can anyone explain to me how to prove that p → q and not q → not p are logically equivalent without truth tables

Printable View

- March 17th 2010, 06:15 PMquestionboylogically equivalent
can anyone explain to me how to prove that p → q and not q → not p are logically equivalent without truth tables

- March 17th 2010, 08:19 PMSoroban
Hello, questionboy!

Quote:

Prove that and are logically equivalent without truth tables

. . I call it "Alternate Definition of Implication" (ADI).

Start with the right side:

. .

- March 18th 2010, 12:32 AMSeppel
- March 18th 2010, 05:55 AMemakarov
This is fine, but expressing as misses an important subtlety. One can

*define*as where denotes a contradiction (e.g., 0 = 1). Indeed, and imply by Modus Ponens, and deriving from proves by Deduction theorem (or implication introduction).

Then becomes . Now, it is surprising that this formula is implied by*without*using the fact that is a contradiction. I.e.,

is the transitivity of , and it is derivable for any using only basic rules about implication.

This suggests that implies in a much stronger sense than just in Boolean logic.