Theorem bianabs 551

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 538	. 2 ⊢ (𝜑 → (𝜒 ↔ (𝜑 ∧ 𝜒)))
3	1, 2	bitr4d 285	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

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:  ceqsrexv  3609  raltpd  4742  opelopab2a  5513  ov  7560  ovg  7581  soseq  8162  ltprord  11064  isfull  18026  isfth  18030  ltsval  27915  axcontlem5  29457  isph  31335  cmbr  32097  cvbr  32795  mdbr  32807  dmdbr  32812  brfldext  34188  brfinext  34195  risc  38801