Show that this argument is valid.

Question

¬p → C; ∴ p. Where C denotes a contradiction.

What does it mean by ¬p → C;?

Also another statement ¬p → F; ∴ p.

Is there any differences between the two statement since from my understanding a contradiction is always false?

Answer 1

In classical logic, in order to prove that :

$¬p → C ∴ p$

where $C$ is a contradiction (i.e. - as you said - always false), we have to show that, for every valuation $v$ :

if $v(¬p → C) = v(¬p → F) =T$, then $v(p)=T$.

But the only way to have $v(¬p → F) =T$, according to the truth-table for $\rightarrow$, is when $v(\lnot p) = F$, which implies : $v(p) = T$.

Thus, we conclude that $p$ is logical consequence of $¬p → F$.

Answer 2

When we find that the assumption of $\lnot p$ implies (leads to) a contradiction $C$, then we can conclude, given the contradiction, that necessarily $\lnot p \rightarrow F$.

Recall that the implication, $\lnot P \rightarrow F$, also denoted $\lnot p \rightarrow \bot$, as a whole, can only be true if $\lnot p$ is false. Hence we can conclude $\lnot \lnot p \equiv p$.

As an illustration: suppose, as part of a proof, we start with the assumption $\lnot p$, and from this assumption and other premises, we fine that we can infer $C = q\land \lnot q$. Since $q \land \lnot q$ is a contradiction, and hence necessarily false, we cite the statement $q\land \lnot q$ and infer $F$ or rather $\bot$.

Then we close out the subproof by concluding that since $\lnot p \rightarrow \bot$, we are justified in concluding $\lnot \lnot p$, by negation introduction, and hence $p$, by double negation.