سوال ۹۲ کنکور ارشد علوم کامپیوتر ۱۴۰۴

Let $S$ = “all students are intelligent,” $T$ = “all teachers are capable,” and $M$ = “financial resources are sufficient.”

Argument (a). Symbolically:

$P_1:\ \neg(S \land T), \qquad P_2:\ M \to T, \qquad \therefore\ S \to \neg M.$

To test the conditional conclusion, assume $S$ is true. From $P_1 = \neg(S \land T) \equiv \neg S \lor \neg T$ and $S$ true, we force $\neg T$ (i.e. $T$ is false). Applying $P_2:\ M \to T$ with $T$ false, the contrapositive $\neg T \to \neg M$ gives $\neg M$. Hence whenever $S$ holds, $\neg M$ holds, so $S \to \neg M$ is true under the premises. The argument is valid (it is essentially a chain of $\neg(S\land T)$ with modus tollens).

Formally, the implication $\big(\neg(S\land T) \land (M\to T)\big) \to (S \to \neg M)$ is a tautology: in the only line where the conclusion could fail ($S$ true and $M$ true), $P_2$ forces $T$ true, which makes $P_1$ false — so the premises are never both true with a false conclusion.

Argument (b). “All students are intelligent; some teachers are intelligent; therefore some students are teachers.” The shared term “intelligent” is the undistributed middle — a classic invalid syllogistic form. Counterexample: let the students be $\{\text{Ali}\}$ and the teachers be $\{\text{Bob}\}$, both intelligent. Both premises hold, yet no student is a teacher, so the conclusion fails. The argument is invalid.

Only (a) is valid, so the answer is option 1.

Exam trick: “All $X$ are $P$; some $Y$ are $P$” never lets you conclude any overlap between $X$ and $Y$ — the middle term must be distributed at least once. Meanwhile $\neg(S\land T)$ plus $M\to T$ chains via modus tollens to $S\to\neg M$.