theorem Th3:
  (FuncAdd(X,Y)).(f,g) = (FuncAdd(X,Y)).(g,f)
proof
  now
    let x be Element of X;
    thus ((FuncAdd(X,Y)).(f,g)).x = g.x + f.x by LOPBAN_1:1
      .= ((FuncAdd(X,Y)).(g,f)).x by LOPBAN_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;
