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Sometimes we get questions like this that essentially ask:

Okay, I know there's at least three different ways of proving an implication, namely:

  • direct proof
  • proof by contraposition
  • proof by contradiction

But what, really, is the difference between them?

As I see it, there's at least two ways of answering this question. One way is to nominate your favorite proof calculus for classical logic and explain how it handles these different proof strategies differently. Another way might be to argue that:

  • direct proof proves $P \rightarrow Q$
  • proof by contraposition proves $\neg Q \rightarrow \neg P$
  • proof by contradiction proves $P \wedge \neg Q \rightarrow \bot$
  • there's another one whereby you prove $\neg P \vee Q$

and that all four of these formulae are intuitionistically inequivalent. Actually, is this even true? Unfortunately, I don't know a thing about intuitionistic logic. So I ask:

Question. Intuitionistically, are each of the following formulae inequivalent? $$P \rightarrow Q,\; \neg Q \rightarrow \neg P,\; P \wedge \neg Q \rightarrow \bot,\; \neg P \vee Q$$

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goblin GONE
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  • The word you're looking for is "intuitively" not "intuitionistically" – David Quinn May 20 '15 at 05:48
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    $\lnot Q \to \lnot P$ and $P \land \lnot Q \to \bot$ are equivalent. The rest are different. – Zhen Lin May 20 '15 at 07:30
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    @DavidQuinn I think the OP meant "intuitionistically". What makes you think he meant "intuitively"? – bof May 20 '15 at 07:50
  • @ZhenLin, so much for distinguishing between proof by contraposition and proof by contradiction in this way. How incredibly disappointing. – goblin GONE May 20 '15 at 08:38
  • There is one slightly subtle point: if $P$ is already negative, say $P = \lnot R$, and $Q$ is already negative, say $Q = \lnot S$, then someone working intuitionistically would try to prove the "stronger" contrapositive $S \to R$, rather than the weaker $\lnot \lnot S \to \lnot \lnot R$. The stronger $S \to R$ does imply $\lnot R \to \lnot S$ intuitionistically, that is, it implies $P \to Q$ as desired. The "two contrapositives" are of course equivalent in classical logic, but someone working in intuitionistic logic has to be a little more careful. – Carl Mummert May 21 '15 at 00:08
  • Similarly, if $Q$ is already negative, say $Q = \lnot S$, then it is intuitionistically fine to prove $\lnot Q$ by assuming $P \land S$ and deriving a contradiction. – Carl Mummert May 21 '15 at 00:14

2 Answers2

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In intuitionistic logic:

  1. $$p \rightarrow q \vdash \neg(p \land \neg q)$$
  2. $$ \neg(p \land \neg q) \nvdash p \rightarrow q$$
  3. $$ \neg p \lor q \vdash p \rightarrow q$$
  4. $$p \rightarrow q \nvdash \neg p \lor q$$
  5. $$p \rightarrow q \vdash \neg q \rightarrow \neg p$$
  6. $$ \neg q \rightarrow \neg p \nvdash p \rightarrow q$$
  7. $$p \rightarrow q \vdash (p \land \neg q) \rightarrow \bot$$

Proof-theoretically speaking, the reason why (2), (4) and (6) do not hold intuitionistically is because you need the double negation elimination to complete the proof. Since in intuitionistic logic this rule is dropped, there are no means to prove those classic entailments. In contrast, the proofs of the remaining (1), (3) and (5) go as usual.

There is a list of other interesting properties of this logic in Van Dalen Logic and Structure, p.156.

This paper and the Stanford Encyclopedia of Philosophy entry on it can also be interesting.

Bruno Bentzen
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There are three strengths in your list: from strongest to weakest,

  1. $\neg P\vee Q$
  2. $P\to Q$
  3. $\neg(P\wedge\neg Q)$, equivalent to $\neg Q\to\neg P$ (and to $P\to\neg\neg Q$)

Note that if (2) implied (1), taking $P=Q$ would yield $\neg Q\vee Q$, which is famously intuitionistically invalid. If (3) implied (2), we could take $P=\neg\neg Q$; then (3) $\neg\neg Q\to\neg\neg Q$ would yield (2) $\neg\neg Q\to Q$, which is again famously intuitionistically invalid. So we definitely expect that these implications do not hold. (Of course this is just some kind of informal reductio... which is not intuitionistically valid.)

For the equivalence of $\neg Q\to\neg P$ and $P\to\neg\neg Q$: the first is $\neg Q\to P\to\bot$; the second is $P\to \neg Q\to\bot$; so their equivalence is an instance of permutation. For the equivalence of $\neg(P\wedge\neg Q)$ and $P\to\neg\neg Q$: the first is $P\wedge\neg Q\to\bot$; the second is $P\to\neg Q\to\bot$; so their equivalence is an instance of the currying principle, that $A\wedge B\to C$ is equivalent to $A\to B\to C$.

As for your goal of distinguishing between contraposition and proof by contradiction: in most systems of sequent calculus, the following derivation is valid both top to bottom and bottom to top: \begin{align*} \Gamma &\vdash \neg Q\to\neg P \\ \Gamma,\neg Q &\vdash \neg P \\ \Gamma,\neg Q &\vdash P\to\bot \\ \Gamma,P,\neg Q &\vdash \bot \\ \Gamma,P\wedge\neg Q &\vdash \bot \\ \Gamma &\vdash P\wedge\neg Q\to\bot \\ \Gamma &\vdash \neg(P\wedge\neg Q) \end{align*}

A logic which doesn't validate this derivation would, I think, have to have one of these properties:

  1. No deduction theorem, or a weak one.
  2. No adjunction, which means a weird notion of $\wedge$.
  3. A definition of $\neg$ other than $\to\bot$.

Not that there aren't such logics, but their existence might not be very compelling for your intended audience, which I assume is a student learning classical logic for the first time.

  • Of course, the logic known as "intuitionistic logic" in proof theory does include the scheme $\bot \vdash \psi$, which is a different sort of "proof by contradiction". In fact, this is how one would derive $P \to Q$ from $(P \to \bot) \lor Q$, that is, derive (2) from (1). That derivation would not work in minimal logic. – Carl Mummert May 21 '15 at 00:11