Theorem elpreqprlem 4867

Description: Lemma for elpreqpr 4868. (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 2733	. . . 4 ⊢ {𝐵, 𝐶} = {𝐵, 𝐶}
2		preq2 4739	. . . . . 6 ⊢ (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
3	2	eqeq2d 2744	. . . . 5 ⊢ (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
4	3	spcegv 3588	. . . 4 ⊢ (𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝐶} → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
5	1, 4	mpi 20	. . 3 ⊢ (𝐶 ∈ V → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})
6	5	a1d 25	. 2 ⊢ (𝐶 ∈ V → (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
7		dfsn2 4642	. . . 4 ⊢ {𝐵} = {𝐵, 𝐵}
8		preq2 4739	. . . . . 6 ⊢ (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
9	8	eqeq2d 2744	. . . . 5 ⊢ (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
10	9	spcegv 3588	. . . 4 ⊢ (𝐵 ∈ 𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
11	7, 10	mpi 20	. . 3 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12		prprc2 4771	. . . . 5 ⊢ (¬ 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
13	12	eqeq1d 2735	. . . 4 ⊢ (¬ 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
14	13	exbidv 1925	. . 3 ⊢ (¬ 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
15	11, 14	imbitrrid 245	. 2 ⊢ (¬ 𝐶 ∈ V → (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
16	6, 15	pm2.61i 182	1 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1542  ∃wex 1782   ∈ wcel 2107  Vcvv 3475  {csn 4629  {cpr 4631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-dif 3952  df-un 3954  df-nul 4324  df-sn 4630  df-pr 4632

This theorem is referenced by:  elpreqpr  4868