Question 11

We are given the compound proposition:

Recall that $p \implies q \equiv \neg p \vee q$. By De Morgan's laws:

Here:

• $\neg(p \implies q) \equiv \neg(\neg p \vee q) \equiv p \wedge \neg q$
• $\neg(\neg r \vee \neg s) \equiv r \wedge s$
  Therefore:

Examine the second disjunct $(r \wedge s)$:

• $r: x$ is a perfect square ($x = k^2$ for some integer $k$).
• $s: x$ is a prime number.
  For any integer $x \ge 2$, if $x$ is a perfect square $k^2$, then $k \ge 2$, so $x$ has proper divisors $1, k, k^2$, which means $x$ is composite. A composite number cannot be prime. Hence, no integer $x \ge 2$ can be both a perfect square and a prime number. Thus, $r \wedge s$ is identically FALSE.

The proposition simplifies to:

• $p: x \in \{8, 9, 10, 11, 12\}$
• $\neg q: x$ is NOT composite
  Among the candidate values in $\{8, 9, 10, 11, 12\}$:
• $8 = 2^3$ (composite)
• $9 = 3^2$ (composite)
• $10 = 2 \times 5$ (composite)
• $11$ (prime, hence not composite)
• $12 = 2^2 \times 3$ (composite)
  Only $x = 11$ satisfies $p \wedge \neg q$.

Thus, the required integer is 11.