Theorem elxpi 5430

Description: Membership in a Cartesian product. Uses fewer axioms than elxp 5431. (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 2782	. . . . . 6 ⊢ (𝑧 = 𝐴 → (𝑧 = ⟨𝑥, 𝑦⟩ ↔ 𝐴 = ⟨𝑥, 𝑦⟩))
2	1	anbi1d 620	. . . . 5 ⊢ (𝑧 = 𝐴 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ (𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
3	2	2exbidv 1883	. . . 4 ⊢ (𝑧 = 𝐴 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
4	3	elabg 3580	. . 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 259	. 2 ⊢ (𝐴 ∈ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))} → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
6		df-xp 5414	. . 3 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
7		df-opab 4993	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
8	6, 7	eqtri 2802	. 2 ⊢ (𝐵 × 𝐶) = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
9	5, 8	eleq2s 2884	1 ⊢ (𝐴 ∈ (𝐵 × 𝐶) → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))

Syntax hints:   → wi 4   ∧ wa 387   = wceq 1507  ∃wex 1742   ∈ wcel 2050  {cab 2758  ⟨cop 4448  {copab 4992   × cxp 5406

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-8 2052  ax-9 2059  ax-10 2079  ax-11 2093  ax-12 2106  ax-ext 2750

This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-tru 1510  df-ex 1743  df-nf 1747  df-sb 2016  df-clab 2759  df-cleq 2771  df-clel 2846  df-nfc 2918  df-opab 4993  df-xp 5414

This theorem is referenced by:  xpdifid  5867  opreuopreu  7548  djuunxp  9146  fmla0xp  32193  rrx2plordisom  44079