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Theorem partfun 29318
 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 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))

Proof of Theorem partfun
StepHypRef Expression
1 mptun 5982 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)))
2 inundif 4018 . . 3 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
3 eqid 2621 . . 3 if(𝑥𝐵, 𝐶, 𝐷) = if(𝑥𝐵, 𝐶, 𝐷)
42, 3mpteq12i 4702 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷))
5 elinel2 3778 . . . . 5 (𝑥 ∈ (𝐴𝐵) → 𝑥𝐵)
65iftrued 4066 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐶)
76mpteq2ia 4700 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐶)
8 eldifn 3711 . . . . 5 (𝑥 ∈ (𝐴𝐵) → ¬ 𝑥𝐵)
98iffalsed 4069 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐷)
109mpteq2ia 4700 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐷)
117, 10uneq12i 3743 . 2 ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
121, 4, 113eqtr3i 2651 1 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
 Colors of variables: wff setvar class Syntax hints:   = wceq 1480   ∈ wcel 1987   ∖ cdif 3552   ∪ cun 3553   ∩ cin 3554  ifcif 4058   ↦ cmpt 4673 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ral 2912  df-v 3188  df-dif 3558  df-un 3560  df-in 3562  df-if 4059  df-opab 4674  df-mpt 4675 This theorem is referenced by: (None)
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