Discussion
Consider the propositional formula
\[ (A \wedge B) \rightarrow \big((\neg C \wedge A) \rightarrow (A \equiv 1)\big) \]
where \( 1 \) denotes the constant true. This formula is
Answer: B
Implication associates to the right, so the formula reads
\[ (A \wedge B) \rightarrow \big((\neg C \wedge A) \rightarrow (A \equiv 1)\big) \]
A formula is valid (a tautology) when it is true under every assignment. Take any assignment and consider the outer implication:
- If \( A \wedge B \) is false, the outer implication is true by vacuous truth - an implication with a false antecedent is always true.
- If \( A \wedge B \) is true, then \( A \) is true, so \( A \equiv 1 \) is true. The inner implication therefore has a true consequent and is true regardless of \( \neg C \wedge A \), which makes the outer implication true as well.
Both cases give true, so the formula holds under all \( 2^{3} = 8 \) assignments of \( A, B, C \). It is valid.
That also rules out the other options: a contradiction is false everywhere, and "satisfiable but not valid" requires at least one assignment making it false. \( C \) never affects the outcome - it appears only inside a subformula whose value is irrelevant once \( A \) is known.
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