theorem
  F is associative implies F.:(F.:(f,g),h) = F.:(f,F.:(g,h))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y, x3 = h.y as Element of X by A2,FUNCT_2:5;
      thus (F.:(F.:(f,g),h)).y = F.((F.:(f,g)).y,h.y) by A2,Th37
        .= F.(F.(f.y,g.y),h.y) by A2,Th37
        .= F.(x1,F.(x2,x3)) by A1
        .= F.(f.y,(F.:(g,h)).y) by A2,Th37;
    end;
    hence thesis by A2,Th38;
  end;
end;
