reserve A for non empty set,
  a,b,x,y,z,t for Element of A,
  f,g,h for Permutation of A;

theorem
  f.a=a implies (f\g).(g.a)=g.a
proof
  assume
A1: f.a=a;
  g".(g.a) = (g"*g).a by FUNCT_2:15
    .= (id A).a by FUNCT_2:61
    .= a;
  hence (f\g).(g.a) = (g*f).a by FUNCT_2:15
    .= g.a by A1,FUNCT_2:15;
end;
