L_P$$,

\begin{aligned} T_1 \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

We write this more suggestively as:

\begin{aligned} (\tau _{\Phi } \tau _{\Theta }=1)_{T_1} \end{aligned}

And $$\tau _{\Theta }$$ is an left inverse of $$\tau _{\Phi }$$ in $$T_2$$ iff, for any $$\beta \in L_Q$$,

\begin{aligned} T_2 \vdash \beta \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\beta )) \end{aligned}

Likewise, we write this more suggestively as:

\begin{aligned} (\tau _{\Theta } \tau _{\Phi } =1)_{T_2} \end{aligned}

The following two lemmas are easy to prove, and yet hold to the key to much that follows. Both lemmas use “invertibility conditions”, of the form:

\begin{aligned} A + d_{\Phi } \models d_{\Theta }\\ T + d_{\Phi } \vdash d_{\Theta }. \end{aligned}

As we see later, these conditions express a very strong constraint on the set $$\Phi$$ of defining $$L_P$$-formulas involved—the property of being a “representation basis” for A (or T) with inverse $$\Theta$$.

### Lemma 9

Suppose that $$A + d_{\Phi } \models d_{\Theta }$$. Then:

 (1) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ for $$\alpha \in L_Q$$. (2) $$A+ d_{\Phi } \models \alpha \leftrightarrow \tau ^{+}_{\Phi }(\alpha )$$ for $$\alpha \in L_{P,Q}$$. (3) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Theta }(\alpha )$$ for $$\alpha \in L_P$$. (4) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$ for $$\alpha \in L_Q$$. (5) $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ for $$\alpha \in L_P$$. (6) $$A + d_{\Phi } \models \tau _{\Phi }(\theta _i) \leftrightarrow \theta _i$$. (7) $$A + d_{\Phi } \models \tau _{\Theta }(\phi _i) \leftrightarrow \phi _i$$. (8) $$A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ for $$\alpha \in L_P$$. (9) $$A+ d_{\Phi } \models \alpha \leftrightarrow \tau ^{+}_{\Theta }(\alpha )$$ for $$\alpha \in L_{P,Q}$$.

### Proof

Claims (1) and (2) are already established in Lemma 7(1,2) and do not need the side condition. (They are included for convenience of reference.)

For (3), the proof is analogous to the proof of Lemma 7(1), but using the fact that $$A + d_{\Phi } \models d_{\Theta }$$. Reasoning by induction, let $$\alpha$$ be an atomic $$L_P$$-sentence, say $$P_i( \overline{x})$$. Then its translation $$\tau _{\Theta }(\alpha )$$ is $$\theta _i$$. And $$A + d_{\Phi } \models \forall \overline{x}(P_i( \overline{x}) \leftrightarrow \theta _i)$$. So $$A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \theta _i$$. That is, $$A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \tau _{\Theta }(P_i( \overline{x}))$$. The other cases are shown by induction on the construction of $$\alpha$$.

For (4), we already have that $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$, for any $$\alpha \in L_Q$$. But $$\tau _{\Phi }(\alpha )$$ is an $$L_P$$-formula. So, by (3), we have: $$A + d_{\Phi } \models \tau _{\Phi }(\alpha ) \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$. So, $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$, as required.

For (5), the reasoning is analogous to that for (4). And (6) and (7) are merely applications of (1) and (3).

For (8), using (5), we have $$A + d_{\Phi } \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$, for any $$\alpha \in L_P$$. But $$A + d_{\Phi }$$ is an expansion of the $$L_P$$-structure A, and $$\alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ is an $$L_P$$-formula. Thus, $$A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$.

For (9), the proof is analogous to the proof of (2), but is applied to the “lift” $$\tau ^{+}_{\Theta }: L_{P,Q} \rightarrow L_Q$$ of $$\tau _{\Theta }$$. Reasoning by induction, let $$\alpha$$ be an atomic $$L_P$$-sentence, say $$P_i( \overline{x})$$. Then its translation $$\tau ^{+}_{\Theta }(\alpha )$$ is $$\theta _i$$. And $$A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \theta _i$$, since $$A + d_{\Phi } \models d_{\Theta }$$. Instead, let $$\alpha$$ be an atomic $$L_Q$$-sentence, say $$Q_i( \overline{x})$$. Then its translation $$\tau ^{+}_{\Theta }(\alpha )$$ is $$Q_i( \overline{x})$$. Trivially, $$A + d_{\Phi } \models Q_i( \overline{x}) \leftrightarrow Q_i( \overline{x})$$. The other cases are shown by induction on the construction of $$\alpha$$. $$\square$$

The following lemma, and the corresponding proofs, is a near repetition of the previous one, except that it deals with theories:

### Lemma 10

Suppose that $$T + d_{\Phi } \vdash d_{\Theta }$$. Then:

 (1) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$ for $$\alpha \in L_Q$$. (2) $$T+ d_{\Phi } \vdash \alpha \leftrightarrow \tau ^{+}_{\Phi }(\alpha )$$ for $$\alpha \in L_{P,Q}$$. (3) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Theta }(\alpha )$$ for $$\alpha \in L_P$$. (4) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$ for $$\alpha \in L_Q$$. (5) $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ for $$\alpha \in L_P$$. (6) $$T + d_{\Phi } \vdash \tau _{\Phi }(\theta _i) \leftrightarrow \theta _i$$. (7) $$T + d_{\Phi } \vdash \tau _{\Theta }(\phi _i) \leftrightarrow \phi _i$$. (8) $$T \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ for $$\alpha \in L_P$$. (9) $$T+ d_{\Phi } \vdash \alpha \leftrightarrow \tau ^{+}_{\Theta }(\alpha )$$ for $$\alpha \in L_{P,Q}$$.

