On Structural Features of the Implication Fragment of Frege’s Grundgesetze

Abstract

We set out the implication fragment of Frege’s Grundgesetze, clarifying the implication rules and showing that this system extends Absolute Implication, or the implication fragment of Intuitionist logic. We set out a sequent calculus which naturally captures Frege’s implication proofs, and draw particular attention to the Cut-like features of his Hypothetical Syllogism rule.

Appendix: : Structural Incompleteness of the System with only 1-rules

Theorem 2

S is independent of the subsystem of $G_{\rightarrow }$ containing only the rules C ₁ , W ₁ , MP ₁ , and HS ₁

Proof

We now present a model, $\mathfrak {M},$ showing that S is independent of $G_{\rightarrow }$, using only the 1-rules. $\mathfrak {M}=\langle W, O, R,v\rangle$ is constructed in ternary relation semantics. It is defined as follows:

• W is a non-empty set (of sets of sentences or worlds).

• $O\subseteq W$ is the set of normal worlds.

• $R\subseteq W^{3}$ is such that

  • If R α β γ and R γ δ 𝜖 then R α β 𝜖

• v is a valuation function $v:\langle A,\alpha \rangle \rightarrow \{1,0\}$ where A is a formula and α∈W. As shorthand for v(A,α)=1 we write $\alpha \vDash A$ and $\alpha \nvDash A$ for v(A,α)=0. Each α∈W is consistent, so for no A is it the case that $\alpha \vDash A$ and $\alpha \nvDash A$. We also demand that $\alpha \vDash A$ or $\alpha \nvDash A$ for all α∈W and A. Beyond this we have a standard valuation for conditional statements:

  • $\alpha \vDash A\rightarrow B$ iff for all β,γ∈W if R α β γ and $\beta \vDash A$ then $\gamma \vDash B$.

A formula A is valid on $\mathfrak {M}$ iff for all α∈O, $\alpha \vDash A$.

We add the following demands to ensure that the rules of $G_{\rightarrow }$ all come out as admissible. Note that these are stated in a non-generalized form, unlike the specific statement of Frege’s rules. The generalized rules are lengthier and more involved, but for the countermodel developed here these rules allow for the similar inferences as would the 1-rules.

• For all α∈O, if $\alpha \vDash A\rightarrow (A\rightarrow B)$ then $\alpha \vDash A\rightarrow B$.

• ∀α∈O if $\alpha \vDash A\rightarrow (B\rightarrow C)$ then $\alpha \vDash B\rightarrow (A\rightarrow C)$.

• ∀α∈O if $\alpha \vDash A\rightarrow B$ and $\alpha \vDash B\rightarrow C$ then $\alpha \vDash A\rightarrow C$.

• ∀α∈O if $\alpha \vDash A\rightarrow B$ and $\alpha \vDash A$ then $\alpha \vDash B$.

We also note the following, as conditions on worlds:

• For all α∈O, if R α β γ then $\beta \vDash A$ implies $\gamma \vDash A$. As shorthand for “$\beta \vDash A$ implies $\gamma \vDash A$” we use β≤γ.

Lemma 1

Any $\mathfrak {M}$ with the frame conditions we have imposed is a model of $G_{\rightarrow }$.

Proof

Suppose that some x∈O were such that $x\nvDash A\rightarrow (B\rightarrow A)$. Thus, there must exist some y,z∈W such that $Rxyz\land y\vDash A\land z\nvDash B\rightarrow A$. Since $z\nvDash B\rightarrow A$ there must exist some u,v such that R z u v and $u\vDash B\land v\nvDash A$. However, given that R x y z and R z u v, we know that R x y v and thus y≤v, and therefore $v\vDash A$, which is impossible. Thus, for all x∈W, $x\vDash A\rightarrow (B\rightarrow A)$.

Similarly, suppose that some x∈O were such that $x\nvDash A\rightarrow A$. Then there must be y,z∈W s.t. R x y z, $y\vDash A$, and $z\nvDash A$. However, since R x y z and x∈O, thus y≤z and hence $z\vDash A$, which is impossible.

