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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" -

Figure 4

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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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