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Theorem sseqin2 3185
Description: A relationship between subclass and intersection. Similar to Exercise 9 of [TakeutiZaring] p. 18. (Contributed by NM, 17-May-1994.)
Assertion
Ref Expression
sseqin2 (𝐴𝐵 ↔ (𝐵𝐴) = 𝐴)

Proof of Theorem sseqin2
StepHypRef Expression
1 dfss1 3170 1 (𝐴𝐵 ↔ (𝐵𝐴) = 𝐴)
Colors of variables: wff set class
Syntax hints:  wb 103   = wceq 1284  cin 2972  wss 2973
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-in 2979  df-ss 2986
This theorem is referenced by:  resabs1  4658  rescnvcnv  4803  frecfnom  6009  nn0supp  8340  uzin  8651  iooval2  8938  fzval2  9032
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