theorem
  (.IB \/ JB.> = IB "\/" JB
proof
  reconsider FB = IB.:, GB = JB.: as Filter of B.:;
  thus (.IB \/ JB.> = <.(IB \/ JB).:.) by Th36
    .= FB"/\"GB by FILTER_0:39
    .= IB "\/" JB by Th44;
end;