It is clear from the construction that $\mathfrak {M}$ validates all the rules of $G_{\rightarrow }$. These rules are simply enforced at every normal world, which is where all validities come out true, so no such world could provide a counterexample. □

Finally, we show that $\mathfrak {M}\nvDash (A\rightarrow (B\rightarrow C))\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$. We shall start from the components and tailor the worlds so that α is a world invalidating this formula. Given these frame conditions restrictions, $\mathfrak {M}$ is constructed as follows as follows:

• W={α,β,γ,δ,𝜖,ζ,η}.

• O={α}

• $R=\{\langle \alpha ,\beta ,\gamma \rangle ,\langle \gamma ,\delta ,\epsilon \rangle ,\langle \epsilon ,\zeta ,\eta \rangle ,\langle \alpha ,\beta ,\epsilon \rangle ,\langle \gamma ,\delta ,\eta \rangle \}\cup \\\{\langle \alpha , x,x\rangle ; x\in W\}\,\cup \,\{\langle \alpha ,\alpha , x\rangle ;x\in W\}$

Let $\zeta \vDash A$ and $\eta \nvDash C$. Given that R 𝜖 ζ η, it follows that $\epsilon \nvDash A\rightarrow C$. Let $\delta \vDash A\rightarrow B$. Given that R γ δ 𝜖, it follows that $\gamma \nvDash (A\rightarrow B)\rightarrow (A\rightarrow C)$. Finally, let $\beta \vDash A\rightarrow (B\rightarrow C)$. Since R α β γ, we know that $\alpha \nvDash (A\rightarrow (B\rightarrow C))\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$. The restrictions on R enforce that R α β 𝜖 and R γ δ η, and so β≤𝜖 and δ≤η. From β≤𝜖 we know that $\epsilon \vDash A\rightarrow (B\rightarrow C)$ and that $\eta \vDash A\rightarrow B$. Since $\epsilon \vDash A\rightarrow (B\rightarrow C)$, R 𝜖 ζ η, and $\zeta \vDash A$, it follows that $\eta \vDash B\rightarrow C$. So, η, s.t. η∉O has $\eta \vDash A$, $\eta \vDash B\rightarrow C$, and $\eta \nvDash C$. □

To expand on this result, consider this same frame with W={α,β,γ,δ,𝜖,ζ,η}, O={α,γ}, and R the same as above. Such a model will invalidate B.

Suppose that $\alpha \nvDash (A\rightarrow B)\rightarrow ((C\rightarrow A)\rightarrow (C\rightarrow B))$. Then there exist β,γ s.t. R α β γ and $\beta \vDash A\rightarrow B$ and $\gamma \nvDash (C\rightarrow A)\rightarrow (C\rightarrow B)$. So there are δ,𝜖 s.t. R γ δ 𝜖, $\delta \vDash C\rightarrow A$, and $\epsilon \nvDash C\rightarrow B$. Again, there are ζ,η s.t. R 𝜖 ζ η, $\zeta \vDash C$, and $\eta \nvDash A$. Since α∈O and R α β γ, β≤γ and $\gamma \vDash A\rightarrow B$. Since R α β γ and R γ δ 𝜖, we have that R α β 𝜖 and thus $\epsilon \vDash A\rightarrow B$. Since R γ δ 𝜖 and γ∈O, we have $\epsilon \vDash C\rightarrow A$. From $\epsilon \vDash C\rightarrow A$, $\zeta \vDash C$ and R 𝜖 ζ η, we know that $\eta \vDash A$. In addition, since R γ δ η, we know that δ≤η, and so $\eta \vDash C\rightarrow A$. So, we end up having that $\eta \vDash C$, $\eta \vDash C\rightarrow A$, $\eta \vDash A$, and $\eta \nvDash B$. This gives a sufficient outline to construct the countermodel for B. So, even given that $G_{\rightarrow }$ with the 1-rules proves K, it does not prove B, and hence it cannot prove S.