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Theorem partfun 30349
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 6487 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)))
2 inundif 4423 . . 3 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
3 eqid 2818 . . 3 if(𝑥𝐵, 𝐶, 𝐷) = if(𝑥𝐵, 𝐶, 𝐷)
42, 3mpteq12i 5150 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷))
5 elinel2 4170 . . . . 5 (𝑥 ∈ (𝐴𝐵) → 𝑥𝐵)
65iftrued 4471 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐶)
76mpteq2ia 5148 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐶)
8 eldifn 4101 . . . . 5 (𝑥 ∈ (𝐴𝐵) → ¬ 𝑥𝐵)
98iffalsed 4474 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐷)
109mpteq2ia 5148 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐷)
117, 10uneq12i 4134 . 2 ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
121, 4, 113eqtr3i 2849 1 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1528  wcel 2105  cdif 3930  cun 3931  cin 3932  ifcif 4463  cmpt 5137
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-ral 3140  df-v 3494  df-dif 3936  df-un 3938  df-in 3940  df-if 4464  df-opab 5120  df-mpt 5138
This theorem is referenced by:  mptprop  30360  cycpm2tr  30688
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