Theorem elxpi 5602

Description: Membership in a Cartesian product. Uses fewer axioms than elxp 5603. (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 2742	. . . . 5 ⊢ (𝑧 = 𝐴 → (𝑧 = ⟨𝑥, 𝑦⟩ ↔ 𝐴 = ⟨𝑥, 𝑦⟩))
2	1	anbi1d 629	. . . 4 ⊢ (𝑧 = 𝐴 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ (𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
3	2	2exbidv 1928	. . 3 ⊢ (𝑧 = 𝐴 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
4		df-xp 5586	. . . 4 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
5		df-opab 5133	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
6	4, 5	eqtri 2766	. . 3 ⊢ (𝐵 × 𝐶) = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))}
7	3, 6	elab2g 3604	. 2 ⊢ (𝐴 ∈ (𝐵 × 𝐶) → (𝐴 ∈ (𝐵 × 𝐶) ↔ ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))))
8	7	ibi 266	1 ⊢ (𝐴 ∈ (𝐵 × 𝐶) → ∃𝑥∃𝑦(𝐴 = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1539  ∃wex 1783   ∈ wcel 2108  {cab 2715  ⟨cop 4564  {copab 5132   × cxp 5578

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-opab 5133  df-xp 5586

This theorem is referenced by:  xpdifid  6060  opreuopreu  7849  djuunxp  9610  fmla0xp  33245  mnringmulrcld  41735  rrx2plordisom  45957