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In the book Gödel's Theorems and Zermelo's Axioms by Halbeisen and Krapf the first chapter contains an excercise that asks us to prove $$\phi\rightarrow \psi \vdash \neg \psi \rightarrow\neg \phi,$$

i.e. contraposition. I can't figure out how to prove it. The axioms that are given are the following (This excercise comes before the deduction theorem):

\begin{align*} L_0:& \phi \vee\neg\phi \\ L_1:& \phi\rightarrow(\psi\rightarrow\phi) \\ L_2:& (\psi\rightarrow(\phi_1\rightarrow\phi_2))\rightarrow((\psi\rightarrow\phi_1)\rightarrow(\psi\rightarrow\phi_2)) \\ L_9:& \neg\phi \rightarrow (\phi \rightarrow \psi). \end{align*}

($L_3$ up to $L_8$ are axioms using $\wedge$ or $\vee$, so they're not relevant for the question I think, even $L_0$ should probably not be used here.)

The back of the book contains the following solution: \begin{align*} \phi_0:& \phi \rightarrow \psi && \text{given} \\ \phi_1:& \phi_0 \rightarrow (\neg \psi \rightarrow \phi_0) && L_1 \\ \phi_2:& \neg\psi\rightarrow\phi_0 && MP \text{ from } \phi_0 \text{ and } \phi_1 \\ \phi_3:& (\phi \rightarrow \psi) \rightarrow ((\phi \rightarrow \neg \psi) \rightarrow \neg\phi) && L_9 \\ \vdots \end{align*} Given $\phi_3$, the rest of the deduction is easy to follow (though I don't see how I could easily come up with it myself). However, I don't see how $\phi_3$ is deduced. Applying $L_9$ but with $\phi \equiv \neg(\phi \rightarrow \psi)$ and $\psi\equiv \neg \phi$ (which is what they appear to be doing) gives $$ (\phi \rightarrow \psi) \rightarrow (\neg(\phi\rightarrow\psi)\rightarrow\neg\phi). $$ Essentially, I have the following two questions:

  1. Is the proof in Halbeisen and Krapf incorrect (i.e. is there a typo), or am I missing something?
  2. How do you actually formally prove contraposition in the given system?
H. de Gracht
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    The antecedent, or left part of the implication of 'L9' is a negation. It has a '¬' symbol as it's primary connective. The left part of the implication of 'ϕ3' is an implication. So, I don't think 'L9' can solely get used to produce 'L9'. Thus, I can't see how their reasoning could be correct. – Doug Spoonwood Sep 04 '25 at 15:30
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    $\phi_3$ above is a tautology, it is a form of Negation Introduction, and it can be useful if we want to mimick Natural Deduction derivations. But we need the Deduction Theorem and we have to derive it from axioms above. – Mauro ALLEGRANZA Sep 04 '25 at 15:45
  • @DougSpoonwood note that the axiom set ${(\neg A \to (A \to B)), ((A \to \neg A) \to \neg A)}$ is sufficient for Intuitionistic Logic over Positive Logic, which is captured by the positive axioms in OP’s set. We can clearly get $((A \to \neg A) \to \neg A)$ from LEM and proper axioms for $\lor$ and $\to$, which we have. The actual instance mentioned in the solution admittedly doesn’t make any sense. – PW_246 Sep 04 '25 at 15:51
  • Is $\neg$ a primitive operation, or is $\neg\phi$ shorthand for $\phi\rightarrow\bot$? – David Knipe Sep 08 '25 at 22:00
  • @DavidKnipe $\neg$ is a primitive operation here, as are $\rightarrow, \wedge$ and $\vee$. – H. de Gracht Sep 10 '25 at 11:53

2 Answers2

1

Yes, there is what is traditionally called saltus in the given proof. Firstly, let us present the mentioned proof in full:

