theorem Th35:
  <%>E in A & n > 0 implies A c= A |^ n
proof
  assume that
A1: <%>E in A and
A2: n > 0;
  consider m such that
A3: m + 1 = n by A2,NAT_1:6;
  <%>E in A |^ m by A1,Th30;
  then A c= A |^ m ^^ A by Th16;
  hence thesis by A3,Th23;
end;
