 Table 1: A team \(X\) in which \(y = x_1 + x_2\). Here \(y\) is a function of \(x_1\) and \(x_2\), and hence \(=\!\!(x_1 x_2, y)\) holds; however, \(y\) is not a function of \(x_1\) alone, so \(=\!\!(x_1, y)\) does not hold.