Theorem elpreqprlem 4797

Description: Lemma for elpreqpr 4798. (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 2739	. . . 4 ⊢ {𝐵, 𝐶} = {𝐵, 𝐶}
2		preq2 4666	. . . . . 6 ⊢ (𝑥 = 𝐶 → {𝐵, 𝑥} = {𝐵, 𝐶})
3	2	eqeq2d 2750	. . . . 5 ⊢ (𝑥 = 𝐶 → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵, 𝐶} = {𝐵, 𝐶}))
4	3	spcegv 3535	. . . 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 4568	. . . 4 ⊢ {𝐵} = {𝐵, 𝐵}
8		preq2 4666	. . . . . 6 ⊢ (𝑥 = 𝐵 → {𝐵, 𝑥} = {𝐵, 𝐵})
9	8	eqeq2d 2750	. . . . 5 ⊢ (𝑥 = 𝐵 → ({𝐵} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝐵}))
10	9	spcegv 3535	. . . 4 ⊢ (𝐵 ∈ 𝑉 → ({𝐵} = {𝐵, 𝐵} → ∃𝑥{𝐵} = {𝐵, 𝑥}))
11	7, 10	mpi 20	. . 3 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵} = {𝐵, 𝑥})
12		prprc2 4698	. . . . 5 ⊢ (¬ 𝐶 ∈ V → {𝐵, 𝐶} = {𝐵})
13	12	eqeq1d 2741	. . . 4 ⊢ (¬ 𝐶 ∈ V → ({𝐵, 𝐶} = {𝐵, 𝑥} ↔ {𝐵} = {𝐵, 𝑥}))
14	13	exbidv 1928	. . 3 ⊢ (¬ 𝐶 ∈ V → (∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥} ↔ ∃𝑥{𝐵} = {𝐵, 𝑥}))
15	11, 14	imbitrrid 247	. 2 ⊢ (¬ 𝐶 ∈ V → (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥}))
16	6, 15	pm2.61i 183	1 ⊢ (𝐵 ∈ 𝑉 → ∃𝑥{𝐵, 𝐶} = {𝐵, 𝑥})

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1547  ∃wex 1786   ∈ wcel 2119  Vcvv 3431  {csn 4555  {cpr 4557

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-v 3433  df-dif 3886  df-un 3888  df-nul 4262  df-sn 4556  df-pr 4558

This theorem is referenced by:  elpreqpr  4798