\leftrightarrow q) \wedge (q \to r) \wedge \sim r] \to (p \wedge \sim q) $$

is false.

So,

$$ (p \leftrightarrow q) \wedge (q \to r) \wedge \sim r $$

is true, and

$$ p \wedge \sim q $$

is false.

Since $\sim r$ is true, we get

$$ r = F $$

Also, $q \to r$ is true. Since $r = F$, the implication $q \to r$ can be true only when

$$ q = F $$

Now $p \leftrightarrow q$ is true, so $p$ and $q$ must have the same truth value. Since $q = F$,

$$ p = F $$

So we have

$$ p = F,\quad q = F,\quad r = F $$

Now check the first statement pattern:

$$ (p \to q) \leftrightarrow (q \to r) $$

Since $p = F$ and $q = F$,

$$ p \to q = F \to F = T $$

Since $q = F$ and $r = F$,

$$ q \to r = F \to F = T $$

So,

$$ (p \to q) \leftrightarrow (q \to r) = T \leftrightarrow T = T $$

Now check the second statement pattern:

$$ \sim(p \vee r) \to (q \wedge p) $$

Since $p = F$ and $r = F$,

$$ p \vee r = F \vee F = F $$

So,

$$ \sim(p \vee r) = T $$

Also,

$$ q \wedge p = F \wedge F = F $$

Thus,

$$ \sim(p \vee r) \to (q \wedge p) = T \to F = F $$

Hence the truth values are

$$ (T, F) $$

So the correct option is Option B.

Mathematical Reasoning

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