### Proof

Essentially, a repetition of the proofs for Lemma 9.$$\square$$

### Lemma 11

Suppose $$T_1$$ is an $$L_P$$-theory and $$T_2$$ is an $$L_Q$$-theory. Then:

 (1) If $$(\tau _{\Phi } \tau _{\Theta }=1)_{T_1}$$ then $$T_1 + d_{\Phi } \vdash d_{\Theta }$$. (2) If $$(\tau _{\Theta } \tau _{\Phi } =1)_{T_2}$$ then $$T_2 + d_{\Theta } \vdash d_{\Phi }$$.

### Proof

For (1), we assume $$\tau _{\Theta }$$ is a right inverse of $$\tau _{\Phi }$$ in $$T_1$$. I.e., for any $$\alpha \in L_P$$, $$T_1 \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$. Thus, $$T_1 \vdash P_i( \overline{x}) \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(P_i( \overline{x})))$$. Since $$\tau _{\Theta }(P_i( \overline{x}))$$ is $$\theta _i$$, we have $$T_1 \vdash P_i( \overline{x}) \leftrightarrow \tau _{\Phi }(\theta _i)$$. Taking the definitional extension, $$T_1 + d_{\Phi } \vdash P_i( \overline{x}) \leftrightarrow \tau _{\Phi }(\theta _i)$$. Now, for any $$L_Q$$-formula $$\beta$$, we have $$T_1 + d_{\Phi } \vdash \beta \leftrightarrow \tau _{\Phi }(\beta )$$ from Lemma 7(2). So, since $$\theta _i \in L_Q$$, $$T_1 + d_{\Phi } \vdash \theta _i \leftrightarrow \tau _{\Phi }(\theta _i)$$. Thus, $$T_1 + d_{\Phi } \vdash P_i( \overline{x}) \leftrightarrow \theta _i$$, as required.

We obtain (2) by relabelling everything ($$T_2$$ is now a theory in $$L_Q$$). $$\square$$

Lemma 11 says that if $$\tau _{\Theta }$$ is a right-inverse of $$\tau _{\Phi }$$, relative to T, then every relation symbol $$P_i$$ from the original language can be explicitly defined from the $$\theta _i$$. In a sense, the original definition system, $$d_{\Phi }$$ is a kind of inverse of $$d_{\Theta }$$.

## 5 Definitional Images

### Definition 23

Let A be an $$L_P$$-structure. Then the $$L_Q$$-structure $$D_{\Phi }A$$ is defined by:

\begin{aligned} D_{\Phi }A := (A + d_{\Phi }) \upharpoonright _{L_Q} \end{aligned}

$$D_{\Phi }A$$ is called the definitional image of A with respect to $$\Phi$$.

Immediately, we see that the following three conditions provide an equivalent characterization of $$D_{\Phi }A$$:

 (1) $$\text {dom}(D_{\Phi }A) = \text {dom}(A)$$ (2) $$x^{D_{\Phi }A} = x^{A}$$, for any variable x (3) $$(Q_i)^{D_{\Phi }A} = \phi _i(A)$$, for each new symbol $$Q_i$$.

### Lemma 12

For any $$\alpha \in L_Q$$: $$\alpha (D_{\Phi }A) = \alpha (A + d_{\Phi })$$.

Although $$A + d_{\Phi }$$ is by construction a definitional expansion of A, it is by no means automatically true that $$A + d_{\Phi }$$ is a definitional expansion of $$D_{\Phi }A$$. The requirement for this to hold is that each primitive relation $$(P_i)^A$$ be definable in $$D_{\Phi }A$$, by some formula, say $$\theta _i$$.

Turning to theories, we introduce analogous concepts:

### Definition 24

The definitional image of T, with respect to $$\Phi$$, is the restriction of the deductive closure of $$T + d_{\Phi }$$ to the new language $$L_Q$$. The definitional image of T with respect to $$\Phi$$ is denoted $$D_{\Phi }T$$. That is,

\begin{aligned} D_{\Phi }T : = \text {DedCl}(T + d_{\Phi })\upharpoonright _{L_Q} = \{\beta \in L_Q \mid T + d_{\Phi } \vdash \beta \} \end{aligned}

The definitional image $$D_{\Phi }T$$ of a theory T in $$L_P$$ is, essentially, the pre-image$$(\tau _{\Phi })^{-1}[T]$$ of T under the translation $$\tau _{\Phi }$$ induced by $$\Phi$$:

### Lemma 13

$$(\tau _{\Phi })^{-1}[T] \equiv D_{\Phi }T$$.

### Proof

Suppose $$D_{\Phi }T \vdash \beta$$, for $$\beta \in L_Q$$. Thus, $$T + d_{\Phi } \vdash \beta$$. But $$T + d_{\Phi } \vdash \beta \leftrightarrow \tau _{\Phi }(\beta )$$, by Lemma 10(1). So, $$T + d_{\Phi } \vdash \tau _{\Phi }(\beta )$$. Since $$T + d_{\Phi }$$ conservatively extends T for $$L_P$$-formulas, $$T \vdash \tau _{\Phi }(\beta )$$. Hence, $$\beta \in (\tau _{\Phi })^{-1}[T]$$. And so $$(\tau _{\Phi })^{-1}[T_1] \vdash \beta$$. Conversely, suppose $$\beta \in (\tau _{\Phi })^{-1}[T]$$. So, $$T \vdash \tau _{\Phi }(\beta )$$. So, $$T + d_{\Phi } \vdash \tau _{\Phi }(\beta )$$. But $$T + d_{\Phi } \vdash \beta \leftrightarrow \tau _{\Phi }(\beta )$$, by Lemma 10(1) again. Thus, $$D_{\Phi }T \vdash \beta$$. $$\square$$

### Definition 25

Let A be an $$L_P$$-structure, B an $$L_Q$$-structure, $$T_1$$ an $$L_P$$-theory and $$T_2$$ an $$L_Q$$-theory. Then we say:

 (1) $$\Phi$$definesB in A iff $$B \cong D_{\Phi }A$$. (2) $$\Phi$$proof-theoretically defines$$T_2$$ in $$T_1$$ iff $$T_2 \equiv D_{\Phi }T_1$$. (3) $$\Phi$$model-theoretically defines$$T_2$$ in $$T_1$$ iff $$\text {Mod}(T_2) = D_{\Phi }[\text {Mod}(T_1)]$$.

