Is the first solution valid? If not, can someone explain to me why the first solution is not valid but the second one is? Both claim that p and not p is true though?
$\lnot p \to p \vdash p$
$1 \hspace{1cm} \lnot p \to p \hspace{1cm} premise$
$\hspace{1cm} 2 \hspace{1cm} \lnot p \hspace{1cm} Assumption$
$\hspace{1cm} 3 \hspace{1cm} p \hspace{1cm} \to e \phantom{B} 1,2$
$4 \hspace{1cm} p \hspace{1cm} \to e \phantom{B} 2-3 $
Or
$1 \hspace{1cm} \lnot p \to p \hspace{1cm} premise$
$\hspace{1cm} 2 \hspace{1cm} \lnot p \hspace{1cm} Assumption$
$\hspace{1cm} 3 \hspace{1cm} p \hspace{1cm} \to e\phantom{BC} 1,2$
$\hspace{1cm} 4 \hspace{1cm} \perp \hspace{1cm} \lnot e \phantom{B} 2,3$
$5 \hspace{1cm} \lnot \lnot \hspace{1cm} \lnot i 2-4$
$6 \hspace{1cm} p \hspace{1cm} \lnot \lnot e \phantom{B}5 \phantom{B} (PBC \phantom{B} 2-4) $