Prove a tautology using truth table How do I prove $(\lnot p \rightarrow F)\rightarrow (p=T)\;$ using a truth table?
(This tautology symbolizes a "proof by contradiction". If p being false leads to a contradiction, then p is true.)
 A: Recall: the only time an implication is FALSE is if the antecedent is true and the consequence is false.
$p\quad \mid \lnot p \quad \mid \lnot p \rightarrow F \mid p = T \mid (\lnot p \rightarrow F)\rightarrow (p = T)$
$T\quad \mid \;F\;\quad \mid \;\;T\quad \;\;\mid \quad T\quad\mid\quad \quad\quad\quad T$
$F \quad \mid \;T\;\quad\mid \;\;F \quad \;\;\mid\quad F\quad \mid\quad\quad\quad\quad T$
Hence, whatever the truth-value of $p$, the statement as a whole is true. Therefore:
$$(\lnot p \rightarrow F)\rightarrow (p = T)$$
is a tautology.
A: 
$(\lnot p \rightarrow F)\rightarrow (p=T)$ ... This tautology symbolizes a "proof by contradiction"

Really??? I don't know any propositional language in which this is even well-formed. Which textbook sets things up so this expression is legitimate? For a start normal propositional languages don't have an identity relation in them.
I suspect you are confusing this with either

$(\neg p \to \bot) \to p$

or 

$(\neg p \to \bot) \to (p \leftrightarrow \top)$

both of which are indeed tautologies. The first of these is the one that more immediately reflects a proof by contradiction. 
