“…S, i.e. there exist/eU(x), g e U(y), heU(z) with (/, g, h) e R. Thus dom/ = dom g = dom h = X, where -Y = (X, D n X 3 ) e e <£(//), x, >, z e X and (/(t), g((), h(t)) e C for any (e X. Further/(x) = co = g(>>) = = h(z). By Lemma 5, we have x # >> # z + x.…”