Metamath Proof Explorer


Theorem condan

Description: Proof by contradiction. (Contributed by NM, 9-Feb-2006) (Proof shortened by Wolf Lammen, 19-Jun-2014)

Ref Expression
Hypotheses condan.1 φ ¬ ψ χ
condan.2 φ ¬ ψ ¬ χ
Assertion condan φ ψ

Proof

Step Hyp Ref Expression
1 condan.1 φ ¬ ψ χ
2 condan.2 φ ¬ ψ ¬ χ
3 1 2 pm2.65da φ ¬ ¬ ψ
4 3 notnotrd φ ψ