Theorem xp11 6167

Description: The Cartesian product of nonempty classes is a one-to-one "function" of its two "arguments". In other words, two Cartesian products, at least one with nonempty factors, are equal if and only if their respective factors are equal. (Contributed by NM, 31-May-2008.)

Assertion

Ref	Expression
xp11	⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Proof of Theorem xp11

Step	Hyp	Ref	Expression
1		xpnz 6150	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2		anidm 575	. . . . . 6 ⊢ (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3		neeq1 3018	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅))
4	3	anbi2d 642	. . . . . 6 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
5	2, 4	bitr3id 288	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
6		eqimss 3989	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7		ssxpb 6166	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
8	6, 7	syl5ibcom 248	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
9		eqimss2 3990	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10		ssxpb 6166	. . . . . . . 8 ⊢ ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))
11	9, 10	syl5ibcom 248	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐶 × 𝐷) ≠ ∅ → (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))
12	8, 11	anim12d 621	. . . . . 6 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵))))
13		an4 669	. . . . . . 7 ⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)))
14		eqss 3946	. . . . . . . 8 ⊢ (𝐴 = 𝐶 ↔ (𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴))
15		eqss 3946	. . . . . . . 8 ⊢ (𝐵 = 𝐷 ↔ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵))
16	14, 15	anbi12i 640	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)))
17	13, 16	bitr4i 281	. . . . . 6 ⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
18	12, 17	imbitrdi 254	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
19	5, 18	sylbid 243	. . . 4 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
20	19	com12 33	. . 3 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
21	1, 20	sylbi 220	. 2 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
22		xpeq12 5676	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
23	21, 22	impbid1 228	1 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ≠ wne 2956   ⊆ wss 3899  ∅c0 4279   × cxp 5649

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-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-xp 5657  df-rel 5658  df-cnv 5659  df-dm 5661  df-rn 5662

This theorem is used by:  xpcan  6168  xpcan2  6169  fseqdom  10098  axcc2lem  10507  degenmgmnfn  19129  lmodfopnelem1  21166  xppss12  43263