theorem
  Z c= dom (f1-f2) & f1 is_differentiable_on Z & f2 is_differentiable_on Z
  implies f1-f2 is_differentiable_on Z &
  for x st x in Z holds ((f1-f2)`|Z).x = diff(f1,x) - diff(f2,x)
proof
A1: f1-f2 = f1+-f2 by Th1;
    assume that
A2: Z c= dom (f1-f2) and
A3: f1 is_differentiable_on Z and
A4: f2 is_differentiable_on Z;
A5: -f2 is_differentiable_on Z by A4,Th14;
    hence f1-f2 is_differentiable_on Z by A1,A2,A3,Th15;
    let x;
    assume
A6: x in Z;
    then
A7: f2 is_differentiable_in x by A4,Th5;
    thus ((f1-f2)`|Z).x = diff(f1,x) + diff(-f2,x) by A1,A2,A3,A5,A6,Th15
    .= diff(f1,x) - diff(f2,x) by A7,Th10;
end;
