Theorem ifbid 3349

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 3348	. 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 98   = wceq 1243  ifcif 3331

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-if 3332

This theorem is referenced by:  ifbieq1d  3350  ifbieq2d  3352  ifbieq12d  3354  sumeq1  9874