\begin{align} \varphi_0 &: \; \varphi \rightarrow \psi && \phi\rightarrow\psi\in\Phi \\ \varphi_1 &: \; \varphi_0 \rightarrow (\neg \psi \rightarrow \varphi_0) && \text{instance of } L_1 \\ \varphi_2 &: \; \neg \psi \rightarrow \varphi_0 && \text{from } \varphi_1 \text{ and } \varphi_0 \text{ by (MP)} \\ \varphi_3 &: \; (\varphi \rightarrow \psi) \rightarrow (\underbrace{(\varphi \rightarrow \neg \psi) \rightarrow \neg \varphi}_{\equiv:\vartheta}) && \text{instance of } L_9 \\ \varphi_4 &: \; \varphi_3 \rightarrow (\neg \psi \rightarrow \varphi_3) && \text{instance of } L_1 \\ \varphi_5 &: \; \neg \psi \rightarrow \varphi_3 && \text{from } \varphi_4 \text{ and } \varphi_3 \text{ by (MP)} \\ \varphi_6 &: \; (\neg\psi\rightarrow(\underbrace{\varphi_0 \rightarrow \vartheta}_{\equiv:\phi_{3}})) \rightarrow ((\neg\psi \rightarrow\varphi_0)\rightarrow (\neg\psi\rightarrow \vartheta)) && \text{instance of } L_2 \\ \varphi_7 &: \; (\neg \psi \rightarrow \varphi_0) \rightarrow (\neg \psi \rightarrow \vartheta) && \text{from } \varphi_6 \text{ and } \varphi_5 \text{ by (MP)} \\ \varphi_8 &: \; \neg \psi \rightarrow \vartheta && \text{from } \varphi_7 \text{ and } \varphi_2 \text{ by (MP)} \\ \varphi_9 &:\;(\neg\psi\rightarrow(\underbrace{(\varphi\rightarrow \neg\psi)\rightarrow\neg\phi)}_{\equiv:\vartheta})\rightarrow ((\neg\psi\rightarrow(\varphi\rightarrow\neg\psi))\rightarrow(\neg\psi \rightarrow \neg\varphi)) && \text{instance of } L_2 \\ \varphi_{10} &:\;(\neg\psi\rightarrow(\varphi\rightarrow\neg\psi))\rightarrow (\neg \psi \rightarrow \neg \varphi) && \text{from } \varphi_9 \text{ and } \varphi_8 \text{ by (MP)} \\ \varphi_{11} &: \; \neg \psi \rightarrow (\varphi \rightarrow \neg \psi) && \text{instance of } L_1 \\ \varphi_{12} &: \; \neg \psi \rightarrow \neg \varphi && \text{from } \varphi_{10} \text{ and } \varphi_11 \text{ by (MP)} \end{align}

Actually, the writers use an alternative axiom

$$(\varphi\rightarrow\psi)\rightarrow((\varphi\rightarrow\neg\psi)\rightarrow\neg\varphi)$$

It is one of the several axioms employed basically to handle negation. Elliott Mendelson defines the system in his Introduction to Mathematical Logic with a variation of it:

$$(\neg\varphi\rightarrow\neg\psi)\rightarrow((\neg\varphi\rightarrow\psi)\rightarrow\varphi)$$

Probably, the most widely used one is

$$(\neg\varphi\rightarrow\neg\psi)\rightarrow(\psi\rightarrow\varphi)$$

These are all interderivable; that is, if one is included in the Hilbert-style set of axioms, the other ones can be derived as theorems. However, the derivations are not trivial. Hence, in the given proof, it seems to occur like by a sleight of hand. See, for example, Meldelson's Lemma 1.11c (in the 6th edition) for the derivation of

$$\neg\varphi\rightarrow(\varphi\rightarrow\psi)$$

and in Halbeisen and Krapf's book cited, the arguments in the solution of the exercise 2.9. related to $\phi_{3}$ what is annotated as "instance of "$L_{9}$ here.

Tankut Beygu
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  • "saltus" means error? – Mauro ALLEGRANZA Sep 04 '25 at 19:01
  • It depends on your appraisal. In the present situation, the formula/line $\varphi_{3}$ stands quite unjustified. – Tankut Beygu Sep 04 '25 at 22:36
  • People may talk about those as interderivable, because of the assumption of other axioms. Maybe inter-derivability works for this text, because of other axioms. However, they are NOT interderivable if they are sole axioms used to derive the other. If CCNxNyCyx is a sole axiom, modus ponens can't even get used to derive anything! CNxNy never unifies with CCNxNyCyx. There is no way to get CCNxNyCyx from the assumption of CCxyCCxNyNx. Even if we assume sufficient relevant axioms to prove the deduction meta-theorem along with CCxyCCxNyNx, we still couldn't prove CCNxNyCyx. – Doug Spoonwood Sep 18 '25 at 22:54
1

L$_9$ is an implication. It's antecedent has the type negation. This implies that L$_9$ could only produce substitution instances of itself. Modus ponens is impossible to use if we considered a sub-logic which had L$_9$ as it's sole axiom. The antecedent of ϕ$_3$ is an implication. But, again, the antecedent of L$_9$ is an negation. Substitution of a negation type of well-formed formula only can produce a negation. So, ϕ$_3$ is not a substitution instance of L$_9$.

For the first part of the above proof, consider a general conditional Cxy (this is in Polish notation... logical types of well-formed formulas are easier to recognize in Polish notation), of the 'C' type. Suppose it our only axiom. Then to derive something using modus ponense, we need two formulas, one of the form 'a', and the other of the form 'Cab'. But, 'a' also has to have to get derived from 'Cxy'. Thus, 'a' has to have the same primary, equivalently 'first', connective as 'Cxy'. This can only happen if 'a' has the 'C' type. But, with a formula of the type 'CNxy', any well-formuled formula for 'Nx' has the N type. Therefore, the rule of detachment (modus ponens) can never get used for any well-formed formula of the type 'CNxy' (where our sole axiom has is of the CNxy type).

So with respect to 1., yes, their argument is incorrect.

Doug Spoonwood
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