Recall that $ p \rightarrow \sim q$ is equivalent to $p \land \sim q$, how can this be used as an explanation for how to use proof by contradiction. Recall that $p \rightarrow \sim q$ is equivalent to $p \land \sim q$, how can this equivalence be used as an explanation for how to use proof by contradiction.
I'm having a hard time answering this question because I do not understand the role of $p \land \sim q$. So i know $p \land \sim q$ means $p$ and not $q$ but how does this play a role in proof by contradiction.
I get $p \rightarrow \sim q$ because that is basically the procedure of how a person uses proof by contradiction.
Any insights?
 A: Just a small comment, you wrote that $p\rightarrow \neg q$ is equivalent to $p \land \neg q$. Didn't you want to say that $p\rightarrow \neg q$ is equivalent to $\neg p \lor \neg q$?
Now suppose you are trying to prove that $p\rightarrow q$ by contradiction. Then you assume that $\neg (p\rightarrow q)$, which is $p\land \neg q$. Now once you have proven that $p\land \neg q$ is false, you have $\neg \neg(p\rightarrow q)$, which is $p\rightarrow q$.
A: The "general form" of a Proof by contradiction is:

we have to prove a proposition $P$ from a set $S$ of axioms or earlier theorems of the theory we are working in. We consider $\lnot P$ in addition to the set $S$ of premises; if this leads to a contradiction $\mathbb F$, then we can conclude that the set of premises $S$ implies the negation of $\lnot P$, i.e. $P$.

In symbols:


if $S \cup \{ \lnot P \} \vdash \mathbb F$, then $S \vdash P$.



Consider the following example of a proof by contradiction of a conditional statement:

Suppose $a \in \mathbb Z$. We have to prove: if $a^2$ is even, then $a$ is even [here the claim we have to prove has the form: $P \to Q$].

Proof. For the sake of contradiction suppose $a^2$ is even and $a$ is not even. [i.e. we are assuming $P \land \lnot Q$, which is the negation of $P \to Q$].
Then $a$ is odd and thus there is an integer $c$ such that $a = 2c +1$.
Then $a^2 = (2c +1)^2 = 4c^2 +4c +1 = 2(2c^2 +2c)+1$, which means that $a^2$ is odd.
Thus $a^2$ is even and $a^2$ is not even, a contradiction [i.e. we have derived $P \land \lnot P$].

Thus, having arrived at a contradiction, our original supposition that $a^2$ is even and $a$ is odd could not be true, and we have proved the theorem.

