In 48 and 49 below, a logical equivalence is derived from
Chapter 2, Problem 48E(choose chapter or problem)
In 48 and 49 below, a logical equivalence is derived from Theorem 2.1.1. Supply a reason for each step.
\((p \wedge \sim q) \vee(p \wedge q) \equiv p \wedge(\sim q \vee q)\) by (a)
\(\equiv p \wedge(q \vee \sim q)\) by (b)
\(\equiv p \wedge \mathbf{t}\) by (c)
\(\equiv p\) by (d)
Therefore , \((p \wedge \sim q) \vee(p \wedge q) \equiv p\).
Text Transcription:
(p wedge sim q) vee(p wedge q) equiv p wedge(sim q vee q)
equiv p wedge(q vee sim q)
equiv p wedge t
equiv p
(p wedge sim q) vee(p wedge q) equiv p
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