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Theorem partfun 28858
 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 5938 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)))
2 inundif 3998 . . 3 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
3 eqid 2610 . . 3 if(𝑥𝐵, 𝐶, 𝐷) = if(𝑥𝐵, 𝐶, 𝐷)
42, 3mpteq12i 4670 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷))
5 inss2 3796 . . . . . 6 (𝐴𝐵) ⊆ 𝐵
65sseli 3564 . . . . 5 (𝑥 ∈ (𝐴𝐵) → 𝑥𝐵)
76iftrued 4044 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐶)
87mpteq2ia 4668 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐶)
9 eldifn 3695 . . . . 5 (𝑥 ∈ (𝐴𝐵) → ¬ 𝑥𝐵)
109iffalsed 4047 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐷)
1110mpteq2ia 4668 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐷)
128, 11uneq12i 3727 . 2 ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
131, 4, 123eqtr3i 2640 1 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
 Colors of variables: wff setvar class Syntax hints:   = wceq 1475   ∈ wcel 1977   ∖ cdif 3537   ∪ cun 3538   ∩ cin 3539  ifcif 4036   ↦ cmpt 4643 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-if 4037  df-opab 4644  df-mpt 4645 This theorem is referenced by: (None)
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