The second of these, (2), amounts to saying that $$T_2 \vdash D_{\Phi }T_1 \text { and } D_{\Phi }T_1 \vdash T_2$$. The third is equivalent to saying that

\begin{aligned} \text {Mod}(T_2) = (\text {Mod}(T_1+d_{\Phi }))\upharpoonright _{L_Q}. \end{aligned}

Note that the restriction $$\upharpoonright _{L_Q}$$ is taken after the models are extracted from the definitional extension $$T_1+d_{\Phi }$$. If the restriction is taken first, we get the rather different set $$\text {Mod}((\text {DedCl}(T_1+d_{\Phi })) \upharpoonright _{L_Q})$$ of models: i.e., $$\text {Mod}(D_{\Phi }T_1)$$. Indeed, this is generally a superset of $$D_{\Phi }[\text {Mod}(T)]$$. As Lemma 15(3) will show, we have: $$D_{\Phi }[\text {Mod}(T)] \subseteq \text {Mod}(D_{\Phi }T)$$.

## 6 Some Book-Keeping Lemmas

We next provide several groups of “book-keeping” lemmas about translations and definitional images. The first, Lemma 14, primarily concerns theories. The second group, Lemma 15, concerns semantics and models. The third group (in particular, Lemma 16) establishes five calculationally useful equivalences for the “definition invertibility condition” on structures:

\begin{aligned} \boxed {A + d_{\Phi } \models d_{\Theta }.} \end{aligned}

The fourth group establishes some analogous results for theories, including the main equivalence (Lemma 19) for the “definition invertibility condition” on theories:

\begin{aligned} \boxed {T + d_{\Phi } \models d_{\Theta }.} \end{aligned}

### Lemma 14

Each of the following is true:

 (1) $$T + d_{\Phi }$$ is a conservative extension of $$D_{\Phi }T$$ for $$L_Q$$-formulas. (2) $$D_{\Phi }T_1 \vdash T_2$$ iff $$T_1 \vdash \tau _{\Phi }[T_2]$$. (3) $$\tau _{\Phi }$$ interprets $$T_2$$ into $$T_1$$ iff $$D_{\Phi }T_1 \vdash T_2$$. (4) $$\tau _{\Phi }$$ interprets $$D_{\Phi }T$$ into T. (5) $$D_{\Phi }T = \{\beta \in L_Q \mid T \vdash \tau _{\Phi }(\beta )\}$$. (6) $$D_{\Phi }T \vdash \beta$$ iff $$T \vdash \tau _{\Phi }(\beta )$$ (for any $$\beta \in L_Q$$). (7) $$T_2 \equiv D_{\Phi }T_1$$ iff, for all $$\beta \in L_Q$$: $$T_2 \vdash \beta \text { iff } T_1 \vdash \tau _{\Phi }(\beta )$$. (8) $$T_2 \equiv D_{\Phi }T_1$$ iff $$\tau _{\Phi }$$ faithfully interprets $$T_2$$ into $$T_1$$.

### Proof

To establish (1), note that $$D_{\Phi }T$$ is simply the restriction of the deductive closure of $$T + d_{\Phi }$$ to $$L_Q$$-sentences. So if $$T + d_{\Phi } \vdash \alpha$$, where $$\alpha \in L_Q$$, then $$D_{\Phi }T \vdash \alpha$$, as required. Statement (2) simply expresses the relationship between images and pre-images. Statement (3) is an immediate corollary of (2). Statement (4) is an immediate corollary of (3).

For (5), note that $$D_{\Phi }T = (\tau _{\Phi })^{-1}[T]$$, from Lemma 13 above. Thus, $$D_{\Phi }T = \{\beta \in L_Q \mid T \vdash \tau _{\Phi }(\beta )\}$$. For (6), let $$\beta \in L_Q$$. Then, from (5), we have $$D_{\Phi }T \vdash \beta \text { iff } T \vdash \tau _{\Phi }(\beta )$$, as required.

For (7), suppose first that $$T_2 \equiv D_{\Phi }T_1$$. So, for all $$\beta \in L_Q$$, $$T_2 \vdash \beta$$ iff $$D_{\Phi }T_1 \vdash \beta$$. But from (6), $$D_{\Phi }T_1 \vdash \beta$$ iff $$T_1 \vdash \tau _{\Phi }(\beta )$$. So, for all $$\beta \in L_Q$$, $$T_2 \vdash \beta$$ iff $$T_1 \vdash \tau _{\Phi }(\beta )$$, as required.

Instead suppose that for all $$\beta \in L_Q$$, we have: $$T_2 \vdash \beta \text { iff } T_1 \vdash \tau _{\Phi }(\beta )$$. Then reason as follows:

 $$T_2\vdash \beta$$ iff $$T_1 \vdash \tau _{\Phi }(\beta )$$ (from the hypothesis) iff $$T_1 + d_{\Phi } \vdash \beta$$ (as $$T_1 + d_{\Phi } \vdash \beta \leftrightarrow \tau _{\Phi }(\beta )$$; Lemma 10(1)) iff $$D_{\Phi }T_1 \vdash \beta$$ (from the definition of $$D_{\Phi }T_1$$; i.e., Definition 24).

So, $$T_2 \equiv D_{\Phi }T_1$$, as required.

