Theorem bibif 699

Description: Transfer negation via an equivalence. (Contributed by NM, 3-Oct-2007.) (Proof shortened by Wolf Lammen, 28-Jan-2013.)

Assertion

Ref	Expression
bibif	⊢ (¬ 𝜓 → ((𝜑 ↔ 𝜓) ↔ ¬ 𝜑))

Proof of Theorem bibif

Step	Hyp	Ref	Expression
1		nbn2 698	. 2 ⊢ (¬ 𝜓 → (¬ 𝜑 ↔ (𝜓 ↔ 𝜑)))
2		bicom 140	. 2 ⊢ ((𝜓 ↔ 𝜑) ↔ (𝜑 ↔ 𝜓))
3	1, 2	bitr2di 197	1 ⊢ (¬ 𝜓 → ((𝜑 ↔ 𝜓) ↔ ¬ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  nbn  700