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Theorem elpreqprlem 4788
Description: Lemma for elpreqpr 4789. (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 2819 . . . 4 {𝐵, 𝐶} = {𝐵, 𝐶}
2 preq2 4662 . . . . . 6 (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
32eqeq2d 2830 . . . . 5 (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
43spcegv 3595 . . . 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 4572 . . . 4 {𝐵} = {𝐵, 𝐵}
8 preq2 4662 . . . . . 6 (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
98eqeq2d 2830 . . . . 5 (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
109spcegv 3595 . . . 4 (𝐵𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
117, 10mpi 20 . . 3 (𝐵𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12 prprc2 4694 . . . . 5 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
1312eqeq1d 2821 . . . 4 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
1413exbidv 1915 . . 3 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
1511, 14syl5ibr 248 . 2 𝐶 ∈ V → (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
166, 15pm2.61i 184 1 (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1530  wex 1773  wcel 2107  Vcvv 3493  {csn 4559  {cpr 4561
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-v 3495  df-dif 3937  df-un 3939  df-nul 4290  df-sn 4560  df-pr 4562
This theorem is referenced by:  elpreqpr  4789
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