Without using truth table prove that (p wedge q) vee (sim p wedge q) vee (p wedge sim q) ≡ p vee q

Question 212 of 227

(p \wedge q) \vee (\sim p \wedge q) \vee (p \wedge \sim q) \equiv p \vee q

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 4mImportance★★★★★

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Group the first two terms to factor out $q$ , then simplify using $p\vee\sim p\equiv T$ .

$(p\wedge q)\vee(\sim p\wedge q)\vee(p\wedge\sim q)$

Group the first two terms (both contain $q$ ):

$= [q\wedge(p\vee\sim p)]\vee(p\wedge\sim q)$

$= [q\wedge T]\vee(p\wedge\sim q) \qquad (\text{since } p\vee\sim p\equiv T)$

$= q\vee(p\wedge\sim q)$

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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p\leftrightarrow q

›Reveal solution

Build the truth table for both statement patterns for every combination of $p,q$ and compare the final columns.

$p$	$q$	$\sim p$	$\sim q$	$p\leftrightarrow q$	$p\wedge q$	$\sim p\wedge\sim q$	$(p\wedge q)\vee(\sim p\wedge\sim q)$
T	T	F	F	T	T	F	T
T	F	F	T	F	F	F	F
F	T	T	F	F	F	F	F
F	F	T	T	T	F	T	T

The column for $p\leftrightarrow q$ is $[T, F, F, T]$ and the column for $(p\wedge q)\vee(\sim p\wedge\sim q)$ is also $[T, F, F, T]$ .

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$p$	$q$	$p\wedge q$	$\sim q$	$(p\wedge q)\vee\sim q$	$p\vee\sim q$
T	T	T	F	T	T
T	F	F	T	T	T
F	T	F	F	F	F
F	F	F	T	T	T
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Q.Without using truth table prove that $(p \wedge q) \vee (\sim p \wedge q) \vee (p \wedge \sim q) \equiv p \vee q$