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The statement $p\rightarrow (q\vee r)$ is not equivalent to
Option: 1
$(p\rightarrow q)\vee (p\rightarrow r)$

Option: 2
$p\wedge (\sim q)\rightarrow r$

Option: 3
$p\wedge (\sim r)\rightarrow q$

Option: 4
$p\wedge q\rightarrow (p\wedge r)\vee (q\wedge r)$

Answers (1)

best_answer

Truth Table of "if-then" -

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Truth table of 'OR' operator -

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$p\rightarrow (q\vee r)\equiv (\sim p)\vee (q\vee r)$

$\equiv(\sim p\vee q)\vee (\sim p\vee r)$

$\equiv( p\rightarrow q)\vee (p\rightarrow r)$

Also , $p\rightarrow (q\vee r)\equiv (\sim p)\vee (q\vee r)\equiv (\sim p\vee q)\vee r$

$\equiv \:\sim (p\: \wedge( \sim q ) )\vee r \equiv p\wedge\:(\sim q)\:\rightarrow r$

Interchanging the roles of q and r in the above paragraph, we find

$p\rightarrow (p\vee r)\equiv p \wedge(\sim q)\rightarrow r\equiv p \:\wedge(\sim r)\rightarrow q$

$for\:\:p=T,q=F,r=F,p\rightarrow (q \vee r)$ is $F$ , but $(p \wedge q)\rightarrow (p \vee r ) \vee (q \wedge r)\:\:\:is\:\:T.$

$\therefore p\rightarrow (q \vee r)$

and $p \wedge q \rightarrow (p \wedge r) \vee (q \wedge r)$

Are not equivalent

Posted by

Gautam harsolia

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