$1\mathrm{.}8$: Proof By Contradiction

Suppose I first assume $"$P AND q AND r$"$ and derive a contradiction. Then I start over, assume $"$P AND

NOT q AND NOT $r",$ and derive another contradiction. Then $P$ itself must be a contradiction, and thus

"NOT p$"$ is a tautology$\mathrm{.}1\mathrm{.}8$: Proof By Contradiction

Suppose I first assume $"$P AND q AND r$"$ and derive a contradiction. Then I start over, assume $"$P AND

NOT q AND NOT $r",$ and derive another contradiction. Then $P$ itself must be a contradiction, and thus

"NOT p$"$ is a tautology.