The statement (P Rightarrow Q) wedge (R Right | vedclass

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Question

(A) Given the expression: $(P$ $\Rightarrow Q) \wedge (R$ $\Rightarrow Q)$We know that the implication $P \Rightarrow Q$ is logically equivalent to $\

Vedclass · Accepted Answer

$(P \vee R) \Rightarrow Q$

Vedclass · Answer

(A) Given the expression: $(P$ $\Rightarrow Q) \wedge (R$ $\Rightarrow Q)$We know that the implication $P \Rightarrow Q$ is logically equivalent to $\sim P \vee Q$.Substituting this into the expression:$(\sim P \vee Q) \wedge (\sim R \vee Q)$Using the distributive law,we can factor out $Q$:$(\sim P \wedge \sim R) \vee Q$By De Morgan's Law,$\sim P \wedge \sim R$ is equivalent to $\sim(P \vee R)$:$\sim(P \vee R) \vee Q$Using the implication rule $\sim A \vee B \equiv A \Rightarrow B$,we get:$(P \vee R) \Rightarrow Q$Therefore,the correct option is $A$.

The statement $(P$ $\Rightarrow Q) \wedge (R$ $\Rightarrow Q)$ is logically equivalent to:

Similar Questions

The conditional statement $(p \wedge q) \implies p$ is:

The logical expression $p \wedge (\sim p \vee \sim q) =$ ?

If $p, q$ and $r$ are three propositions,then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression $\{(p \vee q) \wedge ((\sim p) \vee r)\} \rightarrow ((\sim q) \vee r)$ false?

Among the statements:
$(S1) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow ( p$ $\Rightarrow r )$
$(S2) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow (( p$ $\Rightarrow r ) \vee ( q$ $\Rightarrow r ))$

The negation of $(p \wedge \sim q) \rightarrow (p \vee \sim q)$ is

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