Theorem rbaib 925

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015.)

Hypothesis

Ref	Expression
baib.1	⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))

Assertion

Ref	Expression
rbaib	⊢ (𝜒 → (𝜑 ↔ 𝜓))

Proof of Theorem rbaib

Step	Hyp	Ref	Expression
1		baib.1	. . 3 ⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))
2		ancom 266	. . 3 ⊢ ((𝜓 ∧ 𝜒) ↔ (𝜒 ∧ 𝜓))
3	1, 2	bitri 184	. 2 ⊢ (𝜑 ↔ (𝜒 ∧ 𝜓))
4	3	baib 923	1 ⊢ (𝜒 → (𝜑 ↔ 𝜓))

Syntax hints:   → wi 4   ∧ wa 104   ↔ 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

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  reusv1  4526  opres  4990  cores  5208  fvres  5627  fzsplit2  10214  ablnsg  13837  cnptoprest  14878