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Theorem ssequn1 3089
Description: A relationship between subclass and union. Theorem 26 of [Suppes] p. 27. (Contributed by NM, 30-Aug-1993.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
ssequn1 (AB ↔ (AB) = B)

Proof of Theorem ssequn1
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 bicom 128 . . . 4 ((x B ↔ (x A x B)) ↔ ((x A x B) ↔ x B))
2 pm4.72 724 . . . 4 ((x Ax B) ↔ (x B ↔ (x A x B)))
3 elun 3060 . . . . 5 (x (AB) ↔ (x A x B))
43bibi1i 217 . . . 4 ((x (AB) ↔ x B) ↔ ((x A x B) ↔ x B))
51, 2, 43bitr4i 201 . . 3 ((x Ax B) ↔ (x (AB) ↔ x B))
65albii 1339 . 2 (x(x Ax B) ↔ x(x (AB) ↔ x B))
7 dfss2 2910 . 2 (ABx(x Ax B))
8 dfcleq 2017 . 2 ((AB) = Bx(x (AB) ↔ x B))
96, 7, 83bitr4i 201 1 (AB ↔ (AB) = B)
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   wo 616  wal 1226   = wceq 1228   wcel 1375  cun 2891  wss 2893
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 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1364  ax-ie2 1365  ax-8 1377  ax-10 1378  ax-11 1379  ax-i12 1380  ax-bnd 1381  ax-4 1382  ax-17 1401  ax-i9 1405  ax-ial 1410  ax-i5r 1411  ax-ext 2005
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1629  df-clab 2010  df-cleq 2016  df-clel 2019  df-nfc 2150  df-v 2536  df-un 2898  df-in 2900  df-ss 2907
This theorem is referenced by:  ssequn2  3092  uniop  3965  pwssunim  3994  unisuc  4097  unisucg  4098  rdgisucinc  5887  oasuc  5954  omsuc  5961
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