reserve x for set;
reserve a, b, c for Real;
reserve m, n, m1, m2 for Nat;
reserve k, l for Integer;
reserve p, q for Rational;
reserve s1, s2 for Real_Sequence;

theorem Th26:
  a>=0 & n>=1 & m>=1 implies n -Root a * m -Root a = (n*m) -Root ( a |^ (n+m))
proof
  assume that
A1: a>=0 and
A2: n>=1 and
A3: m>=1 and
A4: n -Root a * m -Root a <> (n*m) -Root (a |^ (n+m));
A5: (n -Root a * m -Root a) |^ (n*m) = (n -Root a) |^ (n*m) * (m -Root a) |^
  (n*m) by NEWTON:7
    .= ((n -Root a) |^ n) |^ m * (m -Root a) |^ (n*m) by NEWTON:9
    .= a |^ m * (m -Root a) |^ (m*n) by A1,A2,Th19
    .= a |^ m * ((m -Root a) |^ m) |^ n by NEWTON:9
    .= a |^ m * a |^ n by A1,A3,Th19
    .= a |^ (n+m) by NEWTON:8;
A6: n*m>=1 by A2,A3,XREAL_1:159;
  per cases by A1;
  suppose
A7: a>0;
    then
A8: m -Root a > 0 by A3,Def2;
    n -Root a > 0 by A2,A7,Def2;
    then
A9: n -Root a * m -Root a > 0 by A8;
A10: a |^ (n+m) > 0 by A7,Th6;
    then
A11: (n*m) -Root (a |^ (n+m)) > 0 by A6,Def2;
    per cases by A4,XXREAL_0:1;
    suppose
      n -Root a * m -Root a < (n*m) -Root (a |^ (n+m));
      hence contradiction by A5,A6,A10,A9,Def2;
    end;
    suppose
      n -Root a * m -Root a > (n*m) -Root (a |^ (n+m));
      hence contradiction by A5,A6,A10,A11,Def2;
    end;
  end;
  suppose
A12: a=0;
    reconsider k=n, k1=1 as Integer;
    reconsider m1=k-k1 as Element of NAT by A2,INT_1:5;
A13: a |^ (n+m) = a |^ (m1+m+1) .= a |^ (m1+m) * a by NEWTON:6
      .= 0 by A12;
    n -Root a = 0 by A2,A12,Def2;
    hence contradiction by A4,A6,A13,Def2;
  end;
end;
