reserve a,b,c,x,y,z for Real;

theorem Th29:
  0 <= a & 0 <= b implies sqrt (a*b) = sqrt a * sqrt b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
A3: 0 <= sqrt a by A1,Def2;
A4: 0 <= sqrt b by A2,Def2;
  (sqrt(a*b))^2 = a*b by A1,A2,Def2
    .= (sqrt a)^2*b by A1,Def2
    .= (sqrt a)^2*(sqrt b)^2 by A2,Def2
    .= (sqrt a*sqrt b)^2;
  hence sqrt(a*b) = sqrt(sqrt a*sqrt b)^2 by A1,A2,Def2
    .= sqrt a*sqrt b by A3,A4,Def2;
end;
