TMUA practice: Necessary Sufficient, problem 2

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

The question

Let

P, Q, R, S, T

be statements such that: if

P

is true then both

Q

and

S

are true; and if both

R

and

S

are true then

T

is false. Which of the following must hold?

\text{If } T \text{ is true then both } P \text{ and } R \text{ are true}

\text{If } T \text{ is true then both } P \text{ and } R \text{ are false}

\text{If } T \text{ is true then at least one of } P, R \text{ is true}

\text{If } T \text{ is true then at least one of } P, R \text{ is false}

The correct answer is highlighted. D

Worked solution

The premises are $P \Rightarrow (Q \wedge S)$ and $(R \wedge S) \Rightarrow \neg T$ .

Suppose $T$ is true. The contrapositive of the second premise gives $\neg(R \wedge S)$ — that is, $\neg R$ or $\neg S$ .

Case $\neg R$ : then $R$ is false, so at least one of $P, R$ is false.
Case $\neg S$ : the contrapositive of the first premise is $\neg(Q \wedge S) \Rightarrow \neg P$ ; since $\neg S$ gives $\neg(Q \wedge S)$ , we get $\neg P$ . So $P$ is false, and again at least one of $P, R$ is false.

In every case, at least one of $P$ and $R$ is false — statement D.

Answer: D.