Theorem ifbid 3681

Description: Equivalence deduction for conditional operators. (Contributed by NM, 18-Apr-2005.)

Hypothesis

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

Assertion

Ref	Expression
ifbid	⊢ (φ → if(ψ, A, B) = if(χ, A, B))

Proof of Theorem ifbid

Step	Hyp	Ref	Expression
1		ifbid.1	. 2 ⊢ (φ → (ψ ↔ χ))
2		ifbi 3680	. 2 ⊢ ((ψ ↔ χ) → if(ψ, A, B) = if(χ, A, B))
3	1, 2	syl 15	1 ⊢ (φ → if(ψ, A, B) = if(χ, A, B))

Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642   ifcif 3663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is referenced by:  ifbieq2d  3683  ifbieq12d  3685  ifan  3702  ifor  3703  enprmaplem5  6081