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Theorem elxpi 5680
Description: Membership in a Cartesian product. Uses fewer axioms than elxp 5681. (Contributed by NM, 4-Jul-1994.)
Assertion
Ref Expression
elxpi (𝐴 ∈ (𝐵 × 𝐶) → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦   𝑥,𝐶,𝑦

Proof of Theorem elxpi
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 eqeq1 2767 . . . . 5 (𝑧 = 𝐴 → (𝑧 = ⟨𝑥, 𝑦⟩ ↔ 𝐴 = ⟨𝑥, 𝑦⟩))
21anbi1d 643 . . . 4 (𝑧 = 𝐴 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ (𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
322exbidv 1959 . . 3 (𝑧 = 𝐴 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
4 df-xp 5664 . . . 4 (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
5 df-opab 5174 . . . 4 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
64, 5eqtri 2786 . . 3 (𝐵 × 𝐶) = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
73, 6elab2g 3637 . 2 (𝐴 ∈ (𝐵 × 𝐶) → (𝐴 ∈ (𝐵 × 𝐶) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
87ibi 270 1 (𝐴 ∈ (𝐵 × 𝐶) → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741  ⟨cop 4840  {copab 5173   × cxp 5656
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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-opab 5174  df-xp 5664
This theorem is used by:  optocl  5753  0xp  5758  xp0  5759  ssrel  5767  xpdifid  6177  xpdifcnvepel  6178  opreuopreu  8080  djuunxp  10039  rngqiprngimfo  21648  fsumdvdsmul  27579  mpofvd  35838  fmla0xp  36356  bj-ssrel  38175  mnringmulrcld  45487  rrx2plordisom  50167
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