Question 63

Option (A): $((a \to b) \land (b \to c)) \to (a \to c)$

• This is the standard rule of Hypothetical Syllogism (transitivity of implication), which is a classical tautology.

Option (B): $(a \leftrightarrow c) \to (\sim b \to (a \land c))$

• Test the truth assignment: $a = \text{False}, c = \text{False}, b = \text{False}$.
  • $a \leftrightarrow c = (\text{False} \leftrightarrow \text{False}) = \text{True}$.
  • $\sim b = \text{True}$.
  • $a \land c = (\text{False} \land \text{False}) = \text{False}$.
  • Consequent: $\sim b \to (a \land c) \equiv \text{True} \to \text{False} = \text{False}$.
  • Entire expression: $\text{True} \to \text{False} = \text{False}$.
• Because the expression evaluates to False for this assignment, it is NOT a tautology.

Option (C): $(a \land b \land c) \to (c \lor a)$

• If $a \land b \land c$ is True, both $a$ and $c$ are True, so $c \lor a$ is True. Always True (tautology).

Option (D): $a \to (b \to a)$

• $\neg a \lor (\neg b \lor a) \equiv (\neg a \lor a) \lor \neg b \equiv \text{True} \lor \neg b \equiv \text{True}$. Always True (tautology).

Conclusion:
Option (B) is NOT a tautology.