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 Cutlike features of his Hypothetical Syllogism rule.
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Notes
See [1] for a modern introduction to combinatory logic.
It should be noted that modern informal interpretations of this system, and the system to be developed later in the paper, are not quite Frege’s, as Frege had a very different conception of the logical vocabulary from that which is common now. We focus only on cases where complex formulae involved are those whose components are propositions, setting aside the additional formalism Frege put in place to ensure that any terms could be connected by the logical vocabulary.
See, for instance, the treatment in Roy Cook’s appendix to [5]
To show fully that C_{2} is Frege’s rule, one would need to go through the proofs given throughout Grundgesetze to ensure that no instances of Permutation are licensed only by C _{3}. This is a daunting task as there might be inferences which resemble those allowed by C_{3} which are made by appeal to other rules. Frege’s notation for showing applications of rules allows for such applications to be chained together in a way which makes this far from an easy task, and for now we shall suppose that C _{2} is the rule.
See the Appendix for details.
Note that from K, S, and the usual modus ponens rule all of the implication principles considered so far beside Peirce’s Law are provable.
We ignore trivial differences in notation.
Section 5.2 in Bimbó [2] is an excellent source on these calculi.
Note that the other instance of B occurring in the upper right formula cannot be the cut formula. This is clear from the statement of Frege’s rule, as the first instance \(B\rightarrow C\) in \((B\rightarrow C)\rightarrow (B\rightarrow C)\) is itself an antecedent.
The implication formula associated with this rule and combinator is \(A\rightarrow ((A\rightarrow B)\rightarrow B)\) which is clearly a kind of permutation principle.
Clearly the intended reading of his conditional is not as inferences or inference tickets, but as the name of a truth value, namely, The True in the case of theorems of the system.
References
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SchroederHeister, P. (2014). Frege’s sequent calculus. In Indrzejczak, A, Kaczmarek, J, & Zawidzki, M (Eds.) Trends in logic XIII: Gentzen’s and Jaśkowski’s heritage – 80 yeas of natural deduction and sequent calculi (pp. 233–245): Łódź University Press.
Acknowledgments
I’d like to thank Katalin Bimbó, Eric Berg, Roy T. Cook, Allen Hazen, Dongwoo Kim, Justin Kuster, David Ripley, Marcus Rossberg, Peter SchroederHeister, Morgan Thomas and Kai Wehmeier for valuable discussion about the topic. In particular, Berg, Cook, Kim, and Kuster provided helpful and detailed commentary on a draft of this paper. In addition, comments from the referee are much appreciated. Thanks are also due to the University of Connecticut Philosophy Department and College of Liberal Arts and Sciences for providing funding during my work on this project.
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The principles we’ve labelled K and S are displayed in §1, and since our focus is on the implication fragment of the Grundgesetze logic, we shall not discuss the negation axioms beyond these introductory remarks.
Appendix: : Structural Incompleteness of the System with only 1rules
Appendix: : Structural Incompleteness of the System with only 1rules
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 1rules. \(\mathfrak {M}=\langle W, O, R,v\rangle \) is constructed in ternary relation semantics. It is defined as follows:

W is a nonempty 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 nongeneralized 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 1rules.

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 1rules proves K, it does not prove B, and hence it cannot prove S.
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Tedder, A. On Structural Features of the Implication Fragment of Frege’s Grundgesetze . J Philos Logic 46, 443–456 (2017). https://doi.org/10.1007/s109920169406x
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DOI: https://doi.org/10.1007/s109920169406x