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Theorem partfun 6690
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 6689 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)))
2 inundif 4473 . . 3 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
3 eqid 2726 . . 3 if(𝑥𝐵, 𝐶, 𝐷) = if(𝑥𝐵, 𝐶, 𝐷)
42, 3mpteq12i 5247 . 2 (𝑥 ∈ ((𝐴𝐵) ∪ (𝐴𝐵)) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷))
5 elinel2 4191 . . . . 5 (𝑥 ∈ (𝐴𝐵) → 𝑥𝐵)
65iftrued 4531 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐶)
76mpteq2ia 5244 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐶)
8 eldifn 4122 . . . . 5 (𝑥 ∈ (𝐴𝐵) → ¬ 𝑥𝐵)
98iffalsed 4534 . . . 4 (𝑥 ∈ (𝐴𝐵) → if(𝑥𝐵, 𝐶, 𝐷) = 𝐷)
109mpteq2ia 5244 . . 3 (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) = (𝑥 ∈ (𝐴𝐵) ↦ 𝐷)
117, 10uneq12i 4156 . 2 ((𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷)) ∪ (𝑥 ∈ (𝐴𝐵) ↦ if(𝑥𝐵, 𝐶, 𝐷))) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
121, 4, 113eqtr3i 2762 1 (𝑥𝐴 ↦ if(𝑥𝐵, 𝐶, 𝐷)) = ((𝑥 ∈ (𝐴𝐵) ↦ 𝐶) ∪ (𝑥 ∈ (𝐴𝐵) ↦ 𝐷))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1533  wcel 2098  cdif 3940  cun 3941  cin 3942  ifcif 4523  cmpt 5224
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-if 4524  df-opab 5204  df-mpt 5225
This theorem is referenced by:  mptiffisupp  32419  mptprop  32424  cycpm2tr  32781  fsuppssindlem2  41703
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