(8) is an immediate corollary of (7) and Lemma 8. $$\square$$

### Lemma 15

Each of the following holds:

 (1) For any $$\alpha \in L_Q$$: $$D_{\Phi }A \models \alpha$$ iff $$A \models \tau _{\Phi }(\alpha )$$. (2) If $$A \models T$$, then $$D_{\Phi }A \models D_{\Phi }T$$. (3) $$D_{\Phi }[\text {Mod}(T)] \subseteq \text {Mod}(D_{\Phi }T)$$. (4) Let $$T_2$$ be a theory in $$L_Q$$. If $$A + d_{\Phi } \models T_2$$ then $$D_{\Phi }A \models T_2$$.

### Proof

(1): By induction. Suppose $$\alpha = Q_i( \overline{x})$$. Thus,

\begin{aligned} D_{\Phi }A \models \alpha\Leftrightarrow & {} ((x_1)^{D_{\Phi }A},\dots , (x_n)^{D_{\Phi }A}) \in (Q_i)^{D_{\Phi }A}\\\Leftrightarrow & {} ((x_1)^{A},\dots , (x_n)^A) \in \phi _i(A)\\\Leftrightarrow & {} A \models \phi _i\\\Leftrightarrow & {} A \models \tau _{\Phi }(Q_i( \overline{x}))\\\Leftrightarrow & {} A \models \tau _{\Phi }(\alpha ). \end{aligned}

The case of equality, and the induction steps are routine.

(2): Let $$A \models T$$ and $$D_{\Phi }T \vdash \alpha$$, for some $$\alpha \in L_Q$$. We want to show $$D_{\Phi }A \models \alpha$$. We do this using the translation $$\tau _{\Phi }$$. Since $$D_{\Phi }T \vdash \alpha$$, it follows, by Lemma 14(4), that $$T \vdash \tau _{\Phi }(\alpha )$$. And so, $$A \models \tau _{\Phi }(\alpha )$$. From (1), $$D_{\Phi }A \models \alpha$$ iff $$A \models \tau _{\Phi }(\alpha )$$. So, $$D_{\Phi }A \models \alpha$$, as required.

(3): Let $$B \in D_{\Phi }[\text {Mod}(T)]$$. So, there is some $$A \models T$$ with $$B \cong D_{\Phi }A$$. Since $$A \models T$$, we have $$D_{\Phi } A \models D_{\Phi }T$$, by (2). So, $$B \in \text {Mod}(D_{\Phi }T)$$.

(4): Suppose that for any $$\beta$$ such that $$T_2 \vdash \beta$$, we have $$A + d_{\Phi } \models \beta$$. We want to show that for any $$\beta$$ such that $$T_2 \vdash \beta$$, we have $$D_{\Phi }A \models \beta$$. So, let $$T_2 \vdash \beta$$ and $$A + d_{\Phi } \models \beta$$. By Lemma 7(1), $$A + d_{\Phi } \models \tau _{\Phi }(\beta )$$. So, $$A \models \tau _{\Phi }(\beta )$$. By (1), $$D_{\Phi }A \models \beta$$. $$\square$$

Neither the converse of Lemma 15(2) nor the converse of the inclusion in Lemma 15(3) is true.

### Lemma 16

The following are equivalent:

 (1) $$A + d_{\Phi } \models d_{\Theta }$$. (2) For all $$P_i$$, for all $$Q_j$$: $$(P_i)^{A + d_{\Phi }} = \theta _i(A + d_{\Phi })$$ & $$(Q_j)^{A + d_{\Phi }} = \phi _j(A + d_{\Phi })$$. (3) $$A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta }$$. (4) $$D_{\Theta }D_{\Phi }A = A$$. (5) $$A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ (with $$\alpha \in L_P$$)

### Proof

For $$(1) \Leftrightarrow (2)$$. This is simply unwinding the definitions.

For $$(1) \Rightarrow (3)$$, suppose $$A + d_{\Phi } \models d_{\Theta }$$. Then, from (2), we have, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = \theta _i(A + d_{\Phi })$$ & $$(Q_j)^{A + d_{\Phi }} = \phi _j(A + d_{\Phi })$$.

So, since $$\theta _i(A + d_{\Phi }) = \theta _i(D_{\Phi }A + d_{\Theta }) = (P_i)^{D_{\Phi }A +d_{\Theta } }$$, we have, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = (P_i)^{D_{\Phi }A +d_{\Theta } }$$ & $$(Q_j)^{A + d_{\Phi }} = \phi _j(A + d_{\Phi })$$.

So, since $$\phi _j(A + d_{\Phi }) = (Q_j)^{D_{\Phi }A + d_{\Theta }}$$, we have, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = (P_i)^{D_{\Phi }A +d_{\Theta } }$$ & $$(Q_j)^{A + d_{\Phi }} = (Q_j)^{D_{\Phi }A + d_{\Theta }}$$.

So, $$A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta }$$, as required.

For $$(3) \Rightarrow (1)$$. This reverses the reasoning $$(1) \Rightarrow (3)$$. Suppose $$A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta }$$. That is, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = (P_i)^{D_{\Phi }A +d_{\Theta } }$$ & $$(Q_j)^{A + d_{\Phi }} = (Q_j)^{D_{\Phi }A + d_{\Theta }}$$.

Now $$(Q_j)^{D_{\Phi }A + d_{\Theta }} = (Q_j)^{D_{\Phi }A} = \phi _j(A) = \phi _j(A + d_{\Phi })$$. So, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = (P_i)^{D_{\Phi }A +d_{\Theta } }$$ & $$(Q_j)^{A + d_{\Phi }} = \phi _j(A + d_{\Theta })$$.

