theorem
  Top NormForm A = {[{},{}]}
proof
  reconsider O = {[{},{}]} as Element of Normal_forms_on A by Th17;
  set sd = StrongImpl(A)[:](diff Bottom NormForm A, Bottom NormForm A);
  set F=M(A).:(pseudo_compl(A), sd);
A1: @(pseudo_compl(A).Bottom NormForm A) = mi(-@Bottom NormForm A) by Def8
    .= mi O by Th13,NORMFORM:57
    .= O by NORMFORM:42;
A2: Bottom NormForm A = {} by NORMFORM:57;
  then (diff Bottom NormForm A).Bottom NormForm A = {} \ {} by Def11
    .= Bottom NormForm A by NORMFORM:57;
  then
A3: @(sd.Bottom NormForm A) = StrongImpl(A).(Bottom NormForm A, Bottom
  NormForm A) by FUNCOP_1:48
    .= mi(@Bottom NormForm A =>> @Bottom NormForm A) by Def9
    .= mi O by A2,Th14
    .= O by NORMFORM:42;
  thus Top NormForm A = (Bottom NormForm A) => Bottom NormForm A by FILTER_0:28
    .= FinJoin(SUB Bottom NormForm A,F) by Th27
    .= J(A)$$(SUB Bottom NormForm A,F) by LATTICE2:def 3
    .= F.Bottom NormForm A by A2,SETWISEO:17,ZFMISC_1:1
    .= M(A).(pseudo_compl(A).Bottom NormForm A, sd.Bottom NormForm A) by
FUNCOP_1:37
    .= mi (O^O) by A1,A3,NORMFORM:def 12
    .= {[{},{}]} by Th3;
end;
