Theorem partfun2 33157

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 6683 and ifmpt2v 7519. (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 6683	. 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 3437	. . . . 5 ⊢ (𝑥 ∈ 𝐷 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
4	3	baib 545	. . . 4 ⊢ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐷 ↔ 𝜑))
5	4	ifbid 4509	. . 3 ⊢ (𝑥 ∈ 𝐴 → if(𝑥 ∈ 𝐷, 𝐵, 𝐶) = if(𝜑, 𝐵, 𝐶))
6	5	mpteq2ia 5204	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐷, 𝐵, 𝐶)) = (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶))
7	2	ssrab3 4033	. . . . 5 ⊢ 𝐷 ⊆ 𝐴
8		sseqin2 4172	. . . . 5 ⊢ (𝐷 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐷) = 𝐷)
9	7, 8	mpbi 233	. . . 4 ⊢ (𝐴 ∩ 𝐷) = 𝐷
10	9	mpteq1i 5200	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) = (𝑥 ∈ 𝐷 ↦ 𝐵)
11	10	uneq1i 4114	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐷) ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))
12	1, 6, 11	3eqtr3i 2793	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝜑, 𝐵, 𝐶)) = ((𝑥 ∈ 𝐷 ↦ 𝐵) ∪ (𝑥 ∈ (𝐴 ∖ 𝐷) ↦ 𝐶))

This proof depends on syntax axioms:   = wceq 1570   ∈ wcel 2145  {crab 3414   ∖ cdif 3899   ∪ cun 3900   ∩ cin 3901   ⊆ wss 3902  ifcif 4485   ↦ cmpt 5190

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-12 2215  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-if 4486  df-opab 5172  df-mpt 5191

This theorem is used by:  extvfvcl  34054