But trivially, $$D_{\Phi }A +d_{\Theta } \models d_{\Theta }$$. So, $$D_{\Phi }A +d_{\Theta } \models P_i( \overline{x}) \leftrightarrow \theta _i$$, for each $$P_i$$. Thus,

\begin{aligned} (P_i)^{D_{\Phi }A +d_{\Theta }} = \theta _i(D_{\Phi }A +d_{\Theta }). \end{aligned}

But we also have

\begin{aligned} D_{\Phi }A +d_{\Theta } \cong A + d_{\Theta } \end{aligned}

Therefore,

\begin{aligned} (P_i)^{D_{\Phi }A +d_{\Theta }} = \theta _i(A + d_{\Theta }) \end{aligned}

Thus, for all $$P_i$$, for all $$Q_j$$,

 $$(P_i)^{A + d_{\Phi }} = \theta _i(A +d_{\Theta })$$ & $$(Q_j)^{A + d_{\Phi }} = \phi _j(A + d_{\Theta })$$.

And therefore, $$A + d_{\Phi } \models d_{\Theta }$$, as required.

For $$(1) \Rightarrow (4)$$. Suppose $$A + d_{\Phi } \models d_{\Theta }$$. So, from (3) above, we have:

\begin{aligned} A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta } \end{aligned}

Therefore, $$D_{\Phi }A + d_{\Theta } \models d_{\Phi }$$. A proof analogous to that of (3) above gives us that if $$B + d_{\Theta } \models d_{\Phi }$$ then $$B + d_{\Theta } \cong D_{\Theta } B + d_{\Phi }$$, and so, if $$D_{\Phi }A + d_{\Theta } \models d_{\Phi } \text { then } D_{\Phi }A + d_{\Theta } \cong D_{\Theta } D_{\Phi }A + d_{\Phi }$$. And so, we may conclude,

\begin{aligned} D_{\Phi }A + d_{\Theta } \cong D_{\Theta } D_{\Phi }A + d_{\Phi } \end{aligned}

And thus,

\begin{aligned} A + d_{\Phi } \cong D_{\Theta } D_{\Phi }A + d_{\Phi } \end{aligned}

And right cancellation gives,

\begin{aligned} A \cong D_{\Theta } D_{\Phi }A \end{aligned}

But since the map $$D_{\Phi }$$ leaves the domain invariant, we have $$A = D_{\Theta } D_{\Phi }A$$.

For $$(4) \Rightarrow (5)$$, suppose $$A = D_{\Theta } D_{\Phi }A$$. We have, applying Lemma 15(1) twice, for any $$\alpha \in L_P$$, $$\beta \in L_Q$$:

\begin{aligned} \begin{array}{rll} A \models \tau _{\Phi }(\beta ) &{} \text { iff } &{} D_{\Phi }A \models \beta \\ D_{\Phi }A \models \tau _{\Theta }(\alpha ) &{} \text { iff } &{} D_{\Theta } D_{\Phi }A \models \alpha \end{array} \end{aligned}

So, for $$\alpha \in L_P$$,

\begin{aligned} A \models \tau _{\Phi }(\tau _{\Theta }(\alpha )) \text { iff } D_{\Theta } D_{\Phi }A \models \alpha \end{aligned}

But $$A = D_{\Theta } D_{\Phi }A$$, and so,

\begin{aligned} A \models \tau _{\Phi }(\tau _{\Theta }(\alpha )) \text { iff } A \models \alpha \end{aligned}

And thus,

\begin{aligned} A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

as required.

For $$(5) \Rightarrow (1)$$, assume $$A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$. So,

\begin{aligned} A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(P_i( \overline{x}))) \end{aligned}

But $$\tau _{\Theta }(P_i( \overline{x})) = \theta _i$$. So,

\begin{aligned} A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \tau _{\Phi }(\theta _i) \end{aligned}

And $$\theta _i \in L_Q$$, and so, from Lemma 7(1),

\begin{aligned} A + d_{\Phi } \models P_i( \overline{x}) \leftrightarrow \theta _i \end{aligned}

and thus,

\begin{aligned} A + d_{\Phi } \models d_{\Theta } \end{aligned}

as required.$$\square$$

Turning next to theories:

### Lemma 17

$$D_{\Phi }T + d_{\Theta } \vdash \alpha \leftrightarrow \tau _{\Theta }(\alpha )$$, for all $$\alpha \in L_P$$.

### Proof

We reason by induction on the construction of $$\alpha$$. Let $$\alpha = P_i( \overline{x})$$. Then we have: $$D_{\Phi }T + d_{\Theta } \vdash P_i( \overline{x}) \leftrightarrow \theta _i$$. But $$\tau _{\Theta }(P_i( \overline{x})) = \theta _i$$. And thus, $$D_{\Phi }T + d_{\Theta } \vdash P_i( \overline{x}) \leftrightarrow \tau _{\Theta }(P_i( \overline{x}))$$. The equality case and compound cases proceed routinely. (Notice this is analogous to Lemma 10(1).) $$\square$$

### Lemma 18

Suppose that $$T + d_{\Phi } \vdash d_{\Theta }$$. Then, for any $$\alpha \in L_Q$$:

 (1) $$D_{\Phi }T \vdash \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$ (2) $$D_{\Phi }T + d_{\Theta } \vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$.

### Proof

For (1), we have, for any $$\beta \in L_Q$$, if $$T + d_{\Phi } \vdash \beta$$, then $$D_{\Phi }T \vdash \beta$$, since $$D_{\Phi }T$$ is simply the restriction of the set of theorems of $$T + d_{\Phi }$$ to $$L_Q$$. By Lemma 10(4), we have $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$, for any $$\alpha \in L_Q$$. Now $$\alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$ is also in $$L_Q$$. Thus, $$D_{\Phi }T \vdash \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$.

