Which one of the following propositional logic formulas is TRUE when exactly two of $p$, $q$, and $r$ are $\text{TRUE}$?

There are three truth assignments where exactly two of $p, q, r$ are TRUE:

1. $p = T, q = T, r = F$

2. $p = T, q = F, r = T$

3. $p = F, q = T, r = T$

Let us evaluate each component of Option B: $(\sim(p \leftrightarrow q) \land r) \lor (p \land q \land \sim r)$:

• Case 1 ($p=T, q=T, r=F$): $p \land q \land \sim r = T \land T \land T = T$. Thus, the formula evaluates to $T$.

• Case 2 ($p=T, q=F, r=T$): $p \leftrightarrow q$ is $F$, so $\sim(p \leftrightarrow q)$ is $T$. Then $\sim(p \leftrightarrow q) \land r = T \land T = T$. Thus, the formula evaluates to $T$.

• Case 3 ($p=F, q=T, r=T$): $p \leftrightarrow q$ is $F$, so $\sim(p \leftrightarrow q)$ is $T$. Then $\sim(p \leftrightarrow q) \land r = T \land T = T$. Thus, the formula evaluates to $T$.

In all cases where exactly two variables are TRUE, the expression in (B) evaluates to TRUE.

• For (A): when $p=T, q=F, r=T$, $(p \leftrightarrow q) \land r = F \land T = F$ and $p \land q \land \sim r = F$, so the overall formula is FALSE.

• For (C): when $p=T, q=F, r=T$, $(p \to q) \land r = F \land T = F$ and $p \land q \land \sim r = F$, so it is FALSE.

• For (D): the conjunction $\land$ between two mutually exclusive cases is always FALSE.

Hence, B is the correct option.