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I am reading "Introduction to Mathematical Logic" by Elliott Mendelson, and I am currently at the axiomization of propositional calculus. Mendelson presents the following three axioms (with modus ponens as the only rule of inference):

$A1: B \implies (C \implies B)$

$A2: (B \implies (C \implies D)) \implies ((B \implies C) \implies (B \implies D))$

$A3: (\lnot C \implies \lnot B) \implies ((\lnot C \implies B) \implies C)$

Immediately after this, Mendelson proves $B \implies B$ and then leaves it as an exercise to prove $(\lnot C \implies \lnot B) \implies (B \implies C)$. Mauro provided a great and detailed proof of the latter in response to the following post: Prove $( \lnot C \implies \lnot B) \implies (B \implies C)$ without the Deduction Theorem

He does so by using several intermediate results, namely:

$T1: (C \implies D) \implies ((B \implies C) \implies (B \implies D))$

$T2: (B \implies (C \implies D)) \implies (C \implies (B \implies D))$

My question is centered around the proof of T2. Mauro proves T2 by proving and then using another intermediate result:

$T3: B \implies ((B \implies C) \implies C)$

However, I believe I found a proof that doesn't require T3. I would like to know if my proof is correct or if there is something I'm missing. Here is my proof (using [ ] occasionally for readability):

$1. \; [B \implies (C \implies D)] \implies [(B \implies C) \implies (B \implies D)]$

---A2

$2. \; [(B \implies C) \implies (B \implies D)] \implies [(C \implies (B \implies C)) \implies (C \implies (B \implies D))]$

---T1

$3. \; [B \implies (C \implies D)] \implies [(C \implies (B \implies C)) \implies (C \implies (B \implies D))]$

---From 1, 2, T1, and MP twice

$4. \; [(B \implies (C \implies D)) \implies (C \implies (B \implies C))] \implies [(B \implies (C \implies D)) \implies (C \implies (B \implies D))]$

---From 3, A2, and MP

$5. \; (C \implies (B \implies C))$

---A1

$6. \; [(B \implies (C \implies D)) \implies (C \implies (B \implies C))]$

---From 5, A1, and MP

$7. \; [(B \implies (C \implies D)) \implies (C \implies (B \implies D))]$

---From 4, 6, and MP

gestory2
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    It looks fine to me. If you’re going to use any theorems, you might as well as just prove a meta-theorem such as Hypothetical Syllogism. It’s typically much easier to use a rule than to use a formula. – PW_246 Jul 15 '24 at 05:13
  • You just say that $A \implies B$ is equivalent with $\neg A \lor B$, the rest is obvious. – Dominique Jul 15 '24 at 06:59
  • $\lambda fxy. fyx$ – Naïm Camille Favier Jul 15 '24 at 08:20
  • If your primary motivation for learning propositional logic is to apply it to writing mathematical proofs in, say, real analysis or group theory, there are far easier, more efficient ways to approach the subject, e.g. some form of natural deduction. If you like a challenge, by all means, play around with the older (?) systems that you find in your textbook. Then you might also enjoy Aristotle's approach to logic with his categorical syllogisms and squares of opposition. – Dan Christensen Jul 15 '24 at 16:32

1 Answers1

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Your proof is correct, but you don’t really need T1 either. It’s easiest, in my opinion, to just prove Hypothetical Syllogism as a meta-theorem, and then proceed.

If you’re unfamiliar with the proof of said meta-theorem, here it is:

  1. $A \to B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$[Premise]
  2. $B \to C~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$[Premise]
  3. $(B \to C) \to (A \to (B \to C))~~~~$ [A1]
  4. $(A \to (B \to C)) \to ((A \to B) \to (A \to C))~~~~~~~~~$ [A2]
  5. $A \to (B \to C)~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ [2,3 MP]
  6. $(A \to B) \to (A \to C)~~~~~~~~~~~~~~~~~$[5,4 MP]
  7. $A \to C~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$[1,6 MP]

Then, we can proceed more naturally. I’m going to leave the citations undone since they’re just axioms, modus ponens, or Hypothetical Syllogism.

  1. $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$

  2. $((A \to B) \to (A \to C)) \to (B \to ((A \to B) \to (A \to C)))$

  3. $(A \to (B \to C)) \to (B \to ((A \to B) \to (A \to C)))$

  4. $(B \to ((A \to B) \to (A \to C))) \to ((B \to (A \to B)) \to (B \to (A \to C)))$

  5. $(A \to (B \to C)) \to ((B \to (A \to B)) \to (B \to (A \to C)))$

  6. $5 \to ((A \to (B \to C)) \to (B \to (A \to B))) \to ((A \to (B \to C)) \to (B \to (A \to C)))$

  7. $((A \to (B \to C)) \to (B \to (A \to B))) \to ((A \to (B \to C)) \to (B \to (A \to C)))$

  8. $B \to (A \to B)$

  9. $8 \to ((A \to (B \to C)) \to (B \to (A \to B)))$

  10. $(A \to (B \to C)) \to (B \to (A \to B))$

  11. $(A \to (B \to C)) \to (B \to (A \to C))$

PW_246
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    As I recall Mendelson's book the meta result that he proves (from A1, A2, and MP) is the Deduction Theorem. And proving the OP's formula using DT is very very easy. – ancient mathematician Jul 15 '24 at 06:44
  • Is Hypothetical Syllogism similar to the Deduction Theorem? The Deduction Theorem is the very next thing Mendelson proves after the contrapositive theorem, and he then proceeds to reprove the contrapositive theorem using the Deduction Theorem. It looks to me like he's doing it both ways to show how much easier it is with the Deduction Theorem? – gestory2 Jul 15 '24 at 13:02
  • @gestory2 it’s similar in that it’s a meta-theorem, but the proof requires induction, so it’s not quite as easy. Still, using the DT makes things so much easier that it’s worth it to prove it. I personally just think that proving 3 theorems before proving the one in question isn’t necessary or helpful for understanding how the actual axiomatic proof works. – PW_246 Jul 15 '24 at 14:56
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    @PW_246 Mendelson's approach was actually helpful to me because it really emphasized the importance of the Deduction Theorem. I definitely prefer the simpler proofs it facilitates. That said, I appreciate your detailed responses and am marking this as the answer. – gestory2 Jul 15 '24 at 15:27