Theorem bianabs 542

Description: Absorb a hypothesis into the second member of a biconditional. (Contributed by FL, 15-Feb-2007.)

Hypothesis

Ref	Expression
bianabs.1	⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜒)))

Assertion

Ref	Expression
bianabs	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Proof of Theorem bianabs

Step	Hyp	Ref	Expression
1		bianabs.1	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜒)))
2		ibar 529	. 2 ⊢ (𝜑 → (𝜒 ↔ (𝜑 ∧ 𝜒)))
3	1, 2	bitr4d 281	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396

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-an 397

This theorem is referenced by:  ceqsrexv  3643  raltpd  4785  opelopab2a  5535  ov  7551  ovg  7571  soseq  8144  ltprord  11024  isfull  17860  isfth  17864  sltval  27147  axcontlem5  28223  isph  30070  cmbr  30832  cvbr  31530  mdbr  31542  dmdbr  31547  brfldext  32721  brfinext  32727  risc  36849