Theorem altopelaltxp 34274

Description: Alternate ordered pair membership in a Cartesian product. Note that, unlike opelxp 5626, 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 34273	. 2 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫)
2		reeanv 3295	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
3		eqcom 2747	. . . . 5 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫)
4		vex 3435	. . . . . 6 ⊢ 𝑥 ∈ V
5		vex 3435	. . . . . 6 ⊢ 𝑦 ∈ V
6	4, 5	altopth 34267	. . . . 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 3181	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
9		risset 3196	. . . 4 ⊢ (𝑋 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 𝑥 = 𝑋)
10		risset 3196	. . . 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 1542   ∈ wcel 2110  ∃wrex 3067  ⟪caltop 34254   ×× caltxp 34255

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-ext 2711  ax-sep 5227  ax-nul 5234  ax-pr 5356

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1545  df-fal 1555  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-ral 3071  df-rex 3072  df-v 3433  df-dif 3895  df-un 3897  df-in 3899  df-ss 3909  df-nul 4263  df-sn 4568  df-pr 4570  df-altop 34256  df-altxp 34257

This theorem is referenced by: (None)