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Theorem elpreqprlem 4819
Description: Lemma for elpreqpr 4820. (Contributed by Scott Fenton, 7-Dec-2020.) (Revised by AV, 9-Dec-2020.)
Assertion
Ref Expression
elpreqprlem (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
Distinct variable groups:   𝑥,𝐵   𝑥,𝐶
Allowed substitution hint:   𝑉(𝑥)

Proof of Theorem elpreqprlem
StepHypRef Expression
1 eqid 2733 . . . 4 {𝐵, 𝐶} = {𝐵, 𝐶}
2 preq2 4688 . . . . . 6 (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
32eqeq2d 2744 . . . . 5 (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
43spcegv 3548 . . . 4 (𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
51, 4mpi 20 . . 3 (𝐶 ∈ V → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
65a1d 25 . 2 (𝐶 ∈ V → (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
7 dfsn2 4590 . . . 4 {𝐵} = {𝐵, 𝐵}
8 preq2 4688 . . . . . 6 (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
98eqeq2d 2744 . . . . 5 (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
109spcegv 3548 . . . 4 (𝐵𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
117, 10mpi 20 . . 3 (𝐵𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12 prprc2 4720 . . . . 5 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
1312eqeq1d 2735 . . . 4 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
1413exbidv 1922 . . 3 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
1511, 14imbitrrid 246 . 2 𝐶 ∈ V → (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
166, 15pm2.61i 182 1 (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1541  wex 1780  wcel 2113  Vcvv 3437  {csn 4577  {cpr 4579
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-v 3439  df-dif 3901  df-un 3903  df-nul 4283  df-sn 4578  df-pr 4580
This theorem is referenced by:  elpreqpr  4820
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