The arithmetic gate holds ×m and ⌊n/m⌋ alike, and the two are not held for the same reason. Legality does two jobs at once — it caps the digits, and it forbids a digit at its cap over a nonzero — and together they make the writing unique. Widen the OUTPUT alone: let the reader emit digits up to ak+1 + s, and up to a1 − 1 + s0 at position 0, drop the below-a-cap rule, and keep feeding it the legal string. The output strings are then a redundant writing — several per integer — with the pair (s, s0) its slack. Read a string's real star Σ ekθk as a real number and not as a point of the circle — two writings of one integer differ by an integer there — and the stars of all admissible strings fill an interval of length 1 + E, the excess E = s0α + s Σk≥1|θk|. At E = 0 — the rule dropped, every cap kept — that interval is exactly [−α, 1 − α], and the tear survives at the address it already had. The pairs that prove the gate agree ever deeper while their images sit on opposite sides of one point, so those images' stars close on the interval's two ENDPOINTS, one image to each; and a string whose star is within δ of an endpoint agrees with (0, a2, 0, a4, …) or with (a1 − 1, 0, a3, …) at every position with |θk| > δ, the difference being a sum of non-negative terms. Whatever the reader chooses, the two sides part where those two strings part — at the lowest admissible digit. Dropping the rule alone buys nothing.
Raise any cap and E > 0, and the writings of one value stop being pinned to the ends: every point of the circle then has a real lift interior to the interval, and consecutive members eLθL + [the tail's range] OVERLAP, by |θL+1| + s Σk>L|θk|. An interval shorter than the overlap sits inside a single member, and at lookahead c what the reader does not yet know about the residual — the part of the image it still owes — is an interval under 2m|θL+c| — so a digit that keeps the star in range exists at every position once 2m|θk+c−1| ≤ (1 + s)|θk| at every k ≥ 1 and 2m|θc| ≤ E — the overlap bound, met at c = O(log(2m/(1 + s))) at every irrational window, raising position 0's cap by one being enough. That reader is the completion's: the map on infinite strings, whose value is a point of the circle. It does not descend to the integers by itself. An integer input must also flush — the residual has to be writable from the position the reader has reached, and every value writable from position L up is 0 or at least qL, which the star cannot see. At the golden window with s = s0 = 1 the residual 3 at position 3 holds its star in range at the digit leaving 3 — below q4 = 5, and unwritable — and at the digit leaving −3, as well as at the one digit that flushes; no tie-break on the star alone repairs that, since 3 there wants the middle digit where a residual of 6 wants the largest. So the integer reader must carry the whole set of integer offsets still alive — and it can. Write the residual in the frame of two consecutive convergents, x qt + y qt−1: the pair (x, y) is an integer pair, one per alive offset, and each digit moves it by a map that reads the one partial quotient at+1 and nothing older; with the partial quotients at most A the game over every such window is ONE finite game, the opponent choosing each quotient from {1, …, A} as it reveals each digit. Every pair the state can still hold while a flush is still possible lies in one box: the star is bounded by the FUTURE, through the telescoping Σk≥t ak+1|θk| = |θt−1| + |θt|, and the residual by the PAST at qt+1 — the digits the reader has already read ahead sit beside the residual and not inside it — which holds |y| under 2(1 + s) + 2m + max(m, 1 + s) and |x| under the same with that maximum multiplied by A + 1, with no lookahead term anywhere in it. So solving the game inside any box CONTAINING that one decides the class both ways: a win transfers from any box at all, a loss only from a containing one. Its winning strategy is a reader that sees the digits through dt+c, the partial quotients through at+c+1, and nothing else of the window: ×2 at s = s0 = 1 reads at lookahead 2 at every irrational window with partial quotients at most 3, and ×3 at s = s0 = 1 reads at EXACTLY 3 over quotients at most 2. At m = 3, s = s0 = 1 and A = 2 that box is (19, 13), and the cell LOSES at lookahead 2 at three boxes each containing it — (19, 13) itself, twice it, and (72, 27); and a reader at lookahead c is simulable at c + 1 by ignoring the extra digit it is handed, so wins rise with c and that loss carries down to 1 and to 0. Every periodic window of the class read so far — every primitive quotient cycle through period 14, 2,538 of them — reads at 2, so what costs ×3 the extra unit is the SWITCH: the one reader that serves the whole class is strictly weaker than the readers those windows have of their own. Whether an integer reader exists at a window with UNBOUNDED partial quotients is open.
