Theorem altopelaltxp 34278

Description: Alternate ordered pair membership in a Cartesian product. Note that, unlike opelxp 5625, there is no sethood requirement here. (Contributed by Scott Fenton, 22-Mar-2012.)

Assertion

Ref	Expression
altopelaltxp	⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))

Proof of Theorem altopelaltxp

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		elaltxp 34277	. 2 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫)
2		reeanv 3294	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
3		eqcom 2745	. . . . 5 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫)
4		vex 3436	. . . . . 6 ⊢ 𝑥 ∈ V
5		vex 3436	. . . . . 6 ⊢ 𝑦 ∈ V
6	4, 5	altopth 34271	. . . . 5 ⊢ (⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
7	3, 6	bitri 274	. . . 4 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
8	7	2rexbii 3182	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
9		risset 3194	. . . 4 ⊢ (𝑋 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 𝑥 = 𝑋)
10		risset 3194	. . . 4 ⊢ (𝑌 ∈ 𝐵 ↔ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌)
11	9, 10	anbi12i 627	. . 3 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
12	2, 8, 11	3bitr4i 303	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))
13	1, 12	bitri 274	1 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))

Syntax hints:   ↔ wb 205   ∧ wa 396   = wceq 1539   ∈ wcel 2106  ∃wrex 3065  ⟪caltop 34258   ×× caltxp 34259

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-sn 4562  df-pr 4564  df-altop 34260  df-altxp 34261

This theorem is referenced by: (None)