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

theorem
  0 <= a & 0 <= b implies 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/(a-b)
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
  per cases;
  suppose
    a <> b;
    hence 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/((sqrt a)^2-(sqrt b)^2) by A1
,A2,Lm5,Th11
      .= (sqrt a + sqrt b)/(a-(sqrt b)^2) by A1,Def2
      .= (sqrt a + sqrt b)/(a-b) by A2,Def2;
  end;
  suppose
A3: a = b;
    then 1/(sqrt a-sqrt b) = 0;
    hence thesis by A3;
  end;
end;
