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Theorem xp11 6173
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 = 𝐶𝐵 = 𝐷)))

Proof of Theorem xp11
StepHypRef Expression
1 xpnz 6156 . . 3 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2 anidm 574 . . . . . 6 (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3 neeq1 3020 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅))
43anbi2d 641 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
52, 4bitr3id 288 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅)))
6 eqimss 3995 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7 ssxpb 6172 . . . . . . . 8 ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
86, 7syl5ibcom 248 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴𝐶𝐵𝐷)))
9 eqimss2 3996 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10 ssxpb 6172 . . . . . . . 8 ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
119, 10syl5ibcom 248 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐶 × 𝐷) ≠ ∅ → (𝐶𝐴𝐷𝐵)))
128, 11anim12d 620 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
13 an4 668 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
14 eqss 3952 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
15 eqss 3952 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1614, 15anbi12i 639 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1713, 16bitr4i 281 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1812, 17imbitrdi 254 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → (𝐴 = 𝐶𝐵 = 𝐷)))
195, 18sylbid 243 . . . 4 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 = 𝐶𝐵 = 𝐷)))
2019com12 33 . . 3 ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
211, 20sylbi 220 . 2 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
22 xpeq12 5686 . 2 ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
2321, 22impbid1 228 1 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400   = wceq 1570  wne 2958  wss 3905  c0 4286   × cxp 5659
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-11 2192  ax-ext 2735  ax-sep 5257  ax-pr 5404
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110  df-opab 5174  df-xp 5667  df-rel 5668  df-cnv 5669  df-dm 5671  df-rn 5672
This theorem is referenced by:  xpcan  6174  xpcan2  6175  fseqdom  10006  axcc2lem  10415  lmodfopnelem1  21019  xppss12  43000
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