For (2), from Lemma 17, we have, for any $$\alpha \in L_Q$$: $$D_{\Phi }T + d_{\Theta } \vdash \tau _{\Phi }(\alpha ) \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$, since $$\tau _{\Phi }(\alpha ) \in L_P$$. By (1), we have $$D_{\Phi }T + d_{\Theta } \vdash \alpha \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\alpha ))$$. Hence, $$D_{\Phi }T + d_{\Theta } \vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha )$$, as required. $$\square$$

The next lemma is the most important result needed for Theorem 2 given in Sect. 8:

### Lemma 19

The following are equivalent

 (1) $$T + d_{\Phi } \vdash d_{\Theta }$$. (2) $$T + d_{\Phi } \equiv D_{\Phi }T + d_{\Theta }$$.

### Proof

$$(1) \Rightarrow (2)$$. Suppose $$T + d_{\Phi } \vdash d_{\Theta }$$. We want to show

 (a) $$T + d_{\Phi } \vdash D_{\Phi }T + d_{\Theta }$$. (b) $$D_{\Phi }T + d_{\Theta } \vdash T + d_{\Phi }$$.

First, for (a), let $$A + d_{\Phi } \models T + d_{\Phi }$$. So, since $$T + d_{\Phi } \vdash d_{\Theta }$$, we have $$A + d_{\Phi } \models d_{\Theta }$$. So, by Lemma 16(3), we have $$A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta }$$. Since $$A \models T$$, we have $$D_{\Phi }A \models D_{\Phi }T$$. So, $$D_{\Phi }A + d_{\Theta } \models D_{\Phi }T + d_{\Theta }$$. Thus, $$A + d_{\Phi } \models D_{\Phi }T + d_{\Theta }$$. And therefore, since A was arbitrary, $$T + d_{\Phi } \vdash D_{\Phi }T + d_{\Theta }$$, as required.

For (b), we want to show that $$D_{\Phi }T + d_{\Theta } \vdash T+d_{\Phi }$$. That is, for for any $$L_{P,Q}$$-formula $$\alpha$$,

\begin{aligned} \text {If } T +d_{\Phi } \vdash \alpha \text { then } D_{\Phi }T + d_{\Theta } \vdash \alpha . \end{aligned}

First, note that we may relabel Lemma 9(2) in terms of some $$L_Q$$-structure B and definition system $$d_{\Theta }$$, rather than A and $$d_{\Phi }$$, to obtain: for any $$\alpha \in L_{P,Q}$$,

 (i) $$B + d_{\Theta } \models \alpha \leftrightarrow \tau ^{+}_{\Theta }(\alpha )$$.

For a contradiction, suppose we have some $$\alpha \in L_{P,Q}$$ such that

 (ii) $$T + d_{\Phi } \vdash \alpha$$ (iii) $$D_{\Phi }T + d_{\Theta } \nvdash \alpha$$.

Since $$T + d_{\Phi } \vdash d_{\Theta }$$, we have, from Lemma 10(9), that $$T + d_{\Phi } \vdash \alpha \leftrightarrow \tau ^{+}_{\Theta }(\alpha )$$ and since $$T + d_{\Phi } \vdash \alpha$$, we have:

 (iv) $$T + d_{\Phi } \vdash \tau ^{+}_{\Theta }(\alpha )$$.

From (iii), there exists an $$L_Q$$-structure $$B \models D_{\Phi }T$$ with $$B + d_{\Theta } \not \models \alpha$$. So, from (i), $$B + d_{\Theta } \not \models \tau ^{+}_{\Theta }(\alpha )$$. And since $$\tau ^{+}_{\Theta }(\alpha ) \in L_Q$$, we have:

 (v) $$B \not \models \tau ^{+}_{\Theta }(\alpha )$$.

Recall that $$D_{\Phi }T = \{\beta \in L_Q \mid T + d_{\Phi } \vdash \beta \}$$. So, since $$B \models D_{\Phi }T$$, it follows that, for any $$\beta \in L_Q$$, if $$T + d_{\Phi } \vdash \beta$$, then $$B \models \beta$$. Thus, if $$T + d_{\Phi } \vdash \tau ^{+}_{\Theta }(\alpha )$$, then $$B \models \tau ^{+}_{\Theta }(\alpha )$$. So, from (iv), we infer:

 (vi) $$B \models \tau ^{+}_{\Theta }(\alpha )$$.

Contradiction.

$$(2) \Rightarrow (1)$$. Let $$T + d_{\Phi } \equiv D_{\Phi }T + d_{\Theta }$$. Then $$T + d_{\Phi } \vdash D_{\Phi }T + d_{\Theta }$$. Thus, $$T + d_{\Phi } \vdash d_{\Theta }$$, as required. $$\square$$

### Lemma 20

$$D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$, for $$\alpha \in L_P$$.

### Proof

If we examine the definitions of $$D_{\Phi }T$$ and $$D_{\Theta }T_2$$ (where $$T_2$$ is in $$L_Q$$), we get

\begin{aligned} D_{\Phi }T= & {} \{\alpha \in L_Q \mid T \vdash \tau _{\Phi }(\alpha )\}\\ D_{\Theta }T_2= & {} \{\alpha \in L_P \mid T_2 \vdash \tau _{\Theta }(\alpha )\} \end{aligned}

Together, these imply that $$D_{\Theta }D_{\Phi }T \vdash \alpha$$ iff $$T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$. $$\square$$

### Lemma 21

Suppose $$T + d_{\Phi } \vdash d_{\Theta }$$. Then

 (1) $$D_{\Phi }T + d_{\Theta } \vdash d_{\Phi }$$. (2) $$D_{\Theta }D_{\Phi }T \equiv T$$.

### Proof

For (1), suppose $$T + d_{\Phi } \vdash d_{\Theta }$$. By Lemma 18(2), for any $$\alpha \in L_Q$$,

\begin{aligned} D_{\Phi }T + d_{\Theta } \vdash \alpha \leftrightarrow \tau _{\Phi }(\alpha ) \end{aligned}

Thus,

\begin{aligned} D_{\Phi }T + d_{\Theta } \vdash Q_i( \overline{x}) \leftrightarrow \tau _{\Phi }(Q( \overline{x})) \end{aligned}

Thus,

\begin{aligned} D_{\Phi }T + d_{\Theta } \vdash Q_i( \overline{x}) \leftrightarrow \phi _i \end{aligned}

as required.

