theorem Th25:
  (ex q st a=q|^d) implies ex b st a=b|^d
proof
  assume ex q st a=q|^d;
  then consider k such that
A1: a=k|^d by Th24;
A2: now
    assume
A3: a=0;
    then d<>0 by A1,NEWTON:4;
    then a=0|^d by A3,NAT_1:14,NEWTON:11;
    hence thesis;
  end;
  per cases;
  suppose d<>0;
    now
      consider e such that
A4:   d=2*e or d=2*e+1 by Lm4;
      consider c being Nat such that
A5:   k=c or k=-c by INT_1:2;
      assume
A6:   not k is Nat;

A7:   now
        assume d=2*e+1;
        then -c|^d=a by A1,A6,A5,Th2;
        hence thesis by A2;
      end;
      now
        assume d=2*e;
        then a=c|^d by A1,A5,Th2;
        hence thesis;
      end;
      hence thesis by A4,A7;
    end;
    hence thesis by A1;
  end;
   suppose
A8: d=0;
    then a=1 by A1,NEWTON:4;
    then a=1|^0;
    hence thesis by A8;
  end;
end;
