Show that $p \Rightarrow (\neg(q \land \neg p))$ is a tautology I just need my solution checked since I'm not sure if it's valid, especially the final statement
Question:
Show $p \Rightarrow (\neg(q \land \neg p))$ is a tautology by assuming:
$u \Rightarrow v$ is logically equivalent to $\neg u \lor v$
My solution:
$\neg(q \land \neg p))$ is logically equivalent to $\neg q \lor p$ (by De Morgan's law),
so $\neg q \lor p$ is logically equivalent to $q \Rightarrow p$ (given by the question)
hence $p \Rightarrow (q \Rightarrow p)$ and therefore $p$ is a tautology?
 A: For your final statement, you want to see if $p\implies(q\implies p)$ is a tautology, not $p$, since $p$ is just an atomic sentence and could take any truth value. If $p\equiv\bot$, then the hypothesis is false, so the whole statement is true. If $p\equiv\top$, then $q\implies p$ is true, since any implication with a true conclusion is true. So $p\implies(q\implies p)\equiv\top$, that is, it is a tautology.
A: Use De Morgan twice. You want to evaluate $A\equiv [p\Rightarrow\lnot(q\land\lnot p)]$. 
(1) For every $r$, $[p\Rightarrow r]\equiv[\lnot p \lor r]$, hence $A\equiv[\lnot p \lor \lnot(q \land \lnot p)]$. 
(2) For every $r$, $[\lnot(q\land r)]\equiv[\lnot q \lor \lnot r]$, hence $A\equiv[\lnot p \lor \lnot q \lor p]$. 
(3) Since $[p \lor \lnot p]\equiv[$true] and [true $\lor r]\equiv[$true] for every $r$, $A\equiv[$true].
Edit A less systematic but shorter path is to note that for every $r$, $[p\Rightarrow\lnot r]\equiv[r\Rightarrow\lnot p]$ and that, for $r\equiv [q\land\lnot p]$, $[r\Rightarrow\lnot p]$ is obviously true.
