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Theorem elpreqprlem 4368
 Description: Lemma for elpreqpr 4369. (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 2621 . . . 4 {𝐵, 𝐶} = {𝐵, 𝐶}
2 preq2 4244 . . . . . 6 (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
32eqeq2d 2631 . . . . 5 (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
43spcegv 3283 . . . 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 4166 . . . 4 {𝐵} = {𝐵, 𝐵}
8 preq2 4244 . . . . . 6 (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
98eqeq2d 2631 . . . . 5 (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
109spcegv 3283 . . . 4 (𝐵𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
117, 10mpi 20 . . 3 (𝐵𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12 prprc2 4276 . . . . 5 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
1312eqeq1d 2623 . . . 4 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
1413exbidv 1847 . . 3 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
1511, 14syl5ibr 236 . 2 𝐶 ∈ V → (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
166, 15pm2.61i 176 1 (𝐵𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   = wceq 1480  ∃wex 1701   ∈ wcel 1987  Vcvv 3189  {csn 4153  {cpr 4155 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-v 3191  df-dif 3562  df-un 3564  df-nul 3897  df-sn 4154  df-pr 4156 This theorem is referenced by:  elpreqpr  4369
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