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