Answer to Question #314297 in Discrete Mathematics for Dreaper

Question

Show, by the use of the truth table (truth matrix), that the (p ∨ q) ∨ [(¬p) ∧ (¬q)] is a contradiction. 

Answer 1

Let us construct the trush table.

$\begin{array}{||c|c||c|c|c|c|c||} \hline \hline p & q & ¬p & ¬q & p ∨ q & (¬p) ∧ (¬q) & (p ∨ q) ∨ [(¬p) ∧ (¬q)]\\ \hline 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ \hline 0 & 1 & 1 & 0 & 1 & 0 & 1\\ \hline 1 & 0 & 0 & 1 & 1 & 0 & 1 \\ \hline 1 & 1 & 0 & 0 & 1 & 0 & 1\\ \hline\hline \end{array}$

It follows that the formula $(p ∨ q) ∨ [(¬p) ∧ (¬q)]$ is a tautology, and it is not a contradiction.