Theorem xp11 6100

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 6084	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2		anidm 565	. . . . . 6 ⊢ (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3		neeq1 3004	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅))
4	3	anbi2d 629	. . . . . 6 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
5	2, 4	bitr3id 284	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
6		eqimss 3987	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7		ssxpb 6099	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
8	6, 7	syl5ibcom 244	. . . . . . 7 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
9		eqimss2 3988	. . . . . . . 8 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10		ssxpb 6099	. . . . . . . 8 ⊢ ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))
11	9, 10	syl5ibcom 244	. . . . . . 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 653	. . . . . . 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 627	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)))
17	13, 16	bitr4i 277	. . . . . 6 ⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
18	12, 17	syl6ib 250	. . . . 5 ⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
19	5, 18	sylbid 239	. . . 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 216	. 2 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
22		xpeq12 5632	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
23	21, 22	impbid1 224	1 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   = wceq 1540   ≠ wne 2941   ⊆ wss 3897  ∅c0 4267   × cxp 5605

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2153  ax-12 2170  ax-ext 2708  ax-sep 5238  ax-nul 5245  ax-pr 5367

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-nf 1785  df-sb 2067  df-mo 2539  df-eu 2568  df-clab 2715  df-cleq 2729  df-clel 2815  df-nfc 2887  df-ne 2942  df-ral 3063  df-rex 3072  df-rab 3405  df-v 3443  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4268  df-if 4472  df-sn 4572  df-pr 4574  df-op 4578  df-br 5088  df-opab 5150  df-xp 5613  df-rel 5614  df-cnv 5615  df-dm 5617  df-rn 5618

This theorem is referenced by:  xpcan  6101  xpcan2  6102  fseqdom  9855  axcc2lem  10265  lmodfopnelem1  20231  xppss12  40407