سوال ۹۷ کنکور ارشد علوم کامپیوتر ۱۴۰۰

Translate the English sentence piece by piece into propositional logic.

Atomic propositions.

The hypothesis (the "if" part).

"Neither the weather is cold nor it is raining" uses the construction neither A nor B, which means not A and not B:

$\text{neither } p \text{ nor } q \;\equiv\; (\sim p) \wedge (\sim q).$

The conclusion (the "then" part).

"Neither a raincoat needs to be worn nor a hat needs to be worn" is again neither/nor. The raincoat is $s$ and the hat is $r$, so:

$\text{neither } s \text{ nor } r \;\equiv\; (\sim s) \wedge (\sim r) \;\equiv\; (\sim r) \wedge (\sim s).$

Assemble the conditional.

"If (hypothesis) then (conclusion)" is the implication

$\big((\sim p) \wedge (\sim q)\big) \Rightarrow \big((\sim r) \wedge (\sim s)\big),$

which is exactly option 3.

Why the others fail.

• Option 1 keeps the conclusion as $r \wedge s$ (positive) — it forgets to negate, saying both items are needed. Wrong.
• Option 2 uses $p \wedge q$ (positive) for the hypothesis instead of $(\sim p) \wedge (\sim q)$. Wrong.
• Option 4 has $(\sim p) \wedge q$ (only $p$ negated, not $q$) and a weaker consequent $\sim(r \wedge s)$, which by De Morgan is $(\sim r) \vee (\sim s)$ — "not both", not "neither". Wrong.

Pitfall: "neither A nor B" is $(\sim A) \wedge (\sim B)$, not $\sim(A \wedge B)$. The latter only says they are not both true. Distinguishing $(\sim r) \wedge (\sim s)$ from $\sim(r \wedge s)$ is precisely what separates the correct option from option 4.