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Theorem onsucsssucr 4288
Description: The subclass relationship between two ordinals is inherited by their predecessors. The converse implies excluded middle, as shown at onsucsssucexmid 4305. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2019.)
Assertion
Ref Expression
onsucsssucr ((𝐴 ∈ On ∧ Ord 𝐵) → (suc 𝐴 ⊆ suc 𝐵𝐴𝐵))

Proof of Theorem onsucsssucr
StepHypRef Expression
1 ordsucim 4279 . . 3 (Ord 𝐵 → Ord suc 𝐵)
2 ordelsuc 4284 . . 3 ((𝐴 ∈ On ∧ Ord suc 𝐵) → (𝐴 ∈ suc 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))
31, 2sylan2 280 . 2 ((𝐴 ∈ On ∧ Ord 𝐵) → (𝐴 ∈ suc 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))
4 ordtr 4168 . . . 4 (Ord 𝐵 → Tr 𝐵)
5 trsucss 4213 . . . 4 (Tr 𝐵 → (𝐴 ∈ suc 𝐵𝐴𝐵))
64, 5syl 14 . . 3 (Ord 𝐵 → (𝐴 ∈ suc 𝐵𝐴𝐵))
76adantl 271 . 2 ((𝐴 ∈ On ∧ Ord 𝐵) → (𝐴 ∈ suc 𝐵𝐴𝐵))
83, 7sylbird 168 1 ((𝐴 ∈ On ∧ Ord 𝐵) → (suc 𝐴 ⊆ suc 𝐵𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  wcel 1434  wss 2984  Tr wtr 3901  Ord word 4152  Oncon0 4153  suc csuc 4155
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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-ral 2358  df-rex 2359  df-v 2614  df-un 2988  df-in 2990  df-ss 2997  df-sn 3428  df-uni 3628  df-tr 3902  df-iord 4156  df-suc 4161
This theorem is referenced by:  nnsucsssuc  6184
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