Which one of the following Boolean expressions is NOT a tautology?

Let us test the truth value of each option:

• Option (A): $((a \rightarrow b) \wedge (b \rightarrow c)) \rightarrow (a \rightarrow c)$ This is the Law of Hypothetical Syllogism (transitivity of implication), which is a well-known tautology.

• Option (B): $(a \leftrightarrow c) \rightarrow (\sim b \rightarrow (a \wedge c))$ Consider the truth assignment: $a = \text{False}$, $b = \text{False}$, $c = \text{False}$.

  • Left side: $a \leftrightarrow c \equiv \text{False} \leftrightarrow \text{False} \equiv \text{True}$.

  • Right side: $\sim b \rightarrow (a \wedge c) \equiv \text{True} \rightarrow (\text{False} \wedge \text{False}) \equiv \text{True} \rightarrow \text{False} \equiv \text{False}$.

  • Overall implication: $\text{True} \rightarrow \text{False} \equiv \text{False}$.

  Since it evaluates to False for this truth assignment, (B) is NOT a tautology.

• Option (C): $(a \wedge b \wedge c) \rightarrow (c \vee a)$ If $a \wedge b \wedge c$ is True, then both $a$ and $c$ are True, which makes $c \vee a$ True. Thus, this is a tautology.

• Option (D): $a \rightarrow (b \rightarrow a)$ Equivalent to $\sim a \vee (\sim b \vee a) \equiv (\sim a \vee a) \vee \sim b \equiv \text{True} \vee \sim b \equiv \text{True}$. Thus, this is a tautology.