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Theorem partfun 30423
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 6496 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)))
2 inundif 4429 . . 3 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
3 eqid 2823 . . 3 if(𝑥𝐵, 𝐶, 𝐷) = if(𝑥𝐵, 𝐶, 𝐷)
42, 3mpteq12i 5161 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷))
5 elinel2 4175 . . . . 5 (𝑥 ∈ (𝐴𝐵) → 𝑥𝐵)
65iftrued 4477 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐶)
76mpteq2ia 5159 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐶)
8 eldifn 4106 . . . . 5 (𝑥 ∈ (𝐴𝐵) → ¬ 𝑥𝐵)
98iffalsed 4480 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐷)
109mpteq2ia 5159 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐷)
117, 10uneq12i 4139 . 2 ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
121, 4, 113eqtr3i 2854 1 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1537  wcel 2114  cdif 3935  cun 3936  cin 3937  ifcif 4469  cmpt 5148
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 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-v 3498  df-dif 3941  df-un 3943  df-in 3945  df-if 4470  df-opab 5131  df-mpt 5149
This theorem is referenced by:  mptprop  30436  cycpm2tr  30763
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