Theorem ceqsex2 3504

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006.)

Hypotheses

Ref	Expression
ceqsex2.1	⊢ Ⅎ𝑥𝜓
ceqsex2.2	⊢ Ⅎ𝑦𝜒
ceqsex2.3	⊢ 𝐴 ∈ V
ceqsex2.4	⊢ 𝐵 ∈ V
ceqsex2.5	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
ceqsex2.6	⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
ceqsex2	⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ 𝜒)

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

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

Proof of Theorem ceqsex2

Step	Hyp	Ref	Expression
1		3anass 1110	. . . . 5 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)))
2	1	exbii 1877	. . . 4 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)))
3		19.42v 1982	. . . 4 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
4	2, 3	bitri 278	. . 3 ⊢ (∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
5	4	exbii 1877	. 2 ⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)))
6		nfv 1943	. . . . 5 ⊢ Ⅎ𝑥 𝑦 = 𝐵
7		ceqsex2.1	. . . . 5 ⊢ Ⅎ𝑥𝜓
8	6, 7	nfan 1928	. . . 4 ⊢ Ⅎ𝑥(𝑦 = 𝐵 ∧ 𝜓)
9	8	nfex 2356	. . 3 ⊢ Ⅎ𝑥∃𝑦(𝑦 = 𝐵 ∧ 𝜓)
10		ceqsex2.3	. . 3 ⊢ 𝐴 ∈ V
11		ceqsex2.5	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
12	11	anbi2d 641	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑦 = 𝐵 ∧ 𝜑) ↔ (𝑦 = 𝐵 ∧ 𝜓)))
13	12	exbidv 1950	. . 3 ⊢ (𝑥 = 𝐴 → (∃𝑦(𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐵 ∧ 𝜓)))
14	9, 10, 13	ceqsex 3501	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ 𝜓))
15		ceqsex2.2	. . 3 ⊢ Ⅎ𝑦𝜒
16		ceqsex2.4	. . 3 ⊢ 𝐵 ∈ V
17		ceqsex2.6	. . 3 ⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))
18	15, 16, 17	ceqsex 3501	. 2 ⊢ (∃𝑦(𝑦 = 𝐵 ∧ 𝜓) ↔ 𝜒)
19	5, 14, 18	3bitri 300	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ 𝜒)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   ∧ w3a 1102   = wceq 1569  ∃wex 1808  Ⅎwnf 1812   ∈ wcel 2142  Vcvv 3454

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-10 2175  ax-11 2191  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-ex 1809  df-nf 1813  df-clel 2837

This theorem is used by: (None)