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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

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

                            \begin{aligned} & \equiv(\sim p \vee q) \vee(\sim p \vee r) \\ & \equiv(p \rightarrow q) \vee(p \rightarrow r) \end{aligned}

also,  p \rightarrow(q \vee r) \equiv(\sim p) \vee(q \vee r) \equiv(\sim p \vee q) \vee r

                                    \begin{aligned} & \equiv \sim(p \wedge(\sim q)) \vee r \\ & \equiv p \wedge(\sim q) \rightarrow r \end{aligned}

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

p \rightarrow(q \vee r) \equiv p \wedge(\sim q) \rightarrow r \equiv p \wedge(\sim r) \rightarrow q
\text { For } p=\mathrm{T}, q=\mathrm{F}, r=\mathrm{F}, p \rightarrow(q \vee r) \text { is } \mathrm{F} \text {, but }
(p \wedge q) \rightarrow(p \vee r) \vee(q \wedge r) \text { is T. }
\therefore \quad p \rightarrow(q \vee r) \text { and }
        p \wedge q \rightarrow(p \wedge r) \vee(q \wedge r)

are not equivalent

Posted by

Devendra Khairwa

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