Theorem rbaibr 874

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

Hypothesis

Ref	Expression
baib.1	⊢ (φ ↔ (ψ ∧ χ))

Assertion

Ref	Expression
rbaibr	⊢ (χ → (ψ ↔ φ))

Proof of Theorem rbaibr

Step	Hyp	Ref	Expression
1		baib.1	. . 3 ⊢ (φ ↔ (ψ ∧ χ))
2		ancom 437	. . 3 ⊢ ((ψ ∧ χ) ↔ (χ ∧ ψ))
3	1, 2	bitri 240	. 2 ⊢ (φ ↔ (χ ∧ ψ))
4	3	baibr 872	1 ⊢ (χ → (ψ ↔ φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ 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:  ssunsn2  3866