For (2), suppose $$T + d_{\Phi } \vdash d_{\Theta }$$. We want to show that, for any $$\alpha \in L_P$$, we have: $$D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \alpha$$. By Lemma 20, we have

\begin{aligned} D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

Since $$T + d_{\Phi } \vdash d_{\Theta }$$, by Lemma 10(8), we have

\begin{aligned} T \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

Hence,

\begin{aligned} T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

And therefore,

\begin{aligned} D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \alpha \end{aligned}

as required. $$\square$$

### Lemma 22

The following are equivalent:

 (1) $$T \equiv D_{\Theta } D_{\Phi } T$$. (2) $$T \vdash \alpha$$ iff $$T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$, with $$\alpha \in L_P$$.

### Proof

$$(1) \Rightarrow (2)$$. From the previous lemma, $$D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$. So, if $$T \equiv D_{\Theta } D_{\Phi } T$$, we infer that $$T \vdash \alpha$$ iff $$T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$, for $$\alpha \in L_P$$.

$$(2) \Rightarrow (1)$$. Suppose, for $$\alpha \in L_P$$,

\begin{aligned} T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

We already have

\begin{aligned} D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

So, for $$\alpha \in L_P$$

\begin{aligned} D_{\Theta }D_{\Phi }T \vdash \alpha \text { iff } T \vdash \alpha \end{aligned}

Thus, $$D_{\Theta }D_{\Phi }T \equiv T$$. $$\square$$

Notice that, in the case of structures, we have:

\begin{aligned} D_{\Theta }D_{\Phi }A = A \text { then } A \models \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

However, unlike that case, for theories, we only have established:

\begin{aligned} \text {If } T \equiv D_{\Theta } D_{\Phi } T \text { then } T \vdash \alpha \text{ iff } T \vdash \tau _{\Phi }(\tau _{\Theta }(\alpha )) \end{aligned}

So, we have not established the provability of the biconditional $$\alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$ inside T itself.Footnote 11

## 7 Definitional Equivalence

We next explain what it means for structures and theories to be definitionally equivalent.Footnote 12 Intuitively, a pair of structures are definitionally equivalent when they have a common definitional expansion. And a pair of theories are definitionally equivalent when they have a common definitional extension.

Throughout the next two definitions, A is an $$L_P$$-structure, $$\Phi = \{\phi _i\}_{i \in I_1}$$ is a set of $$L_P$$-formulas; Q is the new disjoint signature corresponding to $$\Phi$$; and B is an $$L_Q$$-structure. Similarly, $$\Theta = \{\theta _i\}_{i \in I_2}$$ is a set of $$L_Q$$-formulas. $$d_{\Phi }$$ is the definition system of the $$Q_i$$ primitives in terms of the $$\phi _i$$, and $$d_{\Theta }$$ is the definition system the $$P_i$$ primitives in terms of the $$\theta _i$$. Similarly, $$T_1$$ is an $$L_P$$-theory and $$T_2$$ is an $$L_Q$$-theory. Please note that the definitions of these notions given below assume disjoint signatures.

### Definition 26

Structures A and B are definitionally equivalent wrt $$d_{\Phi }$$ and $$d_{\Theta }$$ iff

\begin{aligned} A + d_{\Phi } \cong B + d_{\Theta }. \end{aligned}

If this is so, we write:

\begin{aligned} A \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} B \end{aligned}

### Definition 27

Theories $$T_1$$ and $$T_2$$ are definitionally equivalent wrt $$d_{\Phi }$$ and $$d_{\Theta }$$ iff

\begin{aligned} T_1 + d_{\Phi } \equiv T_2 + d_{\Theta }. \end{aligned}

To express this, we write:

\begin{aligned} T_1 \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} T_2 \end{aligned}

These definitions require that the signatures of A and B (or $$T_1$$ and $$T_2$$) be disjoint. But is not A obviously definitionally equivalent to itself? Is not a theory definitionally equivalent to itself? Well, one can always arrange for a pair of structures A and B in overlapping signatures to be reformulated as copies$$A^c$$ and $$B^c$$ in entirely disjoint signatures (see Definitions 26 above). If $$A^c$$ is a disjoint copy of A in $$L_{P^c}$$, then clearly $$A^c$$ is definitionally equivalent to A with respect to the trivial definition systems:

 $$d^{P^c}_{P}$$ : $$\forall \overline{x}(P^c_i( \overline{x}) \leftrightarrow P_i( \overline{x}))$$ $$d^{P}_{P^c}$$ : $$\forall \overline{x}(P_i( \overline{x}) \leftrightarrow P^c_i( \overline{x}))$$.

That is,

\begin{aligned} A + d^{P^c}_{P} \cong A^c + d^{P}_{P^c} \end{aligned}

A similar copying procedure can be adopted for theories too. If we have a theory T in $$L_P$$, and $$T^c$$ is a disjoint copy of T in $$L_{P^c}$$, then clearly $$T^c$$ is definitionally equivalent to T with respect to the definitions:

\begin{aligned} T + d^{P^c}_{P} \equiv T^c + d^{P}_{P^c} \end{aligned}

So, one can give a more general definition of definitional equivalence by first applying this copying procedure to both structures, and then applying the definition above.Footnote 13

## 8 Main Results

We finally give two main results, which characterize definitional equivalence.

