Theorem partfun2 32775

Description: Rewrite a function defined by parts, using a mapping and an if construct, into a union of functions on disjoint domains. See also partfun 6639 and ifmpt2v 7465. (Contributed by Thierry Arnoux, 25-Jan-2026.)

Hypothesis

Ref	Expression
partfun2.1	⊢ 𝐷 = {𝑥 ∈ 𝐴 ∣ 𝜑}

Assertion

Ref	Expression
partfun2	⊢ (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)   𝐶(𝑥)   𝐷(𝑥)

Proof of Theorem partfun2

Step	Hyp	Ref	Expression
1		partfun 6639	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐷, 𝐵, 𝐶)) = ((𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))
2		partfun2.1	. . . . . 6 ⊢ 𝐷 = {𝑥 ∈ 𝐴 ∣ 𝜑}
3	2	reqabi 3415	. . . . 5 ⊢ (𝑥 ∈ 𝐷 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
4	3	baib 540	. . . 4 ⊢ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐷 ↔ 𝜑))
5	4	ifbid 4485	. . 3 ⊢ (𝑥 ∈ 𝐴 → if(𝑥 ∈ 𝐷, 𝐵, 𝐶) = if(𝜑, 𝐵, 𝐶))
6	5	mpteq2ia 5174	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐷, 𝐵, 𝐶)) = (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶))
7	2	ssrab3 4020	. . . . 5 ⊢ 𝐷 ⊆ 𝐴
8		sseqin2 4159	. . . . 5 ⊢ (𝐷 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐷) = 𝐷)
9	7, 8	mpbi 231	. . . 4 ⊢ (𝐴 ∩ 𝐷) = 𝐷
10	9	mpteq1i 5170	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) = (𝑥 ∈ 𝐷 ↦ 𝐵)
11	10	uneq1i 4101	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))
12	1, 6, 11	3eqtr3i 2771	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))

Syntax hints:   = wceq 1547   ∈ wcel 2119  {crab 3392   ∖ cdif 3887   ∪ cun 3888   ∩ cin 3889   ⊆ wss 3890  ifcif 4461   ↦ cmpt 5160

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-12 2189  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-if 4462  df-opab 5142  df-mpt 5161

This theorem is referenced by:  extvfvcl  33727