Theorem altopelaltxp 36274

Description: Alternate ordered pair membership in a Cartesian product. Note that, unlike opelxp 5676, 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 36273	. 2 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫)
2		reeanv 3228	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
3		eqcom 2763	. . . . 5 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫)
4		vex 3452	. . . . . 6 ⊢ 𝑥 ∈ V
5		vex 3452	. . . . . 6 ⊢ 𝑦 ∈ V
6	4, 5	altopth 36267	. . . . 5 ⊢ (⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
7	3, 6	bitri 277	. . . 4 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
8	7	2rexbii 3132	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
9		risset 3231	. . . 4 ⊢ (𝑋 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 𝑥 = 𝑋)
10		risset 3231	. . . 4 ⊢ (𝑌 ∈ 𝐵 ↔ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌)
11	9, 10	anbi12i 636	. . 3 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
12	2, 8, 11	3bitr4i 305	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))
13	1, 12	bitri 277	1 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))

Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1554   ∈ wcel 2136  ∃wrex 3080  ⟪caltop 36254   ×× caltxp 36255

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-8 2138  ax-9 2146  ax-ext 2728  ax-sep 5240  ax-pr 5384

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-tru 1557  df-ex 1794  df-sb 2085  df-clab 2735  df-cleq 2748  df-clel 2831  df-ral 3071  df-rex 3081  df-v 3450  df-un 3904  df-ss 3916  df-sn 4577  df-pr 4579  df-altop 36256  df-altxp 36257

This theorem is referenced by: (None)