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Theorem ceqsex2 3275
 Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006.)
Hypotheses
Ref Expression
ceqsex2.1 𝑥𝜓
ceqsex2.2 𝑦𝜒
ceqsex2.3 𝐴 ∈ V
ceqsex2.4 𝐵 ∈ V
ceqsex2.5 (𝑥 = 𝐴 → (𝜑𝜓))
ceqsex2.6 (𝑦 = 𝐵 → (𝜓𝜒))
Assertion
Ref Expression
ceqsex2 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ 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
StepHypRef Expression
1 3anass 1059 . . . . 5 ((𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
21exbii 1814 . . . 4 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
3 19.42v 1921 . . . 4 (∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
42, 3bitri 264 . . 3 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
54exbii 1814 . 2 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
6 nfv 1883 . . . . 5 𝑥 𝑦 = 𝐵
7 ceqsex2.1 . . . . 5 𝑥𝜓
86, 7nfan 1868 . . . 4 𝑥(𝑦 = 𝐵𝜓)
98nfex 2192 . . 3 𝑥𝑦(𝑦 = 𝐵𝜓)
10 ceqsex2.3 . . 3 𝐴 ∈ V
11 ceqsex2.5 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
1211anbi2d 740 . . . 4 (𝑥 = 𝐴 → ((𝑦 = 𝐵𝜑) ↔ (𝑦 = 𝐵𝜓)))
1312exbidv 1890 . . 3 (𝑥 = 𝐴 → (∃𝑦(𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑦 = 𝐵𝜓)))
149, 10, 13ceqsex 3272 . 2 (∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)) ↔ ∃𝑦(𝑦 = 𝐵𝜓))
15 ceqsex2.2 . . 3 𝑦𝜒
16 ceqsex2.4 . . 3 𝐵 ∈ V
17 ceqsex2.6 . . 3 (𝑦 = 𝐵 → (𝜓𝜒))
1815, 16, 17ceqsex 3272 . 2 (∃𝑦(𝑦 = 𝐵𝜓) ↔ 𝜒)
195, 14, 183bitri 286 1 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ 𝜒)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 383   ∧ w3a 1054   = wceq 1523  ∃wex 1744  Ⅎwnf 1748   ∈ wcel 2030  Vcvv 3231 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-ext 2631 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1056  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-v 3233 This theorem is referenced by:  ceqsex2v  3276
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