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Theorem copsex2gd 38027
Description: Implicit substitution inference for ordered pairs. (Contributed by NM, 28-May-1995.) Use a similar proof to copsex4g 5467 to reduce axiom usage. (Revised by SN, 1-Sep-2024.) Adapt copsex2g 5465 $p to deduction form. (Revised by BJ, 28-Mar-2026.) Do not use copsex2g 5465. (Proof modification is discouraged.)
Hypothesis
Ref Expression
copsex2gd.is ((𝜑 ∧ (𝑥 = 𝐴 ∧ 𝑦 = 𝐵)) → (𝜓 ↔ 𝜒))
Assertion
Ref Expression
copsex2gd ((𝜑 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊)) → (∃𝑥∃𝑦(⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ 𝜒))
Distinct variable groups:   𝑥,𝑦,𝜑   𝜒,𝑥,𝑦   𝑥,𝐴,𝑦   𝑥,𝐵,𝑦
Allowed substitution hints:   𝜓(𝑥, 𝑦)   𝑉(𝑥, 𝑦)   𝑊(𝑥, 𝑦)

Proof of Theorem copsex2gd
StepHypRef Expression
1 eqcom 2768 . . . . 5 (⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ↔ ⟨𝑥, 𝑦⟩ = ⟨𝐴, 𝐵⟩)
2 vex 3455 . . . . . 6 𝑥 ∈ V
3 vex 3455 . . . . . 6 𝑦 ∈ V
42, 3opth 5445 . . . . 5 (⟨𝑥, 𝑦⟩ = ⟨𝐴, 𝐵⟩ ↔ (𝑥 = 𝐴 ∧ 𝑦 = 𝐵))
51, 4bitri 278 . . . 4 (⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ↔ (𝑥 = 𝐴 ∧ 𝑦 = 𝐵))
65anbi1i 636 . . 3 ((⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵) ∧ 𝜓))
762exbii 1882 . 2 (∃𝑥∃𝑦(⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑥∃𝑦((𝑥 = 𝐴 ∧ 𝑦 = 𝐵) ∧ 𝜓))
8 simpr 490 . . 3 ((𝜑 ∧ (𝑥 = 𝐴 ∧ 𝑦 = 𝐵)) → (𝑥 = 𝐴 ∧ 𝑦 = 𝐵))
9 copsex2gd.is . . 3 ((𝜑 ∧ (𝑥 = 𝐴 ∧ 𝑦 = 𝐵)) → (𝜓 ↔ 𝜒))
108, 9cgsex2gd 38026 . 2 ((𝜑 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊)) → (∃𝑥∃𝑦((𝑥 = 𝐴 ∧ 𝑦 = 𝐵) ∧ 𝜓) ↔ 𝜒))
117, 10bitrid 286 1 ((𝜑 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊)) → (∃𝑥∃𝑦(⟨𝐴, 𝐵⟩ = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ 𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2145  ⟨cop 4590
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733  ax-sep 5249  ax-pr 5391
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591
This theorem is used by: (None)
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