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Theorem sseqin2 3192
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 3177 1 (𝐴𝐵 ↔ (𝐵𝐴) = 𝐴)
Colors of variables: wff set class
Syntax hints:  wb 103   = wceq 1285  cin 2973  wss 2974
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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-v 2604  df-in 2980  df-ss 2987
This theorem is referenced by:  resabs1  4668  rescnvcnv  4813  frecfnom  6050  nn0supp  8407  uzin  8732  iooval2  9014  fzval2  9108
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