User:Lord Farin/Tableau Proof Rules/Proof by Contradiction

Proof by Contradiction
Given $p \vdash \bot$, we have $\neg p$ (without assuming $p$).


 * align="right" | $j$ ||
 * align="right" | $I,$ but NOT $i$
 * $\neg p$
 * $\neg \mathcal I$
 * $i..i'$
 * $i..i'$