Theorem partfun 6580

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 6579	. 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 4412	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
3		eqid 2738	. . 3 ⊢ if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = if(𝑥 ∈ 𝐵, 𝐶, 𝐷)
4	2, 3	mpteq12i 5180	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))
5		elinel2 4130	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐵)
6	5	iftrued 4467	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐶)
7	6	mpteq2ia 5177	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶)
8		eldifn 4062	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → ¬ 𝑥 ∈ 𝐵)
9	8	iffalsed 4470	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
10	9	mpteq2ia 5177	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷)
11	7, 10	uneq12i 4095	. 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 2774	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))

Syntax hints:   = wceq 1539   ∈ wcel 2106   ∖ cdif 3884   ∪ cun 3885   ∩ cin 3886  ifcif 4459   ↦ cmpt 5157

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-if 4460  df-opab 5137  df-mpt 5158

This theorem is referenced by:  mptprop  31031  cycpm2tr  31386  fsuppssindlem2  40281