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Theorem xp11 6010
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 5994 . . 3 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
2 anidm 568 . . . . . 6 (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
3 neeq1 3073 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅))
43anbi2d 631 . . . . . 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 3998 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
7 ssxpb 6009 . . . . . . . 8 ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
86, 7syl5ibcom 248 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴𝐶𝐵𝐷)))
9 eqimss2 3999 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
10 ssxpb 6009 . . . . . . . 8 ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
119, 10syl5ibcom 248 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐶 × 𝐷) ≠ ∅ → (𝐶𝐴𝐷𝐵)))
128, 11anim12d 611 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
13 an4 655 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
14 eqss 3957 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
15 eqss 3957 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1614, 15anbi12i 629 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1713, 16bitr4i 281 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1812, 17syl6ib 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 32 . . 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 5557 . 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 399   = wceq 1538  wne 3011  wss 3908  c0 4265   × cxp 5530
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2178  ax-ext 2794  ax-sep 5179  ax-nul 5186  ax-pr 5307
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-mo 2622  df-eu 2653  df-clab 2801  df-cleq 2815  df-clel 2894  df-nfc 2962  df-ne 3012  df-ral 3135  df-rex 3136  df-v 3471  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4266  df-if 4440  df-sn 4540  df-pr 4542  df-op 4546  df-br 5043  df-opab 5105  df-xp 5538  df-rel 5539  df-cnv 5540  df-dm 5542  df-rn 5543
This theorem is referenced by:  xpcan  6011  xpcan2  6012  fseqdom  9441  axcc2lem  9847  lmodfopnelem1  19661  xppss12  39356
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