Theorem bianabs 544

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 531	. 2 ⊢ (𝜑 → (𝜒 ↔ (𝜑 ∧ 𝜒)))
3	1, 2	bitr4d 284	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398

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 209  df-an 399

This theorem is referenced by:  ceqsrexv  3648  raltpd  4709  opelopab2a  5414  ov  7288  ovg  7307  ltprord  10446  isfull  17174  isfth  17178  axcontlem5  26748  isph  28593  cmbr  29355  cvbr  30053  mdbr  30065  dmdbr  30070  brfldext  31032  brfinext  31038  soseq  33091  sltval  33149  risc  35258