Question

Use a truth table to determine whether the following statement is a contradiction, a tautology or neither. If it is a contradiction or a tautology, verify your answer using logical equivalences. ((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q

Answer 1

Answer:

The statement $((p \lor q) \land (p \implies r) \land (\neg r)) \implies q$ is a tautology.

Step-by-step explanation:

A truth table shows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it's constructed.

We can see from the truth table that the last column contains only true values. Therefore, the statement is a tautology.

Logical equivalences are a type of relationship between two statements or sentences in propositional logic. To simplify an equivalency, start with one side of the equation and attempt to replace sections of it with equivalent expressions. Continue doing this until you have achieved the desired statement form.

$((p \lor q) \land (p \implies r) \land (\neg r)) \implies q \\\equiv \neg[(p \lor q) \land (p \implies r) \land (\neg r)] \lor q$ by implication law

$\equiv \neg[(p \lor q) \land (\neg p \lor r) \land (\neg r))] \lor q$ by implication law

$\equiv \neg(p \lor q) \lor \neg (\neg p \lor r) \lor \neg(\neg r) \lor q$ by de Morgan’s law

$\equiv \neg(p \lor q) \lor \neg (\neg p \lor r) \lor r \lor q$ by Double Negative

$\equiv [(\neg p \land \neg q) \lor (p \land \neg r)] \lor r \lor q$ by de Morgan’s law

$\equiv [(\neg p \land \neg q) \lor q] \lor [(p \land \neg r) \lor r]$ by commutative and associative laws

$\equiv [(\neg p \lor q) \land (\neg q \lor q)] \lor [(p \lor r) \land (\neg r \lor r)]$ by distributive laws

$\equiv (\neg p \lor q) \lor (p \lor r)$ by negation and identity laws

$\equiv (\neg p \lor p) \lor (q \lor r)$ by communicative and associative laws

$\equiv T$ by negation and domination laws

Therefore, the statement is a tautology.