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Theorem elpreqprb 4588
Description: A set is an element of an unordered pair iff there is another (maybe the same) set which is an element of the unordered pair. (Proposed by BJ, 8-Dec-2020.) (Contributed by AV, 9-Dec-2020.)
Assertion
Ref Expression
elpreqprb (𝐴𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥}))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐶   𝑥,𝑉

Proof of Theorem elpreqprb
StepHypRef Expression
1 elpreqpr 4587 . 2 (𝐴 ∈ {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥})
2 prid1g 4484 . . . 4 (𝐴𝑉𝐴 ∈ {𝐴, 𝑥})
3 eleq2 2867 . . . 4 ({𝐵, 𝐶} = {𝐴, 𝑥} → (𝐴 ∈ {𝐵, 𝐶} ↔ 𝐴 ∈ {𝐴, 𝑥}))
42, 3syl5ibrcom 239 . . 3 (𝐴𝑉 → ({𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
54exlimdv 2029 . 2 (𝐴𝑉 → (∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
61, 5impbid2 218 1 (𝐴𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥}))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198   = wceq 1653  wex 1875  wcel 2157  {cpr 4370
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-v 3387  df-dif 3772  df-un 3774  df-nul 4116  df-sn 4369  df-pr 4371
This theorem is referenced by: (None)
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