سوال ۹۶ کنکور ارشد علوم کامپیوتر ۱۴۰۱

An argument is valid when every truth assignment that makes all premises true also makes the conclusion true. Equivalently, the conjunction of premises must logically entail the conclusion.

Argument (a): $\neg a \wedge (a \to b) \;\therefore\; b$

Try to break it: choose $a = \text{F}$ and $b = \text{F}$.

• Premise: $\neg a \wedge (a \to b) = \text{T} \wedge (\text{F} \to \text{F}) = \text{T} \wedge \text{T} = \text{T}$.
• Conclusion: $b = \text{F}$.

The single premise is true while the conclusion is false, so (a) is invalid. Knowing $\neg a$ tells you nothing about $b$: with $a$ false, the implication $a \to b$ is vacuously true regardless of $b$. (This is the fallacy of denying the antecedent; valid modus ponens would need $a$ and $a\to b$ to conclude $b$.)

Argument (b): $p \to q,\; q \to r \;\therefore\; p \to r$

This is the hypothetical syllogism (transitivity of implication), a standard valid rule. To confirm, suppose both premises hold and $p$ is true. Then $q$ is true (from $p\to q$), hence $r$ is true (from $q\to r$); so $p \to r$ holds. If $p$ is false, $p \to r$ is vacuously true. In every case the conclusion follows, so (b) is valid.

Conclusion

Only (b) is valid, giving option 2.

Pitfall: $\neg a \wedge (a\to b)$ looks like it should yield something about $b$, but $\neg a$ makes $a\to b$ true for free. Valid elimination of an implication is modus ponens ($a,\ a\to b \vdash b$), not $\neg a$.