The floors do not follow, at any slack. Two tails of different residue send one converging input to images a fixed nonzero rotation apart, while the cells a common output prefix pins those images to shrink as the prefix grows — so no reader at any lookahead can hand them one: ⌊n/m⌋ stays torn at every slack and every irrational window. The two tears are therefore different objects. ×m's was the WRITING's — z ↦ mz is continuous on the circle, and it was legality's uniqueness that forced the images apart — while the floors' is the MAP's: x ↦ x/m is m-valued on the circle and the branch is n mod m, a function of every digit, which no alphabet widens away. The successor sits with the multiplications: n + 1 reads at lookahead 0 the moment the rule is dropped, at zero slack, so its lookahead 1 was the normalization's and never the carry's.
How much delay the integer reader needs is read exactly at a periodic window, by the same finite automaton the limit of the lookahead is read from, run as a game: the reader emits its digit before the next input digit shows, the state carries the offsets still alive, and it must reach zero offset under zero input. Over six windows — [1], [2], [3], [1, 2] and two of period 4, bracketing the block of partial quotients the window repeats — and ×m for m = 2..5, raising position 0's cap alone already reads every one of them, at 2 to 6, and every one of those cells sits at or under the overlap bound. And the lookahead is never 1 — a RULE, proved at every periodic window and every m. A cell reads at 0 exactly where the digitwise writing ek = m dk is admissible at every position past the first — s ≥ (m − 1)amax, amax the largest partial quotient; position 0 needs no slack of its own, its overflow spilling up into q1 = a1 with the greedy cap on d1 over a nonzero d0 leaving the room — and at 2 or more everywhere else. What forbids 1 is that the FLUSHED state is the exposed one. A reader of the integer n has, from some position on, nothing left to write; append a late digit at its cap, ak+1 qk over a zero, at a k past that position. The reader's outputs on the two inputs agree through position k − c − 1, so on the second it now owes m ak+1 qk from position k − c, with nothing below. From position k − 1 that integer has exactly one capped writing, ek = m ak+1 — writing qk+j = Aj qk + Bj qk−1, the qk−1-coefficient of any string from position k − 1 is a multiple of qk, hence 0 once qk−1 > m ak+1 — and it breaks the cap ak+1 + s exactly when s < (m − 1)ak+1. An overflow at position k spills into the two positions below it, qk = ak qk−1 + qk−2, and a reader that has flushed at lookahead 0 or 1 has already written both. So at those lookaheads every state the game reaches with nothing left to write is lost, and the flush requirement does not trim the winning set, it empties it — at all 115 cells below the digitwise line, of the 120 grid cells that read at all and the ten band cells next. The digitwise line and the bound's own grant of lookahead 1 — 1 + s ≥ 2m, its second half 2m|θ1| ≤ E, met at every cell read — CROSS at amax > 2 + 1/(m − 1), and between them lies a band, 2m − 1 ≤ s < (m − 1)amax, where the bound grants the completion's reader lookahead 1 and the digitwise writing does not fit. The grid misses it by both coordinates — its windows stop at amax = 3, where the band opens at m = 3, s = 5, and its slacks stop at 3 — and ten of its cells, the constant window with every quotient 4 at ×2 from s = 3, the one with every quotient 3 at ×3 from s = 5, up to 1 + s = 10 against 2m = 6, read at 2, strategies certified: the rule's sharpest exhibit, the residual deliverable one digit ahead and the integer reader refused it anyway.
