Theorem partfun 6647

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 6646	. 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 4433	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
3		eqid 2737	. . 3 ⊢ if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = if(𝑥 ∈ 𝐵, 𝐶, 𝐷)
4	2, 3	mpteq12i 5197	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))
5		elinel2 4156	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐵)
6	5	iftrued 4489	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐶)
7	6	mpteq2ia 5195	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶)
8		eldifn 4086	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → ¬ 𝑥 ∈ 𝐵)
9	8	iffalsed 4492	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
10	9	mpteq2ia 5195	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷)
11	7, 10	uneq12i 4120	. 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 2768	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))

Syntax hints:   = wceq 1542   ∈ wcel 2114   ∖ cdif 3900   ∪ cun 3901   ∩ cin 3902  ifcif 4481   ↦ cmpt 5181

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-if 4482  df-opab 5163  df-mpt 5182

This theorem is referenced by:  partfun2  32766  mptiffisupp  32783  mptprop  32788  cycpm2tr  33213  redvmptabs  42730  fsuppssindlem2  42950