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Theorem xp11m 5226
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 5209 . . 3 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
2 anidm 400 . . . . . 6 ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
3 eleq2 2302 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝑧 ∈ (𝐴 × 𝐵) ↔ 𝑧 ∈ (𝐶 × 𝐷)))
43exbidv 1878 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)))
54anbi2d 468 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
62, 5bitr3id 194 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
7 eqimss 3302 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
8 ssxpbm 5223 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
97, 8syl5ibcom 155 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → (𝐴𝐶𝐵𝐷)))
10 eqimss2 3303 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
11 ssxpbm 5223 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
1210, 11syl5ibcom 155 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → (𝐶𝐴𝐷𝐵)))
139, 12anim12d 335 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
14 an4 592 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
15 eqss 3263 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
16 eqss 3263 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1715, 16anbi12i 464 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1814, 17bitr4i 187 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1913, 18imbitrdi 161 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → (𝐴 = 𝐶𝐵 = 𝐷)))
206, 19sylbid 150 . . . 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 121 . 2 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
23 xpeq12 4793 . 2 ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
2422, 23impbid1 142 1 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables:    wff set class
This proof depends on syntax axioms:  wi 4  wa 104  wb 105   = wceq 1402  wex 1545  wcel 2209  wss 3220   × cxp 4772
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-pow 4311  ax-pr 4346
This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-br 4131  df-opab 4193  df-xp 4780  df-rel 4781  df-cnv 4782  df-dm 4784  df-rn 4785
This theorem is used by:  cc2lem  7632  lmodfopnelem1  14661
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