Theorem elxpi 4800

Description: Membership in a cross product. Uses fewer axioms than elxp 4801. (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 2359	. . . . . 6 ⊢ (z = A → (z = ⟨x, y⟩ ↔ A = ⟨x, y⟩))
2	1	anbi1d 685	. . . . 5 ⊢ (z = A → ((z = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C)) ↔ (A = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C))))
3	2	2exbidv 1628	. . . 4 ⊢ (z = A → (∃x∃y(z = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C)) ↔ ∃x∃y(A = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C))))
4	3	elabg 2986	. . 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 232	. 2 ⊢ (A ∈ {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C))} → ∃x∃y(A = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C)))
6		df-xp 4784	. . 3 ⊢ (B × C) = {⟨x, y⟩ ∣ (x ∈ B ∧ y ∈ C)}
7		df-opab 4623	. . 3 ⊢ {⟨x, y⟩ ∣ (x ∈ B ∧ y ∈ C)} = {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C))}
8	6, 7	eqtri 2373	. 2 ⊢ (B × C) = {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C))}
9	5, 8	eleq2s 2445	1 ⊢ (A ∈ (B × C) → ∃x∃y(A = ⟨x, y⟩ ∧ (x ∈ B ∧ y ∈ C)))

Syntax hints:   → wi 4   ∧ wa 358  ∃wex 1541   = wceq 1642   ∈ wcel 1710  {cab 2339  ⟨cop 4561  {copab 4622   × cxp 4770

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-opab 4623  df-xp 4784

This theorem is referenced by: (None)