2023 CSAT — Q78

Locate. On a conditional-logic free-CR item, the answer is never a sentence you can point to — it is derived. The repeatable move is: write each premise as an arrow (A → B), and write each conclusion as an arrow too, then ask whether the conclusion’s arrow is forced by the premises through only two legal moves — chaining (A→B, B→C ⟹ A→C) and contraposition (A→B ⟹ ¬B→¬A). The two illegal moves that trap students are the converse (A→B does not give B→A) and the inverse.

Test (derive each conclusion; accept only chaining + contraposition). Premise I: P → (Q ∧ S). Premise II: (R ∧ S) → ¬T. Conclusion 1 says T → (¬P ∨ ¬R). Contrapose II: T → ¬(R ∧ S), i.e. T → (¬R ∨ ¬S). If ¬R, conclusion holds outright. If ¬S, use I’s consequence P → S (its contrapositive ¬S → ¬P), so ¬S forces ¬P, and the conclusion holds again. Forced in both branches ⟹ valid. Conclusion 2 says Q → P — but Premise I only gives P → Q. Asserting Q → P is the converse error; Q can hold while P is false. Invalid.

Eliminate by anatomy. (b) and (c) both endorse Conclusion 2, a step the text doesn’t license — reading a conditional backwards (P→Q taken as Q→P), the single most common logic trap. (d) wrongly rejects Conclusion 1, the same kind of unlicensed step in reverse — failing to run the contrapositive of II and the chained ¬S → ¬P. The transferable rule: a conditional licenses its contrapositive and nothing else; treat “if A then B” as one-directional and most CSAT logic traps disarm themselves. Key: (a).

Conclusion 1: Premise II is “(R ∧ S) → ¬T”; its contrapositive is “T → ¬(R ∧ S) = (¬R ∨ ¬S)”. So if T is true, then R is false OR S is false. If R is false, “at least one of P, R is false” already holds. If instead S is false, Premise I “P → (Q ∧ S)” gives “P → S” (contrapositive ¬S → ¬P), so S false forces P false — and again “at least one of P, R is false” holds. Either branch forces it → Conclusion 1 is VALID. Conclusion 2 asserts “Q → P”, which is the CONVERSE of Premise I’s “P → (Q ∧ S)”; a conditional never entails its converse (Q can be true with P false), so Conclusion 2 is INVALID. Only 1 → (a).

Premises as arrows: I. $P \rightarrow (Q \wedge S)$. II. $(R \wedge S) \rightarrow \neg T$.

Conclusion 1 — If T is true, then at least one of P and R is false. Contrapositive of II: $T \rightarrow \neg(R \wedge S)$, that is $T \rightarrow (\neg R \vee \neg S)$. Suppose T true. Then $\neg R$ or $\neg S$.

• If $\neg R$: “at least one of P, R false” holds (R is false).
• If $\neg S$: from I, $P \rightarrow S$, so its contrapositive $\neg S \rightarrow \neg P$ forces P false; “at least one of P, R false” holds (P is false).

Both branches force the conclusion. Valid.

Conclusion 2 — If Q is true, then P is true. Premise I gives $P \rightarrow Q$ (among other things), not $Q \rightarrow P$. The converse does not follow: Q may be true while P is false. Invalid.

Only Conclusion 1 can be drawn. Answer: (a) 1 only.

This item is a pure test of two legal moves and one forbidden one. The legal moves — chaining and contraposition — get you Conclusion 1: contrapose II, then for the $\neg S$ branch lean on I’s hidden $P \rightarrow S$. The forbidden move — the converse — is the entire trap of Conclusion 2: $P \rightarrow Q$ never yields $Q \rightarrow P$. The aspirant who internalises “a conditional points one way; I may only flip-and-negate it, never reverse it” converts a scary-looking five-variable puzzle into two short derivations and lands on (a).