“…We have shown that π meets both (β2)-tips of π· am , in particular, π β π· am . Now π β
π
am β©½ 1 by formula (12), so in fact π β
π
am = 1 by condition (7), as claimed. β‘ Proposition 6.1 shows that, under our assumptions, π min is β(1, 2, 3), β(1, 1, 2), or β 2 .…”