Theorem xp11 6148

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 6132	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2		anidm 564	. . . . . 6 ⊢ (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3		neeq1 2987	. . . . . . 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 4005	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7		ssxpb 6147	. . . . . . . 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 4006	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10		ssxpb 6147	. . . . . . . 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 3962	. . . . . . . 8 ⊢ (𝐴 = 𝐶 ↔ (𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴))
15		eqss 3962	. . . . . . . 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 5663	. 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 1540   ≠ wne 2925   ⊆ wss 3914  ∅c0 4296   × cxp 5636

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-11 2158  ax-12 2178  ax-ext 2701  ax-sep 5251  ax-nul 5261  ax-pr 5387

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3406  df-v 3449  df-dif 3917  df-un 3919  df-ss 3931  df-nul 4297  df-if 4489  df-sn 4590  df-pr 4592  df-op 4596  df-br 5108  df-opab 5170  df-xp 5644  df-rel 5645  df-cnv 5646  df-dm 5648  df-rn 5649

This theorem is referenced by:  xpcan  6149  xpcan2  6150  fseqdom  9979  axcc2lem  10389  lmodfopnelem1  20804  xppss12  42217