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 )