Exercise · ZW-CONTRADICTION-011 · ★★★ Advanced
Let \(Q=(P\Rightarrow R)\land P\land\neg R\). Is \(Q\) a contradiction?
Choose one option.
Show answer
- D. Yes, it is false under all four assignments
Explanation
With \(P=F\) the conjunct \(P\) fails; with \(P=T\) and \(R=F\), \(P\Rightarrow R\) fails; with \(R=T\), \(\neg R\) fails, so \(Q\) is false in all four rows.
Review the concept: Contradiction
A compound proposition \(P\) is a contradiction if it is false under every assignment of its components.
\[P\land\neg P\equiv\bot\]
More practice problems: contradiction
- ZW-CONTRADICTION-001 A contradiction is false...
- ZW-CONTRADICTION-002 Which one is a contradiction?
- ZW-CONTRADICTION-003 In the table of a contradiction, the final column contains...
- ZW-CONTRADICTION-004 If P is true, P ∧ ¬P turns out...
- ZW-CONTRADICTION-005 A contradiction cannot be true...
- ZW-CONTRADICTION-006 Under which assignments is a contradiction false?
- ZW-CONTRADICTION-007 Which of these formulas is a contradiction?
- ZW-CONTRADICTION-008 The final column of a contradiction contains...