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Lemma
Mathematical Reasoning (Paper 2) Difficulty 4

TMUA practice: Necessary Sufficient, problem 3

A TMUA-calibre necessary sufficient problem at difficulty 4 of 5, with the full worked solution.

The question
Let P,Q,R,SP, Q, R, S be statements such that: if PP is true then QQ is true; if QQ is true then RR is true; and if SS is true then at least one of QQ and RR is false. Which of the following follows?
A If S is false then both Q and R are true\text{If } S \text{ is false then both } Q \text{ and } R \text{ are true}
B If at least one of Q,R is true then S is false\text{If at least one of } Q, R \text{ is true then } S \text{ is false}
C If P is true then S is false\text{If } P \text{ is true then } S \text{ is false}
D If Q is true then S is true\text{If } Q \text{ is true then } S \text{ is true}

The correct answer is highlighted. C

Worked solution

The premises are P⇒QP \Rightarrow Q, Q⇒RQ \Rightarrow R, and S⇒(¬Q∨¬R)S \Rightarrow (\neg Q \vee \neg R).

Suppose PP is true. Then P⇒QP \Rightarrow Q gives QQ true, and Q⇒RQ \Rightarrow R gives RR true. So Q∧RQ \wedge R holds.

The contrapositive of the third premise is ¬(¬Q∨¬R)⇒¬S\neg(\neg Q \vee \neg R) \Rightarrow \neg S, i.e. (Q∧R)⇒¬S(Q \wedge R) \Rightarrow \neg S. Since Q∧RQ \wedge R holds, SS is false.

Therefore PP true ⇒S\Rightarrow S false — statement C.

(B fails: one of Q,RQ, R being true does not force both, so the third premise need not fire. A and D are not derivable either.)

Answer: C.