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 Th22:
  a>=0 & b>=0 & n>=1 implies n -Root (a*b) = n -Root a * n -Root b
proof
  assume that
A1: a>=0 and
A2: b>=0 and
A3: n>=1 and
A4: n -Root (a*b) <> n -Root a * n -Root b;
A5: (n -Root a * n -Root b) |^ n = (n -Root a) |^ n * (n -Root b) |^ n by
NEWTON:7
    .= a * (n -Root b) |^ n by A1,A3,Th19
    .= a * b by A2,A3,Th19;
  per cases by A1,A2;
  suppose
A6: a>0 & b>0;
    then
A7: n -Root b > 0 by A3,Def2;
A8: n -Root a > 0 by A3,A6,Def2;
    a*b>0 by A6;
    then
A9: n -Root (a*b) > 0 by A3,Def2;
    per cases by A4,XXREAL_0:1;
    suppose
      n -Root (a*b) < n -Root a * n -Root b;
      hence contradiction by A3,A5,A9,Th19;
    end;
    suppose
      n -Root (a*b) > n -Root a * n -Root b;
      hence contradiction by A3,A5,A8,A7,Th19;
    end;
  end;
  suppose
A10: a=0 & b>0;
    then n -Root (a*b) = 0 by A3,Def2;
    hence contradiction by A4,A10;
  end;
  suppose
A11: a>0 & b=0;
    then n -Root (a*b) = 0 by A3,Def2;
    hence contradiction by A4,A11;
  end;
  suppose
A12: a=0 & b=0;
    then n -Root (a*b) = 0 by A3,Def2;
    hence contradiction by A4,A12;
  end;
end;
