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Theorem ssequn1 3091
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 pm4.72 725 . . . 4 ((x Ax B) ↔ (x B ↔ (x A x B)))
3 elun 3062 . . . . 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 1338 . 2 (x(x Ax B) ↔ x(x (AB) ↔ x B))
7 dfss2 2912 . 2 (ABx(x Ax B))
8 dfcleq 2016 . 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 1314   = wceq 1373   wcel 1375  cun 2893  wss 2895
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 1315  ax-7 1316  ax-gen 1317  ax-ie1 1362  ax-ie2 1363  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 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1329  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2535  df-un 2900  df-in 2902  df-ss 2909
This theorem is referenced by:  ssequn2  3094  uniop  3944  pwssunim  3974  unisuc  4073  rdgisuc2  5858  oasuc  5923
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