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Theorem xp11m 4779
Description: The cross product of inhabited classes is one-to-one. (Contributed by Jim Kingdon, 13-Dec-2018.)
Assertion
Ref Expression
xp11m ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Distinct variable groups:   𝑥,𝐴   𝑦,𝐵
Allowed substitution hints:   𝐴(𝑦)   𝐵(𝑥)   𝐶(𝑥,𝑦)   𝐷(𝑥,𝑦)

Proof of Theorem xp11m
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 xpm 4765 . . 3 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
2 anidm 388 . . . . . 6 ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
3 eleq2 2142 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝑧 ∈ (𝐴 × 𝐵) ↔ 𝑧 ∈ (𝐶 × 𝐷)))
43exbidv 1746 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)))
54anbi2d 451 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
62, 5syl5bbr 192 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
7 eqimss 3051 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
8 ssxpbm 4776 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
97, 8syl5ibcom 153 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → (𝐴𝐶𝐵𝐷)))
10 eqimss2 3052 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
11 ssxpbm 4776 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
1210, 11syl5ibcom 153 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → (𝐶𝐴𝐷𝐵)))
139, 12anim12d 328 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
14 an4 550 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
15 eqss 3014 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
16 eqss 3014 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1715, 16anbi12i 447 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1814, 17bitr4i 185 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1913, 18syl6ib 159 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → (𝐴 = 𝐶𝐵 = 𝐷)))
206, 19sylbid 148 . . . 4 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → (𝐴 = 𝐶𝐵 = 𝐷)))
2120com12 30 . . 3 (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
221, 21sylbi 119 . 2 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
23 xpeq12 4382 . 2 ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
2422, 23impbid1 140 1 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103   = wceq 1284  wex 1421  wcel 1433  wss 2973   × cxp 4361
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-14 1445  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-sep 3896  ax-pow 3948  ax-pr 3964
This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-eu 1944  df-mo 1945  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-v 2603  df-un 2977  df-in 2979  df-ss 2986  df-pw 3384  df-sn 3404  df-pr 3405  df-op 3407  df-br 3786  df-opab 3840  df-xp 4369  df-rel 4370  df-cnv 4371  df-dm 4373  df-rn 4374
This theorem is referenced by: (None)
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