Difficulty understanding proof by contradiction Here's my understanding of proof by contradiction based on what I've read and I've been taught.
We show $\neg P \implies (c \land \neg c)$ is always true. This is done by assuming $\neg P$ is true. Then, we realize that $(c \land \neg c)$ is logically equivalent to $F$ so we've just shown that $\neg P \implies F$ is always true. But, examining the $\implies$ truth table, we realize that $\neg P$ must be false to ensure $\neg P \implies F$ is always true. Therefore, we conclude that $\neg P$ is false, and thus $P$ is true. 
The conundrum I have is that our analysis is based on the assumption that $\neg P$ is true. Then later we conclude that $\neg P$ is false. So why do we accept that $\neg P$ is false even though we assume $\neg P$ is true?

Suppose you've proved $\lnot P\to F$.
\begin{align*}
\text{Then:}\;\;\;
&1.\qquad \lnot P\to F\\[4pt]
&2.\qquad \lnot P\to (P\land\lnot P)\\[4pt]
&3.\qquad \lnot P\to P\\[4pt]
&4.\qquad P\to P\qquad\text{(tautologically)}\\[4pt]
&5.\qquad (P\lor\lnot P)\to P\\[4pt]
&6.\qquad T\to P\\[4pt]
&7.\qquad P\\[8pt]
\text{Or even simpler:}\;\;\;
&1.\qquad \lnot P\to F\\[4pt]
&2.\qquad \lnot F\to\lnot (\lnot P)\qquad\text{(contrapositive)}\\[4pt]
&3.\qquad T\to P\\[4pt]
&4.\qquad P
\end{align*}
Thus, once you've proved $\lnot P\to F$, you can prove $P$ (it's not an assumption).
