theorem
  f\(g*h) = (f\h)\g
proof
  thus f\(g*h) = ((g*h)*f)*(h"*g") by FUNCT_1:44
    .= (((g*h)*f)*h")*g" by RELAT_1:36
    .= ((g*(h*f))*h")*g" by RELAT_1:36
    .= (f\h)\g by RELAT_1:36;
end;
