Theorem ceqsex2v 2698

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

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		nfv 1491	. 2 ⊢ Ⅎ𝑥𝜓
2		nfv 1491	. 2 ⊢ Ⅎ𝑦𝜒
3		ceqsex2v.1	. 2 ⊢ 𝐴 ∈ V
4		ceqsex2v.2	. 2 ⊢ 𝐵 ∈ V
5		ceqsex2v.3	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6		ceqsex2v.4	. 2 ⊢ (𝑦 = 𝐵 → (𝜓 ↔ 𝜒))
7	1, 2, 3, 4, 5, 6	ceqsex2 2697	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝜑) ↔ 𝜒)

Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 945   = wceq 1314  ∃wex 1451   ∈ wcel 1463  Vcvv 2657

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-ext 2097

This theorem depends on definitions:  df-bi 116  df-3an 947  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-v 2659

This theorem is referenced by:  ceqsex3v  2699  ceqsex4v  2700