Answer to Question #262665 in Discrete Mathematics for Arianne

Question #262665

p ^   ~ (q ^ r)


1
Expert's answer
2021-11-08T19:47:13-0500

Let us construct the trush table for the formula "p \\land \\sim (q \\land r):"


"\\begin{array}{||c|c|c||c|c|c||}\n\\hline\\hline\np & q & r & q \\land r & \\sim(q \\land r) &p \\land \\sim (q \\land r)\\\\\n\\hline\\hline\n0 & 0 & 0 & 0 & 1 & 0\\\\\n\\hline\n0 & 0 & 1 & 0 & 1 & 0\\\\\n\\hline\n0 & 1 & 0 & 0 & 1 & 0\\\\\n\\hline\n0 & 1 & 1 & 1 & 0 & 0\\\\\n\\hline\n1 & 0 & 0 & 0 & 1 & 1\\\\\n\\hline\n1 & 0 & 1 & 0 & 1 & 1\\\\\n\\hline\n1 & 1 & 0 & 0 & 1 & 1\\\\\n\\hline\n1 & 1 & 1 & 1 & 0 &0\\\\\n\\hline\\hline\n\\end{array}"


We conclude that this formula is neither tautology nor contradiction.


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