If q is false and p wedge q leftrightarrow r | vedclass

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(B) Given that $q$ is false and $p \wedge q \leftrightarrow r$ is true.Since $q \equiv F$,then $p \wedge q \equiv F$.For the biconditional $p \wedge q

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that $q$ is false and $p \wedge q \leftrightarrow r$ is true.Since $q \equiv F$,then $p \wedge q \equiv F$.For the biconditional $p \wedge q \leftrightarrow r$ to be true,$r$ must have the same truth value as $p \wedge q$.Therefore,$r \equiv F$.Now,let us evaluate the options:$(A)$ $p \vee r \equiv p \vee F \equiv p$,which is not a tautology.$(B)$ $(p \wedge r)$ $\rightarrow (p \vee r) \equiv (p \wedge F)$ $\rightarrow (p \vee F) \equiv F$ $\rightarrow p$. Since $F \rightarrow p$ is always true for any truth value of $p$,this is a tautology.$(C)$ $(p \vee r)$ $\rightarrow (p \wedge r) \equiv (p \vee F)$ $\rightarrow (p \wedge F) \equiv p$ $\rightarrow F$,which is not a tautology.$(D)$ $p \wedge r \equiv p \wedge F \equiv F$,which is a contradiction.

If $q$ is false and $p \wedge q \leftrightarrow r$ is true,then which of the following is a tautology?

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