theorem Th42:
  mi K = K
proof
  thus mi K c= K by Th40;
  now
    let a;
    assume
A1: a in K;
    then for b st b in K & b c= a holds b = a by Th32;
    hence a in mi K by A1,Th39;
  end;
  hence thesis by Lm5;
end;
