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Theorem ceqsex2 2611
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 900 . . . . 5 ((𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
21exbii 1512 . . . 4 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
3 19.42v 1802 . . . 4 (∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
42, 3bitri 177 . . 3 (∃𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ (𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
54exbii 1512 . 2 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
6 nfv 1437 . . . . 5 𝑥 𝑦 = 𝐵
7 ceqsex2.1 . . . . 5 𝑥𝜓
86, 7nfan 1473 . . . 4 𝑥(𝑦 = 𝐵𝜓)
98nfex 1544 . . 3 𝑥𝑦(𝑦 = 𝐵𝜓)
10 ceqsex2.3 . . 3 𝐴 ∈ V
11 ceqsex2.5 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
1211anbi2d 445 . . . 4 (𝑥 = 𝐴 → ((𝑦 = 𝐵𝜑) ↔ (𝑦 = 𝐵𝜓)))
1312exbidv 1722 . . 3 (𝑥 = 𝐴 → (∃𝑦(𝑦 = 𝐵𝜑) ↔ ∃𝑦(𝑦 = 𝐵𝜓)))
149, 10, 13ceqsex 2609 . 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 2609 . 2 (∃𝑦(𝑦 = 𝐵𝜓) ↔ 𝜒)
195, 14, 183bitri 199 1 (∃𝑥𝑦(𝑥 = 𝐴𝑦 = 𝐵𝜑) ↔ 𝜒)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  w3a 896   = wceq 1259  wnf 1365  wex 1397  wcel 1409  Vcvv 2574
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-3an 898  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-v 2576
This theorem is referenced by:  ceqsex2v  2612
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