Mr Davis 97
Say I have the theorem ##p \rightarrow q##. What is the difference between proving that ##\neg q \rightarrow \neg p## is true and showing that ##\neg (p \rightarrow q) = \neg p \wedge q## leads to a contradiction?

Mentor
##\neg (p \rightarrow q) = \neg p \wedge q##
That should be ##\neg (p \rightarrow q) = \neg q \wedge p##. If you are careful with statements like "for all" and "there exists", then they are all the same thing.

Homework Helper
Gold Member
That last formula should be ##p\wedge \neg q##.

Subject to that, the two approaches are logically equivalent in classical first-order predicate logic, which is all that mathematicians that don't specialise in logic worry about.

In intuitionist logic and other logics where some of the basic axioms such as ##\neg\neg p\leftrightarrow p## are not accepted, the approaches may give different results.

Edgardo
You are asking about the difference between "Proof by contraposition" and "Proof by contradiction", and here is an example.

To prove $p \rightarrow q$:

- In proof by contraposition you start by assuming that $\neg q$ is true and derive the statement $\neg p$. Here, the path is clear, i.e. you start at $\neg q$ and arrive at $\neg p$.

- In proof by contradiction your start by assuming that the opposite of $p \rightarrow q$ is true. So you assume that $p \wedge \neg q$ is true and derive some contradiction. Here the path is not clear, nobody is going to tell you what the contradiction is and what it looks like.