The same argument prices every lookahead. Drop the flush requirement alone from the same finite game — the reader released from finishing, which must write forever inside the box and need never land on the value — and the automaton prints a lookahead per cell; separately, the completion's reader has a minimum of its own and carries no offsets at all, since at each position it owes the unseen tail scaled by m, an interval, to be placed whole inside one member — at a periodic window one real coordinate with the phase running round the period. The two are EQUAL at every periodic window, every m and every slack pair the box is derived for (s0 ≤ s + 1) — a THEOREM, the 120 grid cells that read and the ten band cells its check. Both directions are one fact about the released reader's state. The completion's target is a point of the circle, and an integer lift chooses which real number stands for it; the state holds one alive offset per lift still possible, each as its frame pair, and a digit moves every pair to exactly one successor, dropping it once it leaves the box. So the offsets alive along a play form a forest with finitely many at each depth, and a play that keeps the set non-empty forever — a safe one, which is all the released reader is asked for — carries one offset that never dies, by König's lemma (a forest with finitely many nodes at each depth and some node at every depth has an infinite path). The box holds a star under a fixed multiple of |θt| at depth t, and that is all it does here — its width never enters — so the surviving offset's star tends to 0, and the digits written are a writing of the target lifted by M, its lift: a coding of the target. Wherever the completion's reader is constrained at all — a depth at which the members are narrower than a full turn — the continuations of one seen prefix have residuals filling an interval, and an interval inside a disjoint union of closed members lies in one of them; so the lifts agree across the prefix, and the safe strategy, a function of the seen digits, wins the completion's game at the same lookahead. Conversely a strategy that wins the completion's game at lookahead c, run in the integer game, keeps the offset of each input's true lift alive: that offset's residual at depth t is the output's tail from t less m times the input's tail past t + c, under (1 + s)(|θt−1| + |θt|) + m(|θt| + |θt+1|) by the tail bound Σj≥t|θj| ≤ |θt−1| + |θt| — an equality at golden — which sits inside the reader's own box with a unit of θ to spare, and its lift inside the initial set, at all 154 cells — the grid's 144, reading or not, and the band's ten. So the offset frame and the box confining it cost NOTHING in lookahead beyond the flush itself, and the whole distance from the integer reader down to the completion's IS the flush's own price — the extra lookahead that one requirement costs the game. That price, cell by cell: over the grid it is 0 at 91 of the 120, 1 at 28 and 2 at one, and over the band 1 at 29 cells and 2 at four — every price-2 cell a period-4 window whose released reader needs no lookahead at all. And the late digit says where it lands. Let L* be the least number of positions below k that m times the value of τ needs to be a capped writing, over every legal tail τ that can arrive at a k where the reader has nothing left to write. That is the lookahead of a third reader, the one released from IGNORANCE — shown the whole input — from a state with nothing left to write, and it is computed exactly: the set of residuals any choice of output can reach, pruned to the box, is a finite automaton on the tail's digits, so no tail length enters (a first reading truncated tails at P + 2 digits, P the period, and agrees at 128 of 130 cells, rising from 2 to 4 at two where the released reader needs 5). L* and the released reader's lookahead are both forced lower bounds on the integer reader's, and the integer reader reads at EXACTLY the larger of the two at all 153 cells decided — equal to the released reader's at 91 of them and to L* at 117. So a nonzero price lands on 2 at 60 cells and on 4 at two, ×5 at the period-4 window (1, 1, 1, 2) at the slacks (1, 0) and (1, 1), where a late tail rather than a late digit needs four positions.
