Theorem partfun 6715

Description: Rewrite a function defined by parts, using a mapping and an if construct, into a union of functions on disjoint domains. (Contributed by Thierry Arnoux, 30-Mar-2017.)

Assertion

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

Proof of Theorem partfun

Step	Hyp	Ref	Expression
1		mptun 6714	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)))
2		inundif 4479	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
3		eqid 2737	. . 3 ⊢ if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = if(𝑥 ∈ 𝐵, 𝐶, 𝐷)
4	2, 3	mpteq12i 5248	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))
5		elinel2 4202	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐵)
6	5	iftrued 4533	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐶)
7	6	mpteq2ia 5245	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶)
8		eldifn 4132	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → ¬ 𝑥 ∈ 𝐵)
9	8	iffalsed 4536	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
10	9	mpteq2ia 5245	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷)
11	7, 10	uneq12i 4166	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))
12	1, 4, 11	3eqtr3i 2773	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))

Syntax hints:   = wceq 1540   ∈ wcel 2108   ∖ cdif 3948   ∪ cun 3949   ∩ cin 3950  ifcif 4525   ↦ cmpt 5225

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-if 4526  df-opab 5206  df-mpt 5226

This theorem is referenced by:  mptiffisupp  32702  mptprop  32707  cycpm2tr  33139  redvmptabs  42390  fsuppssindlem2  42602