Question

$\sim q$ can be deduced from $\sim(p \supset (q \lor r))$ by using rules of propositional logic in the following sequence(s):

• A. Material Implication $>$ De Morgan's Theorem $>$ Commutation $>$ Simplification $>$ De Morgan's Theorem $>$ Simplification
• B. Conjunction $>$ Addition $>$ Modus Ponens $>$ De Morgan's Theorem
• C. Transposition $>$ Material Implication $>$ Double Negation $>$ De Morgan's Theorem $>$ Simplification $>$ De Morgan's Theorem $>$ Simplification
• D. Modus Tollens $>$ Conjunction $>$ Hypothetical Syllogism $>$ Simplification

Hint:

Look for ways to apply De Morgan's Theorem after transforming implications — it helps break down disjunctions into conjunctive negations, making simplification straightforward.

Answer 1

The Correct Option is A, C

Route (A) – as listed: \[ \begin{aligned} 1.\ &\sim(p \supset (q \lor r)) 2.\ &\sim(\sim p \lor (q \lor r)) \text{(Material Implication)} 3.\ &\sim\sim p \ \cdot\ \sim(q \lor r) \text{(De Morgan)} 4.\ &\sim(q \lor r)\ \cdot\ \sim\sim p \text{(Commutation)} 5.\ &\sim(q \lor r) \text{(Simplification)} 6.\ &\sim q \ \cdot\ \sim r \text{(De Morgan)} 7.\ &\sim q \text{(Simplification).} \end{aligned} \] 

Route (C) – an alternative valid chain: \[ \begin{aligned} 1.\ &\sim(p \supset (q \lor r)) 2.\ &\sim(\sim(q \lor r) \supset \sim p) \text{(Transposition on the embedded implication)} 3.\ &\sim(\sim\sim(q \lor r) \lor \sim p) \text{(Material Implication)} 4.\ &\sim((q \lor r) \lor \sim p) \text{(Double Negation)} 5.\ &\sim(q \lor r)\ \cdot\ \sim\sim p \text{(De Morgan)} 6.\ &\sim(q \lor r) \text{(Simplification)} 7.\ &\sim q \ \cdot\ \sim r \text{(De Morgan)} 8.\ &\sim q \text{(Simplification).} \end{aligned} \] Therefore both sequences in (A) and (C) correctly derive $\boxed{\sim q}$.