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Theorem mptun 6682
Description: Union of mappings which are mutually compatible. (Contributed by Mario Carneiro, 31-Aug-2015.)
Assertion
Ref Expression
mptun (𝑥 ∈ (𝐴𝐵) ↦ 𝐶) = ((𝑥𝐴𝐶) ∪ (𝑥𝐵𝐶))

Proof of Theorem mptun
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 df-mpt 5197 . 2 (𝑥 ∈ (𝐴𝐵) ↦ 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶)}
2 df-mpt 5197 . . . 4 (𝑥𝐴𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐶)}
3 df-mpt 5197 . . . 4 (𝑥𝐵𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦 = 𝐶)}
42, 3uneq12i 4128 . . 3 ((𝑥𝐴𝐶) ∪ (𝑥𝐵𝐶)) = ({⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐶)} ∪ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦 = 𝐶)})
5 elun 4115 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 635 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝑦 = 𝐶))
7 andir 1024 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝑦 = 𝐶) ↔ ((𝑥𝐴𝑦 = 𝐶) ∨ (𝑥𝐵𝑦 = 𝐶)))
86, 7bitri 278 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶) ↔ ((𝑥𝐴𝑦 = 𝐶) ∨ (𝑥𝐵𝑦 = 𝐶)))
98opabbii 5182 . . . 4 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶)} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥𝐴𝑦 = 𝐶) ∨ (𝑥𝐵𝑦 = 𝐶))}
10 unopab 5195 . . . 4 ({⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐶)} ∪ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦 = 𝐶)}) = {⟨𝑥, 𝑦⟩ ∣ ((𝑥𝐴𝑦 = 𝐶) ∨ (𝑥𝐵𝑦 = 𝐶))}
119, 10eqtr4i 2795 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶)} = ({⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐶)} ∪ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦 = 𝐶)})
124, 11eqtr4i 2795 . 2 ((𝑥𝐴𝐶) ∪ (𝑥𝐵𝐶)) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝑦 = 𝐶)}
131, 12eqtr4i 2795 1 (𝑥 ∈ (𝐴𝐵) ↦ 𝐶) = ((𝑥𝐴𝐶) ∪ (𝑥𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wa 400  wo 860   = wceq 1567  wcel 2149  cun 3911  {copab 5177  cmpt 5196
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-v 3465  df-un 3918  df-opab 5178  df-mpt 5197
This theorem is referenced by:  partfun  6683  fmptap  7169  fmptapd  7170  fmptunsnop  32985  esumrnmpt2  34402  ptrest  38157  fsuppssind  43216  dftpos5  49536
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