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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 )