If the truth value of the compound statement $[(p \vee q) \wedge (q \to r) \wedge (\sim r)] \to (p \wedge q)$ is false, then the truth values of $p \to q$ and $q \to p$ are respectively...

  • A
    $(F, T)$
  • B
    $(T, F)$
  • C
    $(T, T)$
  • D
    $(F, F)$

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Similar Questions

The given circuit is equivalent to:

The negation of the statement "If an integer is greater than $4$ and less than $5$, then it is a multiple of $3$" is:

Consider the two statements:
$(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$ is a tautology.
$(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$ is a fallacy.
Then:

$(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ is equivalent to $(\sim p)$ when $r$ is.

Simplify $(p \vee q) \wedge (p \vee \sim q)$

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