Logic Questions and Answers

Answer: C

Correct answer: Logically equivalent to p ^ q

Distribute: \( p \land (\lnot p \lor q) \equiv (p \land \lnot p) \lor (p \land q) \), and the first disjunct is a contradiction, so the whole reduces to \( p \land q \). It is therefore neither a tautology nor a contradiction - it is true exactly when both p and q are true.

Answer: B

Correct answer: a ^ b → b v c

If the antecedent \( a \land b \) is true then b is true, so \( b \lor c \) is true and the implication holds; if the antecedent is false the implication holds vacuously. The other two fail: the first when a is true and b false, and the third when a is false, b true and c false.

Answer: B

Implication associates to the right, so the formula reads

\[ (A \wedge B) \rightarrow \big((\neg C \wedge A) \rightarrow (A \equiv 1)\big) \]

A formula is valid (a tautology) when it is true under every assignment. Take any assignment and consider the outer implication:

• If \( A \wedge B \) is false, the outer implication is true by vacuous truth - an implication with a false antecedent is always true.
• If \( A \wedge B \) is true, then \( A \) is true, so \( A \equiv 1 \) is true. The inner implication therefore has a true consequent and is true regardless of \( \neg C \wedge A \), which makes the outer implication true as well.

Both cases give true, so the formula holds under all \( 2^{3} = 8 \) assignments of \( A, B, C \). It is valid.

That also rules out the other options: a contradiction is false everywhere, and "satisfiable but not valid" requires at least one assignment making it false. \( C \) never affects the outcome - it appears only inside a subformula whose value is irrelevant once \( A \) is known.

Answer: D

Correct answer: Q ^ (P v R)

Work forward from the premises. From \( P \to M \) and \( \lnot M \), modus tollens gives \( \lnot P \). With \( P \lor Q \) that gives Q, and \( Q \to R \) then gives R. So Q and R are both true, and since R is true so is \( P \lor R \) - which makes the conjunction the valid conclusion.

Answer: A

Correct answer: True

The first equivalence makes a equivalent to \( b \lor \lnot b \), which is a tautology, so a is always true. Now suppose the antecedent \( a \land b \) holds; then b is true, and since \( b \leftrightarrow c \), c is true too, so \( a \land c \) is true and the consequent holds. The implication is therefore true in every case.

Answer: B

Correct answer: He says

A declarative sentence makes a statement and can be assigned a truth value. “It’s right”, “Two may not be an even integer” and “I love you” are all declarative in form. “He says” is a fragment with no complement, asserts nothing and cannot be called true or false, so it is not a statement at all.

Answer: A

Correct answer: (P ^ Q) → R

Expand both sides into disjunctive form. The left is \( \lnot P \lor (\lnot Q \lor R) \), which is \( \lnot P \lor \lnot Q \lor R \); the right is \( \lnot(P \land Q) \lor R \), which by De Morgan is the same expression. The result is the exportation law, and it is why a nested implication can be flattened into one with a conjunctive antecedent.

Answer: D

Correct answer: Both (a) & (b)

The first is the absorption law: \( (P \lor Q) \land Q \) simplifies to Q, so the biconditional is \( Q \leftrightarrow Q \) and is always true. The second is the disjunctive syllogism: given \( P \lor Q \) and \( \lnot P \), Q must hold. The third fails when P is true and Q false, since the antecedent is then true and the consequent false.

Answer: B

Correct answer: The conjuction F1 ^ F2 is not satisfiable

If both implications are tautologies then whenever \( F_1 \land F_2 \) is true, F3 and its negation are both true - which is impossible. So \( F_1 \land F_2 \) can never be true, and it is unsatisfiable. Nothing follows about F1 or F2 individually; each may be satisfiable on its own.

Answer: C

Correct answer: F1 is satisfiable, F2 is valid

F1 is \( P \to \lnot P \), which reduces to \( \lnot P \); that is true when P is false, so F1 is satisfiable though not valid. F2 is \( (P \to \lnot P) \lor (\lnot P \to P) \), which reduces to \( \lnot P \lor P \) and is therefore true in every case, so F2 is valid.