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Theorem ceqsex2 3535
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 1094 . . . . 5 ((𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
21exbii 1845 . . . 4 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
3 19.42v 1951 . . . 4 (∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
42, 3bitri 275 . . 3 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
54exbii 1845 . 2 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
6 nfv 1912 . . . . 5 𝑥 𝑦 = 𝐵
7 ceqsex2.1 . . . . 5 𝑥𝜓
86, 7nfan 1897 . . . 4 𝑥(𝑦 = 𝐵𝜓)
98nfex 2323 . . 3 𝑥𝑦(𝑦 = 𝐵𝜓)
10 ceqsex2.3 . . 3 𝐴 ∈ V
11 ceqsex2.5 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
1211anbi2d 630 . . . 4 (𝑥 = 𝐴 → ((𝑦 = 𝐵𝜑) ↔ (𝑦 = 𝐵𝜓)))
1312exbidv 1919 . . 3 (𝑥 = 𝐴 → (∃𝑦(𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑦 = 𝐵𝜓)))
149, 10, 13ceqsex 3528 . 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 3528 . 2 (∃𝑦(𝑦 = 𝐵𝜓) ↔ 𝜒)
195, 14, 183bitri 297 1 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ 𝜒)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  w3a 1086   = wceq 1537  wex 1776  wnf 1780  wcel 2106  Vcvv 3478
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-10 2139  ax-11 2155  ax-12 2175
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-ex 1777  df-nf 1781  df-clel 2814
This theorem is referenced by: (None)
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