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If we can prove that $¬Y \implies ¬X,$ then is it always possible to prove that $X \implies Y$ without first proving that $¬Y \implies ¬X$ ?

Motivation: when studying analysis, there are some problems/theorems where we must prove $X\implies Y$ and then I try to prove it directly, spend hours stuck without being able to prove it and finally when I use the contrapositive, the theorem is easily proved but it seemed that If I kept trying to prove $X\implies Y$ instead of $¬Y \implies ¬X$, I would never be able to do so.

ryang
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Red Banana
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  • Related https://math.stackexchange.com/questions/838184/can-one-prove-by-contraposition-in-intuitionistic-logic – Damian Jul 10 '24 at 08:25
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    Not very clear... In Classical Logic $X \to Y$ and $\lnot Y \to \lnot X$ are equivalent. Thus, having proved one of them, we immediately know that also the other one holds. – Mauro ALLEGRANZA Jul 10 '24 at 09:42
  • It has its reasons why this technique exists. Sometimes , the proof becomes much much easier with it. Not sure whether we always would be succeful without it at all. – Peter Jul 10 '24 at 10:39
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    Not marking as a duplicate, but probably of interest: https://math.stackexchange.com/questions/638756/when-to-use-the-contrapositive-to-prove-a-statment – Andrew D. Hwang Jul 10 '24 at 12:43
  • @MauroALLEGRANZA I know. The trouble is: In actual mathematical practice, finding the latter without the former can be very hard. – Red Banana Jul 10 '24 at 16:06

1 Answers1

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A direct proof of $P$ is a (finite) sequence of implications

$P_1\implies P_2\implies ...\implies P,$

where $P_i$ is a true sentence for every $i=1,...,n$.

Here we suppose that there is a proof for $\neg Y\implies \neg X$, so there is a finite set of true sentences $\{P_i\}_{i=1,...,n}$ such that

$\neg Y\implies P_1\implies P_2\implies...\implies P_n\implies \neg X.$

Starting from right to left we have $X\implies \neg P_n$, $\neg P_i\implies \neg P_j,$ for $i>j$ and $\neg P_1\implies Y$, i.e.

$X\implies \neg P_n\implies ...\implies\neg P_1\implies Y,$

which is a direct proof for $X\implies Y.$

This is an answer from a typical point of view, which gives a direct proof apparently depending on the proof of contrapositive: It is not necessary to find first the proof of contrapositive to find then the direct one; you can find first the direct proof. However from an empirical point of view it is frequently hard to give a direct proof, while the proof of contrapositive is more approachable. (That' s why we do prefer the second one.)

Maximus
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    This "definition" of a direct proof is circular: surely you must also require that $P_i \implies P_{i+1}$ has a direct proof. – Naïm Camille Favier Jul 16 '24 at 12:06
  • @NaïmFavier Why circular and not inductive? – Maximus Jul 16 '24 at 18:15
  • Induction on what? What's your base case? You could try defining "direct proofs of length n" by induction on n, but then what's a direct proof of length 1 of $P \implies Q$? – Naïm Camille Favier Jul 16 '24 at 18:51
  • The only mathematically objective definition of a "direct proof" is "a proof that doesn't use double negation elimination/proof by contradiction". In particular, every constructive proof is direct. (Which is not to say that one can't use "direct proof" in the hand-wavey sense of "doesn't introduce additional assumptions" or something like that.) – Naïm Camille Favier Jul 16 '24 at 18:55
  • See this question: https://math.stackexchange.com/questions/3977000/what-is-a-direct-proof-formally – Naïm Camille Favier Jul 16 '24 at 19:00