Theorem rbaibr 547

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015.) (Proof shortened by Wolf Lammen, 19-Jan-2020.)

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	1	biancomi 468	. 2 ⊢ (𝜑 ↔ (𝜒 ∧ 𝜓))
3	2	baibr 546	1 ⊢ (𝜒 → (𝜓 ↔ 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401

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

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by:  rbaib  548  exintrbi  1924  moeuex  2612  sssseq  3956  ssunsn2  4795  sdrgacs  20956  cmpfi  23617  fimgmcyc  43362  nanorxor  45075