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Theorem xp11m 4786
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 4772 . . 3 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
2 anidm 382 . . . . . 6 ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
3 eleq2 2117 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝑧 ∈ (𝐴 × 𝐵) ↔ 𝑧 ∈ (𝐶 × 𝐷)))
43exbidv 1722 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)))
54anbi2d 445 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵)) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
62, 5syl5bbr 187 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ↔ (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷))))
7 eqimss 3024 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷))
8 ssxpbm 4783 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
97, 8syl5ibcom 148 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐴 × 𝐵) → (𝐴𝐶𝐵𝐷)))
10 eqimss2 3025 . . . . . . . 8 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵))
11 ssxpbm 4783 . . . . . . . 8 (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶𝐴𝐷𝐵)))
1210, 11syl5ibcom 148 . . . . . . 7 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (∃𝑧 𝑧 ∈ (𝐶 × 𝐷) → (𝐶𝐴𝐷𝐵)))
139, 12anim12d 322 . . . . . 6 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → ((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵))))
14 an4 528 . . . . . . 7 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
15 eqss 2987 . . . . . . . 8 (𝐴 = 𝐶 ↔ (𝐴𝐶𝐶𝐴))
16 eqss 2987 . . . . . . . 8 (𝐵 = 𝐷 ↔ (𝐵𝐷𝐷𝐵))
1715, 16anbi12i 441 . . . . . . 7 ((𝐴 = 𝐶𝐵 = 𝐷) ↔ ((𝐴𝐶𝐶𝐴) ∧ (𝐵𝐷𝐷𝐵)))
1814, 17bitr4i 180 . . . . . 6 (((𝐴𝐶𝐵𝐷) ∧ (𝐶𝐴𝐷𝐵)) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
1913, 18syl6ib 154 . . . . 5 ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((∃𝑧 𝑧 ∈ (𝐴 × 𝐵) ∧ ∃𝑧 𝑧 ∈ (𝐶 × 𝐷)) → (𝐴 = 𝐶𝐵 = 𝐷)))
206, 19sylbid 143 . . . 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 118 . 2 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
23 xpeq12 4391 . 2 ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷))
2422, 23impbid1 134 1 ((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102   = wceq 1259  wex 1397  wcel 1409  wss 2944   × cxp 4370
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-14 1421  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038  ax-sep 3902  ax-pow 3954  ax-pr 3971
This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-eu 1919  df-mo 1920  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-rex 2329  df-v 2576  df-un 2949  df-in 2951  df-ss 2958  df-pw 3388  df-sn 3408  df-pr 3409  df-op 3411  df-br 3792  df-opab 3846  df-xp 4378  df-rel 4379  df-cnv 4380  df-dm 4382  df-rn 4383
This theorem is referenced by: (None)
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