Theorem partfun2 32661

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 6633 and ifmpt2v 7454. (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 6633	. 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 3419	. . . . 5 ⊢ (𝑥 ∈ 𝐷 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
4	3	baib 535	. . . 4 ⊢ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐷 ↔ 𝜑))
5	4	ifbid 4498	. . 3 ⊢ (𝑥 ∈ 𝐴 → if(𝑥 ∈ 𝐷, 𝐵, 𝐶) = if(𝜑, 𝐵, 𝐶))
6	5	mpteq2ia 5188	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐷, 𝐵, 𝐶)) = (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶))
7	2	ssrab3 4031	. . . . 5 ⊢ 𝐷 ⊆ 𝐴
8		sseqin2 4172	. . . . 5 ⊢ (𝐷 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐷) = 𝐷)
9	7, 8	mpbi 230	. . . 4 ⊢ (𝐴 ∩ 𝐷) = 𝐷
10	9	mpteq1i 5184	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) = (𝑥 ∈ 𝐷 ↦ 𝐵)
11	10	uneq1i 4113	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))
12	1, 6, 11	3eqtr3i 2764	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))

Syntax hints:   = wceq 1541   ∈ wcel 2113  {crab 3396   ∖ cdif 3895   ∪ cun 3896   ∩ cin 3897   ⊆ wss 3898  ifcif 4474   ↦ cmpt 5174

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-12 2182  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-if 4475  df-opab 5156  df-mpt 5175

This theorem is referenced by:  extvfvcl  33587