Theorem bianabs 545

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 532	. 2 ⊢ (𝜑 → (𝜒 ↔ (𝜑 ∧ 𝜒)))
3	1, 2	bitr4d 285	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399

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 210  df-an 400

This theorem is referenced by:  ceqsrexv  3563  raltpd  4697  opelopab2a  5416  ov  7353  ovg  7373  ltprord  10644  isfull  17417  isfth  17421  axcontlem5  27059  isph  28903  cmbr  29665  cvbr  30363  mdbr  30375  dmdbr  30380  brfldext  31436  brfinext  31442  soseq  33540  sltval  33587  risc  35881