Use a Karnaugh map to simplify the following Boolean expression:
(x'"\\land" y'"\\land" z') v (x'"\\land" y'"\\land" z') v (x'"\\land" y'"\\land" z') v (x'"\\land" y'"\\land" z') v (x'"\\land" y'"\\land" z')
Let use a Karnaugh map to simplify the following Boolean expression:
"(x'\\land y'\\land z') \\lor (x'\\land y'\\land z') \\lor (x'\\land y'\\land z') \\lor (x'\\land y'\\land z') \\lor (x'\\land y'\\land z')"
A Karnaugh map is the following:
We conclude that this expression is equivalent to "x'\\land y'\\land z'." It follows also from the idempotent law: "p\\lor p=p."
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