### Theorem 1

The following are equivalent:

 (1) $$A \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} B$$ (2) $$D_{\Phi }A \cong B$$ and $$D_{\Theta }B \cong A$$

### Proof

$$(1) \Rightarrow (2)$$. Let $$A \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} B$$. Thus, $$A + d_{\Phi } \cong B + d_{\Theta }$$. So,

\begin{aligned} A + d_{\Phi } \models d_{\Theta }\\ B + d_{\Theta } \models d_{\Phi } \end{aligned}

Thus, by Lemma 16(3) (switching labels in the second case), we have:

\begin{aligned} A + d_{\Phi } \cong D_{\Phi }A + d_{\Theta }\\ B + d_{\Theta } \cong D_{\Theta }B + d_{\Phi } \end{aligned}

Thus,

\begin{aligned} B + d_{\Theta } \cong D_{\Phi }A + d_{\Theta }\\ A + d_{\Phi } \cong D_{\Theta }B + d_{\Phi } \end{aligned}

And by right cancellation,

\begin{aligned} B \cong D_{\Phi }A\\ A \cong D_{\Theta }B \end{aligned}

$$(2) \Rightarrow (1)$$. Let $$D_{\Phi }A \cong B$$ and $$D_{\Theta }B \cong A$$. Since we have two isomorphisms, we can arrange, without loss of generality, for $$\text {dom}(D_{\Phi }A) = \text {dom}(B)$$, and $$\text {dom}(D_{\Theta }B) = \text {dom}(A)$$. We want to show

 (a) $$(P_i)^{A + d_{\Phi }} = (P_i)^{B+ d_{\Theta }}$$ (b) $$(Q_i)^{A + d_{\Phi }} = (Q_i)^{B + d_{\Theta }}$$

First, we work out the extensions of $$P_i$$ and $$Q_i$$ in $$A + d_{\Phi }$$. For $$Q_i$$, we have $$(Q_i)^{A + d_{\Phi }} = \phi _i(A)$$. Using the assumption $$D_{\Theta }B = A$$, we have $$(P_i)^{A + d_{\Phi }} = (P_i)^{D_{\Theta }B+ d_{\Phi }}$$. So, $$(P_i)^{A + d_{\Phi }} = \theta _i(B)$$. And using the assumption $$D_{\Phi }A = B$$, we have $$(P_i)^{A+ d_{\Phi }} = \theta _i(D_{\Phi }A)$$. Summarizing:

 (i) $$(P_i)^{A+ d_{\Phi }} = \theta _i(D_{\Phi }A)$$. (ii) $$(Q_i)^{A + d_{\Phi }} = \phi _i(A)$$.

Next working out the extensions in $$B + d_{\Theta }$$, and using the assumption $$D_{\Phi }A = B$$, we have $$(P_i)^{B+ d_{\Theta }} = (P_i)^{D_{\Phi }A + d_{\Theta }}$$ and $$(Q_i)^{B + d_{\Theta }} = (Q_i)^{D_{\Phi }A + d_{\Theta }}$$. I.e.,

 (iii) $$(P_i)^{B+ d_{\Theta }} = \theta _i(D_{\Phi }A)$$. (iv) $$(Q_i)^{B + d_{\Theta }} = \phi _i(A)$$.

And (i) and (iii) imply (a), while (ii) and (iv) imply (b). $$\square$$

### Theorem 2

The following are equivalent:

 (1) $$T_1 \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} T_2$$. (2) $$(\tau _{\Phi } \tau _{\Theta }=1)_{T_1}$$ and $$T_2 \equiv D_{\Phi }T_1$$.

### Proof

For $$(1) \Rightarrow (2)$$, suppose $$T_1 \underset{\Theta }{\overset{\Phi }{\longleftrightarrow }} T_2$$. Thus, $$T_1 + d_{\Phi } \equiv T_2 + d_{\Theta }$$. Thus, $$T_1 + d_{\Phi } \vdash d_{\Theta }$$.

Then, from Lemma 10(8), we have $$T_1 \vdash \alpha \leftrightarrow \tau _{\Phi }(\tau _{\Theta }(\alpha ))$$, for $$\alpha \in L_P$$. I.e., $$(\tau _{\Phi } \tau _{\Theta }=1)_{T_1}$$. And secondly, since $$T_1 + d_{\Phi } \vdash d_{\Theta }$$, we have $$T_1 + d_{\Phi } \equiv D_{\Phi }T_1 + d_{\Theta }$$, by Lemma 19. So, $$T_2 + d_{\Theta } \equiv D_{\Phi }T_1 + d_{\Theta }$$, and by right cancellation, $$T_2 \equiv D_{\Phi }T_1$$.

For $$(2) \Rightarrow (1)$$, suppose $$(\tau _{\Phi } \tau _{\Theta }=1)_{T_1}$$ and $$T_2 \equiv D_{\Phi }T_1$$. From Lemma 11(1), we may conclude that $$T_1 + d_{\Phi } \vdash d_{\Theta }$$, and from this, we may conclude that $$T_1 + d_{\Phi } \equiv D_{\Phi }T_1 + d_{\Theta }$$, via Lemma 19. But since $$T_2 \equiv D_{\Phi }T_1$$, we may conclude that $$T_1 + d_{\Phi } \equiv T_2 + d_{\Theta }$$, as required.$$\square$$

Indeed, from $$T_1 + d_{\Phi } \equiv T_2 + d_{\Theta }$$, we may also conclude $$(\tau _{\Phi } \tau _{\Theta }=1)_{T_1}$$, $$(\tau _{\Theta } \tau _{\Phi } =1)_{T_2}$$, $$T_2 \equiv D_{\Phi }T_1$$ and $$T_1 \equiv D_{\Theta }T_2$$. For example, if $$T_2 + d_{\Theta } \vdash d_{\Phi }$$, from Lemma 10(8), by relabelling, we have $$T_2 \vdash \beta \leftrightarrow \tau _{\Theta }(\tau _{\Phi }(\beta ))$$, for $$\beta \in L_Q$$, and thus $$(\tau _{\Theta } \tau _{\Phi } =1)_{T_2}$$.