Theorem elpreqprb 4825

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	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥}))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐶   𝑥,𝑉

Proof of Theorem elpreqprb

Step	Hyp	Ref	Expression
1		elpreqpr 4824	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥})
2		prid1g 4720	. . . 4 ⊢ (𝐴 ∈ 𝑉 → 𝐴 ∈ {𝐴, 𝑥})
3		eleq2 2852	. . . 4 ⊢ ({𝐵, 𝐶} = {𝐴, 𝑥} → (𝐴 ∈ {𝐵, 𝐶} ↔ 𝐴 ∈ {𝐴, 𝑥}))
4	2, 3	syl5ibrcom 250	. . 3 ⊢ (𝐴 ∈ 𝑉 → ({𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
5	4	exlimdv 1967	. 2 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥} → 𝐴 ∈ {𝐵, 𝐶}))
6	1, 5	impbid2 229	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ ∃𝑥{𝐵, 𝐶} = {𝐴, 𝑥}))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  ∃wex 1811   ∈ wcel 2146  {cpr 4589

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3455  df-dif 4049  df-un 4153  df-nul 4281  df-sn 4588  df-pr 4590

This theorem is used by: (None)