Theorem ifbid 3370

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 3369	. 2 ⊢ ((𝜓 ↔ 𝜒) → if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐴, 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐴, 𝐵))

Syntax hints:   → wi 4   ↔ wb 103   = wceq 1284  ifcif 3351

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-if 3352

This theorem is referenced by:  ifbieq1d  3371  ifbieq2d  3373  ifbieq12d  3375  sumeq1  10192