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 4438	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
3		eqid 2729	. . 3 ⊢ if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = if(𝑥 ∈ 𝐵, 𝐶, 𝐷)
4	2, 3	mpteq12i 5199	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))
5		elinel2 4161	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐵)
6	5	iftrued 4492	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐶)
7	6	mpteq2ia 5197	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶)
8		eldifn 4091	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → ¬ 𝑥 ∈ 𝐵)
9	8	iffalsed 4495	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
10	9	mpteq2ia 5197	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷)
11	7, 10	uneq12i 4125	. 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 2760	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))

Syntax hints:   = wceq 1540   ∈ wcel 2109   ∖ cdif 3908   ∪ cun 3909   ∩ cin 3910  ifcif 4484   ↦ cmpt 5183

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 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3446  df-dif 3914  df-un 3916  df-in 3918  df-if 4485  df-opab 5165  df-mpt 5184

This theorem is referenced by:  mptiffisupp  32666  mptprop  32671  cycpm2tr  33091  redvmptabs  42341  fsuppssindlem2  42573