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 Th25:
  a>=0 & n>=1 & m>=1 implies n -Root (m -Root a) = (n*m) -Root a
proof
  assume that
A1: a>=0 and
A2: n>=1 and
A3: m>=1 and
A4: n -Root (m -Root a) <> (n*m) -Root a;
  per cases by A1;
  suppose
A5: a>0;
    then
A6: m -Root a > 0 by A3,Def2;
    then
A7: n -Root (m -Root a) > 0 by A2,Def2;
A8: (n -Root (m -Root a)) |^ (n*m) = ((n -Root (m -Root a)) |^ n) |^ m by
NEWTON:9
      .= (m -Root a) |^ m by A2,A6,Lm2
      .= a by A1,A3,Th19;
A9: n*m>=1 by A2,A3,XREAL_1:159;
    then
A10: (n*m) -Root a > 0 by A5,Def2;
    per cases by A4,XXREAL_0:1;
    suppose
      n -Root (m -Root a) < (n*m) -Root a;
      hence contradiction by A5,A8,A9,A7,Def2;
    end;
    suppose
      (n*m) -Root a < n -Root (m -Root a);
      hence contradiction by A5,A8,A9,A10,Def2;
    end;
  end;
  suppose
A11: a=0;
    n*m>=1 by A2,A3,XREAL_1:159;
    then
A12: (n*m) -Root a = 0 by A11,Def2;
    m -Root a = 0 by A3,A11,Def2;
    hence contradiction by A2,A4,A12,Def2;
  end;
end;
