does not satisfy (EP), because for x = 0.3, y = 0.5 and z = 0.5 we get I (x, I (y, z)) = 0.7 = 0.5 = I (y, I (x, z)). The corrected last two rows are provided in Table 1 here. In both of these cases the natural negation NI is the classical negation NC(x) = 1 − x. One can easily check that the first function satisfies (I1) and it does not satisfy (EP), because for x = 0.3, y = 1 and z = 0, we ge...