Theorem condan 769

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

Hypotheses

Ref	Expression
condan.1	⊢ ((φ ∧ ¬ ψ) → χ)
condan.2	⊢ ((φ ∧ ¬ ψ) → ¬ χ)

Assertion

Ref	Expression
condan	⊢ (φ → ψ)

Proof of Theorem condan

Step	Hyp	Ref	Expression
1		condan.1	. . 3 ⊢ ((φ ∧ ¬ ψ) → χ)
2		condan.2	. . 3 ⊢ ((φ ∧ ¬ ψ) → ¬ χ)
3	1, 2	pm2.65da 559	. 2 ⊢ (φ → ¬ ¬ ψ)
4		notnot2 104	. 2 ⊢ (¬ ¬ ψ → ψ)
5	3, 4	syl 15	1 ⊢ (φ → ψ)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by: (None)