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Theorem xp11m 4903
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 4886 . . 3 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
2 anidm 389 . . . . . 6 ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
3 eleq2 2158 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝑧 ∈ (𝐴 × 𝐵) ↔ 𝑧 ∈ (𝐶 × 𝐷)))
43exbidv 1760 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)))
54anbi2d 453 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
62, 5syl5bbr 193 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
7 eqimss 3093 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
8 ssxpbm 4900 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
97, 8syl5ibcom 154 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → (𝐴𝐶𝐵𝐷)))
10 eqimss2 3094 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
11 ssxpbm 4900 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
1210, 11syl5ibcom 154 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → (𝐶𝐴𝐷𝐵)))
139, 12anim12d 329 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
14 an4 554 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
15 eqss 3054 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
16 eqss 3054 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1715, 16anbi12i 449 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1814, 17bitr4i 186 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1913, 18syl6ib 160 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → (𝐴 = 𝐶𝐵 = 𝐷)))
206, 19sylbid 149 . . . 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 120 . 2 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
23 xpeq12 4486 . 2 ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
2422, 23impbid1 141 1 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104   = wceq 1296  wex 1433  wcel 1445  wss 3013   × cxp 4465
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-14 1457  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077  ax-sep 3978  ax-pow 4030  ax-pr 4060
This theorem depends on definitions:  df-bi 116  df-3an 929  df-tru 1299  df-nf 1402  df-sb 1700  df-eu 1958  df-mo 1959  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-ral 2375  df-rex 2376  df-v 2635  df-un 3017  df-in 3019  df-ss 3026  df-pw 3451  df-sn 3472  df-pr 3473  df-op 3475  df-br 3868  df-opab 3922  df-xp 4473  df-rel 4474  df-cnv 4475  df-dm 4477  df-rn 4478
This theorem is referenced by: (None)
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