Theorem partfun 6637

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 6636	. 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 4429	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
3		eqid 2734	. . 3 ⊢ if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = if(𝑥 ∈ 𝐵, 𝐶, 𝐷)
4	2, 3	mpteq12i 5193	. 2 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷))
5		elinel2 4152	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐵)
6	5	iftrued 4485	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐶)
7	6	mpteq2ia 5191	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶)
8		eldifn 4082	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → ¬ 𝑥 ∈ 𝐵)
9	8	iffalsed 4488	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) → if(𝑥 ∈ 𝐵, 𝐶, 𝐷) = 𝐷)
10	9	mpteq2ia 5191	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷)
11	7, 10	uneq12i 4116	. 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 2765	1 ⊢ (𝑥 ∈ 𝐴 ↦ if(𝑥 ∈ 𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴 ∩ 𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴 ∖ 𝐵) ↦ 𝐷))

Syntax hints:   = wceq 1541   ∈ wcel 2113   ∖ cdif 3896   ∪ cun 3897   ∩ cin 3898  ifcif 4477   ↦ cmpt 5177

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2713  df-cleq 2726  df-clel 2809  df-v 3440  df-dif 3902  df-un 3904  df-in 3906  df-if 4478  df-opab 5159  df-mpt 5178

This theorem is referenced by:  partfun2  32704  mptiffisupp  32721  mptprop  32726  cycpm2tr  33150  redvmptabs  42557  fsuppssindlem2  42777