Theorem xp11 6197

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 6181	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2		anidm 564	. . . . . 6 ⊢ (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3		neeq1 3001	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅))
4	3	anbi2d 630	. . . . . 6 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
5	2, 4	bitr3id 285	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
6		eqimss 4054	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7		ssxpb 6196	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
8	6, 7	syl5ibcom 245	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
9		eqimss2 4055	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10		ssxpb 6196	. . . . . . . 8 ⊢ ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))
11	9, 10	syl5ibcom 245	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐶 × 𝐷) ≠ ∅ → (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))
12	8, 11	anim12d 609	. . . . . 6 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵))))
13		an4 656	. . . . . . 7 ⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)))
14		eqss 4011	. . . . . . . 8 ⊢ (𝐴 = 𝐶 ↔ (𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴))
15		eqss 4011	. . . . . . . 8 ⊢ (𝐵 = 𝐷 ↔ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵))
16	14, 15	anbi12i 628	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)))
17	13, 16	bitr4i 278	. . . . . 6 ⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
18	12, 17	imbitrdi 251	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
19	5, 18	sylbid 240	. . . 4 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
20	19	com12 32	. . 3 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
21	1, 20	sylbi 217	. 2 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
22		xpeq12 5714	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
23	21, 22	impbid1 225	1 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537   ≠ wne 2938   ⊆ wss 3963  ∅c0 4339   × cxp 5687

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-11 2155  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-br 5149  df-opab 5211  df-xp 5695  df-rel 5696  df-cnv 5697  df-dm 5699  df-rn 5700

This theorem is referenced by:  xpcan  6198  xpcan2  6199  fseqdom  10064  axcc2lem  10474  lmodfopnelem1  20913  xppss12  42247