Theorem elxpi 5164

Description: Membership in a Cartesian product. Uses fewer axioms than elxp 5165. (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.

Step	Hyp	Ref	Expression
1		eqeq1 2655	. . . . . 6 ⊢ (𝑧 = 𝐴 → (𝑧 = ⟨𝑥, 𝑦⟩ ↔ 𝐴 = ⟨𝑥, 𝑦⟩))
2	1	anbi1d 741	. . . . 5 ⊢ (𝑧 = 𝐴 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ (𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
3	2	2exbidv 1892	. . . 4 ⊢ (𝑧 = 𝐴 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
4	3	elabg 3383	. . 3 ⊢ (𝐴 ∈ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))} → (𝐴 ∈ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))} ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
5	4	ibi 256	. 2 ⊢ (𝐴 ∈ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))} → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
6		df-xp 5149	. . 3 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
7		df-opab 4746	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
8	6, 7	eqtri 2673	. 2 ⊢ (𝐵 × 𝐶) = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
9	5, 8	eleq2s 2748	1 ⊢ (𝐴 ∈ (𝐵 × 𝐶) → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))

Syntax hints:   → wi 4   ∧ wa 383   = wceq 1523  ∃wex 1744   ∈ wcel 2030  {cab 2637  ⟨cop 4216  {copab 4745   × cxp 5141

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-13 2282  ax-ext 2631

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-v 3233  df-opab 4746  df-xp 5149

This theorem is referenced by:  xpdifid  5597