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Theorem ssequn1 3107
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 pm4.72 735 . . . 4 ((x Ax B) ↔ (x B ↔ (x A x B)))
3 elun 3078 . . . . 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 1356 . 2 (x(x Ax B) ↔ x(x (AB) ↔ x B))
7 dfss2 2928 . 2 (ABx(x Ax B))
8 dfcleq 2031 . 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 628  wal 1240   = wceq 1242   wcel 1390  cun 2909  wss 2911
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-v 2553  df-un 2916  df-in 2918  df-ss 2925
This theorem is referenced by:  ssequn2  3110  uniop  3982  pwssunim  4011  unisuc  4115  unisucg  4116  rdgisucinc  5909  oasuc  5976  omsuc  5983
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