Theorem ceqsex2v 3521

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006.) Avoid ax-10 2130 and ax-11 2147. (Revised by GG, 20-Aug-2023.)

Hypotheses

Ref	Expression
ceqsex2v.1	⊢ 𝐴 ∈ V
ceqsex2v.2	⊢ 𝐵 ∈ V
ceqsex2v.3	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
ceqsex2v.4	⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
ceqsex2v	⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ 𝜒)

Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦   𝜓,𝑥   𝜒,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑦)   𝜒(𝑥)

Proof of Theorem ceqsex2v

Step	Hyp	Ref	Expression
1		3anass 1092	. . . . 5 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)))
2	1	exbii 1843	. . . 4 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)))
3		19.42v 1950	. . . 4 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
4	2, 3	bitri 274	. . 3 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
5	4	exbii 1843	. 2 ⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
6		ceqsex2v.1	. . 3 ⊢ 𝐴 ∈ V
7		ceqsex2v.3	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
8	7	anbi2d 628	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑦 = 𝐵 ∧ 𝜓)))
9	8	exbidv 1917	. . 3 ⊢ (𝑥 = 𝐴 → (∃𝑦(𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐵 ∧ 𝜓)))
10	6, 9	ceqsexv 3515	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ 𝜓))
11		ceqsex2v.2	. . 3 ⊢ 𝐵 ∈ V
12		ceqsex2v.4	. . 3 ⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))
13	11, 12	ceqsexv 3515	. 2 ⊢ (∃𝑦(𝑦 = 𝐵 ∧ 𝜓) ↔ 𝜒)
14	5, 10, 13	3bitri 296	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ 𝜒)

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 394   ∧ w3a 1084   = wceq 1534  ∃wex 1774   ∈ wcel 2099  Vcvv 3462

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101

This theorem depends on definitions:  df-bi 206  df-an 395  df-3an 1086  df-ex 1775  df-clel 2803

This theorem is referenced by:  ceqsex3v  3522  ceqsex4v  3523  ispos  18334  elfuns  35752  brimg  35774  brapply  35775  brsuccf  35778  brrestrict  35786  dfrdg4  35788  diblsmopel  40883