Theorem altopelaltxp 36711

Description: Alternate ordered pair membership in a Cartesian product. Note that, unlike opelxp 5687, 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 36710	. 2 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫)
2		reeanv 3235	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
3		eqcom 2768	. . . . 5 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫)
4		vex 3455	. . . . . 6 ⊢ 𝑥 ∈ V
5		vex 3455	. . . . . 6 ⊢ 𝑦 ∈ V
6	4, 5	altopth 36704	. . . . 5 ⊢ (⟪𝑥, 𝑦⟫ = ⟪𝑋, 𝑌⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
7	3, 6	bitri 278	. . . 4 ⊢ (⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
8	7	2rexbii 3139	. . 3 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝑥 = 𝑋 ∧ 𝑦 = 𝑌))
9		risset 3238	. . . 4 ⊢ (𝑋 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 𝑥 = 𝑋)
10		risset 3238	. . . 4 ⊢ (𝑌 ∈ 𝐵 ↔ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌)
11	9, 10	anbi12i 640	. . 3 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ↔ (∃𝑥 ∈ 𝐴 𝑥 = 𝑋 ∧ ∃𝑦 ∈ 𝐵 𝑦 = 𝑌))
12	2, 8, 11	3bitr4i 306	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 ⟪𝑋, 𝑌⟫ = ⟪𝑥, 𝑦⟫ ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))
13	1, 12	bitri 278	1 ⊢ (⟪𝑋, 𝑌⟫ ∈ (𝐴 ×× 𝐵) ↔ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∃wrex 3087  ⟪caltop 36691   ×× caltxp 36692

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733  ax-sep 5249  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-v 3453  df-un 3904  df-ss 3916  df-sn 4585  df-pr 4587  df-altop 36693  df-altxp 36694

This theorem is used by: (None)