Theorem bifald 36172

Description: Infer the equivalence to a contradiction from a negation, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)

Hypothesis

Ref	Expression
bifald.1	⊢ (𝜑 → ¬ 𝜓)

Assertion

Ref	Expression
bifald	⊢ (𝜑 → (𝜓 ↔ ⊥))

Proof of Theorem bifald

Step	Hyp	Ref	Expression
1		bifald.1	. 2 ⊢ (𝜑 → ¬ 𝜓)
2		id 22	. . 3 ⊢ (𝜓 → 𝜓)
3		falim 1556	. . 3 ⊢ (⊥ → 𝜓)
4	2, 3	pm5.21ni 378	. 2 ⊢ (¬ 𝜓 → (𝜓 ↔ ⊥))
5	1, 4	syl 17	1 ⊢ (𝜑 → (𝜓 ↔ ⊥))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205  ⊥wfal 1551

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 206  df-tru 1542  df-fal 1552

This theorem is referenced by: (None)