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Theorem eqssi 2955
Description: Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)
Hypotheses
Ref Expression
eqssi.1 AB
eqssi.2 BA
Assertion
Ref Expression
eqssi A = B

Proof of Theorem eqssi
StepHypRef Expression
1 eqssi.1 . 2 AB
2 eqssi.2 . 2 BA
3 eqss 2954 . 2 (A = B ↔ (AB BA))
41, 2, 3mpbir2an 848 1 A = B
Colors of variables: wff set class
Syntax hints:   = wceq 1242  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-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  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-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-in 2918  df-ss 2925
This theorem is referenced by:  inv1  3247  unv  3248  undifabs  3294  intab  3635  intid  3951  find  4265  limom  4279  dmv  4494  0ima  4628  rnxpid  4698  dftpos4  5819  dfuzi  8104  unirnioo  8592
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