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  3612  raltpd  4745  opelopab2a  5517  ov  7560  ovg  7581  soseq  8160  ltprord  11040  isfull  18003  isfth  18007  ltsval  27879  axcontlem5  29409  isph  31287  cmbr  32049  cvbr  32747  mdbr  32759  dmdbr  32764  brfldext  34140  brfinext  34147  risc  38721