Lemma 14.4, Case I: the even lower endpoint s = 2 #
When the successor depth M is even, the formal predecessor endpoint 1 is
not in the open odd parity domain. It is nevertheless not an index of the
actual finite recurrence: every summand has p < D^(1/2), hence its inherited
coordinate log D / log p - 1 is strictly larger than 1.
Thus the endpoint is a strict-carrier boundary phenomenon. The actual natural-ceiling recurrence remains exact, and both coordinates used by the induction hypothesis lie in the predecessor parity domain. No estimate for the desired Lemma-14.4 conclusion is assumed below.
At even depth the odd source-cutoff premise of the exact recurrence is
vacuous. Therefore suzukiActualT_caseI_recurrence_strict applies at the
literal natural ceiling without any (generally false) ceiling-square claim.
Every actual recurrence index at the even endpoint has inherited
coordinate strictly inside the predecessor's open odd parity domain. This is
the strict inequality which the formal endpoint 2 - 1 = 1 itself lacks.
The ceiling-recursive coordinate is also legal, and Proposition 9.3 gives
exactly the finite-source-layer comparison needed by the pointwise IH. This
uniform packet uses only actual recurrence indices; it has no hypothesis that
1 belongs to the predecessor parity domain.