Theorem altopelaltxp 36183

Description: Alternate ordered pair membership in a Cartesian product. Note that, unlike opelxp 5661, 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 36182	. 2 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫)
2		reeanv 3209	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
3		eqcom 2744	. . . . 5 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫)
4		vex 3445	. . . . . 6 ⊢ 𝑥 ∈ V
5		vex 3445	. . . . . 6 ⊢ 𝑦 ∈ V
6	4, 5	altopth 36176	. . . . 5 ⊢ (⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
7	3, 6	bitri 275	. . . 4 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
8	7	2rexbii 3113	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
9		risset 3212	. . . 4 ⊢ (𝑋 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 𝑥 = 𝑋)
10		risset 3212	. . . 4 ⊢ (𝑌 ∈ 𝐵 ↔ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌)
11	9, 10	anbi12i 629	. . 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 275	1 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1542   ∈ wcel 2114  ∃wrex 3061  ⟪caltop 36163   ×× caltxp 36164

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5242  ax-nul 5252  ax-pr 5378

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3062  df-v 3443  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4287  df-sn 4582  df-pr 4584  df-altop 36165  df-altxp 36166

This theorem is referenced by: (None)