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Theorem elpreqprb 4335
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 4334 . 2 (𝐴 ∈ {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥})
2 prid1g 4239 . . . 4 (𝐴𝑉𝐴 ∈ {𝐴, 𝑥})
3 eleq2 2677 . . . 4 ({𝐵, 𝐶} = {𝐴, 𝑥} → (𝐴 ∈ {𝐵, 𝐶} ↔ 𝐴 ∈ {𝐴, 𝑥}))
42, 3syl5ibrcom 236 . . 3 (𝐴𝑉 → ({𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
54exlimdv 1848 . 2 (𝐴𝑉 → (∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
61, 5impbid2 215 1 (𝐴𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥}))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 195   = wceq 1475  wex 1695  wcel 1977  {cpr 4127
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-dif 3543  df-un 3545  df-nul 3875  df-sn 4126  df-pr 4128
This theorem is referenced by: (None)
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