سوال ۹۶ کنکور ارشد علوم کامپیوتر ۱۴۰۰

A statement is a theorem (tautology) only if both sides of the biconditional are logically equivalent — i.e., agree in every row of the truth table. A single counterexample row falsifies the claim.

Statement (a): $((p \land q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \land (q \Rightarrow r))$

Try $p = T,\ q = F,\ r = F$:

• LHS: $p \land q = T \land F = F$, so $F \Rightarrow r = T$.
• RHS: $(p \Rightarrow r) = (T \Rightarrow F) = F$ and $(q \Rightarrow r) = (F \Rightarrow F) = T$, so $F \land T = F$.

LHS $= T$ but RHS $= F$, so the biconditional is false here. Statement (a) is not a tautology.

(The correct identity for the left side is $((p \land q) \Rightarrow r) \equiv ((p \Rightarrow r) \lor (q \Rightarrow r))$ — note the $\lor$, not $\land$.)

Statement (b): $((p \lor q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \lor (q \Rightarrow r))$

Try the same $p = T,\ q = F,\ r = F$:

• LHS: $p \lor q = T \lor F = T$, so $T \Rightarrow F = F$.
• RHS: $(p \Rightarrow r) = (T \Rightarrow F) = F$ and $(q \Rightarrow r) = (F \Rightarrow F) = T$, so $F \lor T = T$.

LHS $= F$ but RHS $= T$, so the biconditional is false here. Statement (b) is not a tautology.

(The correct identity for the left side is $((p \lor q) \Rightarrow r) \equiv ((p \Rightarrow r) \land (q \Rightarrow r))$ — note the $\land$, not $\lor$.)

Conclusion

The two valid distribution laws for implication over a disjunctive/conjunctive hypothesis are:

$((p \lor q) \Rightarrow r) \equiv (p \Rightarrow r) \land (q \Rightarrow r)$ $((p \land q) \Rightarrow r) \equiv (p \Rightarrow r) \lor (q \Rightarrow r)$

Both printed statements pair the wrong connective on the right side, so neither is a theorem. The answer is option 4.

Pitfall: When the hypothesis of an implication is a disjunction, it splits into a conjunction of implications (and vice versa). The connective flips — a single $p=T,\ q=F,\ r=F$ row catches the error every time.