Theorem elpreqprlem 4826

Description: Lemma for elpreqpr 4827. (Contributed by Scott Fenton, 7-Dec-2020.) (Revised by AV, 9-Dec-2020.)

Assertion

Ref	Expression
elpreqprlem	⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})

Distinct variable groups:   𝑥,𝐵   𝑥,𝐶

Allowed substitution hint:   𝑉(𝑥)

Proof of Theorem elpreqprlem

Step	Hyp	Ref	Expression
1		eqid 2763	. . . 4 ⊢ {𝐵, 𝐶} = {𝐵, 𝐶}
2		preq2 4697	. . . . . 6 ⊢ (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
3	2	eqeq2d 2774	. . . . 5 ⊢ (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
4	3	spcegv 3555	. . . 4 ⊢ (𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
5	1, 4	mpi 21	. . 3 ⊢ (𝐶 ∈ V → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
6	5	a1d 26	. 2 ⊢ (𝐶 ∈ V → (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
7		dfsn2 4597	. . . 4 ⊢ {𝐵} = {𝐵, 𝐵}
8		preq2 4697	. . . . . 6 ⊢ (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
9	8	eqeq2d 2774	. . . . 5 ⊢ (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
10	9	spcegv 3555	. . . 4 ⊢ (𝐵 ∈ 𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
11	7, 10	mpi 21	. . 3 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12		prprc2 4729	. . . . 5 ⊢ (¬ 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
13	12	eqeq1d 2765	. . . 4 ⊢ (¬ 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
14	13	exbidv 1956	. . 3 ⊢ (¬ 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
15	11, 14	imbitrrid 249	. 2 ⊢ (¬ 𝐶 ∈ V → (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
16	6, 15	pm2.61i 184	1 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   = wceq 1570  ∃wex 1812   ∈ wcel 2147  Vcvv 3454  {csn 4588  {cpr 4590

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-dif 4050  df-un 4154  df-nul 4282  df-sn 4589  df-pr 4591

This theorem is used by:  elpreqpr  4827