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Theorem ordelsuc 4197
Description: A set belongs to an ordinal iff its successor is a subset of the ordinal. Exercise 8 of [TakeutiZaring] p. 42 and its converse. (Contributed by NM, 29-Nov-2003.)
Assertion
Ref Expression
ordelsuc ((A 𝐶 Ord B) → (A B ↔ suc AB))

Proof of Theorem ordelsuc
StepHypRef Expression
1 ordsucss 4196 . . 3 (Ord B → (A B → suc AB))
21adantl 262 . 2 ((A 𝐶 Ord B) → (A B → suc AB))
3 sucssel 4127 . . 3 (A 𝐶 → (suc ABA B))
43adantr 261 . 2 ((A 𝐶 Ord B) → (suc ABA B))
52, 4impbid 120 1 ((A 𝐶 Ord B) → (A B ↔ suc AB))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98   wcel 1390  wss 2911  Ord word 4065  suc csuc 4068
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-v 2553  df-un 2916  df-in 2918  df-ss 2925  df-sn 3373  df-uni 3572  df-tr 3846  df-iord 4069  df-suc 4074
This theorem is referenced by:  onsucmin  4198  onsucelsucr  4199  onsucsssucr  4200  onsucsssucexmid  4212  ordgt0ge1  5957  nnsucsssuc  6010
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