Scope. Theorem at every irrational window for the separation — the tear surviving at zero excess, the completion's reader at the overlap bound wherever the excess is positive, and the floors torn at every slack — periodicity used nowhere. That the completion's reader does not descend on its own is carried by two computed exhibits, both of them negatives: the golden residual above, exact arithmetic at one window, and the band cells, where the same failure is isolated from the star. That the flush's price is the WHOLE distance to that reader is a THEOREM at every periodic window, every m and every slack pair the box is derived for, s0 ≤ s + 1 — the one place the box's derivation enters — the 120 grid cells that read and the ten band cells its verification, each decided by a certificate, and two of the three containments the converse rests on — the real coordinate's and the lift's — printed at all 154 cells, the third the reader's own conjugate formula read. The one-line argument once offered for the easier half — that a strategy held inside the box already meets the completion reader's requirement — is FALSE and stays so, the box being far wider than the member a residual must sit in to be completable, and at one window, √3 − 1 at ×3 and slack (1, 0), a safe strategy reaches a state none of whose alive offsets sits, by its present star, inside a member at its depth. That state completes all the same, the surviving offset's star carrying the unseen input tail beside the output's until both vanish: no reading of a state's present offsets against the members is the criterion, and the box forcing every surviving star to 0 is. Not claimed: any window off periodicity, where there is no frame and no box; and nothing about the finite game's size — the theorem moves the completion reader's minimum onto that game and says which one. Ten of the grid's 120 cells fall between that reader's two certificates, and what decides them is that the completion map is a map to the circle: each input takes its own integer lift, at the price of a wrap at the levels whose range is a full turn or more, which is a finite prefix game feeding the periodic one — five of the ten needed it, and five were losses the game already held and did not read. The INTEGER reader is a rule at two scopes. At the six windows above, ×2..×5 and the five raised-cap slacks of the grid — 120 cells, and the unraised one where the theorem answers — each winning cell's strategy run on the integers below 1500 with every value and every cap checked, and the legal alphabet through the same game reproducing the gate's own verdicts as the control. And over the bounded-quotient classes — ×2 at s = s0 = 1 at lookahead 2 with quotients at most 3, ×3 at s = s0 = 1 at exactly 3 with quotients at most 2 — the winning strategy extracted and run on the integers below 1000 at twelve random windows of sixty quotients over the alphabet and at the alphabet's periodic windows of period at most 2, value, caps and flush checked; the one-letter alphabets reproducing the grid's [1], [2] and [3] cells for ×2 and ×3 at the slacks (1, 1) and (2, 2) as the control. The EXACTLY is the losing half and rests on computation: an exhaustive solve at lookahead 2 at three boxes each containing the box above, with 1 and 0 ruled out by the simulation rather than by runs of their own. That box is proved here and then read back off the trajectories that actually flush, at six winning cells each solved in a box several times it so a violation had room; for ×3 at a multi-letter alphabet the reading is silent, both such cells needing more memory than the runs allow. Its existence at unbounded partial quotients is open, the game's state being finite only under a bound on them. That no cell reads at lookahead 1 is a RULE, proved above at every periodic window and every m, and past periodicity its two halves carry on their own hypotheses — the zero half wherever s ≥ (m − 1)ak at every k, the floor wherever (m − 1)ak+1 > s at infinitely many k with qk−1 > m ak+1, which bounded quotients supply; that the integer reader reads at exactly the larger of the released reader's lookahead and L* is an OBSERVATION at 153 cells — the grid's 120 and 33 of the band, each decided exactly by the automaton — with both sides proved lower bounds and no sufficiency proof under the equality: what a proof must rule out is a COMMITMENT, two continuations of the seen digits demanding two different pending writings of one residual that the box does not tell apart, and the delay has no value law beyond it. The band is not exhausted: 33 were decided of the 42 the hunt found under s ≤ 12, itself a count that cap sets, three of those bands still open at 12; the nine undecided are [5] at ×4 and ×5 from s = 9 up and the period-4 window (2, 1, 3, 1) at ×5 from s = 10, whose games do not fit the memory this work runs under. The positional redundant purchase is a LEADING-end door (Redundant arithmetic); this is the trailing end, where the borrow runs against the read.
verifiers:
explore_redundant_ostrowski.py,
explore_universal_reader.py,
explore_reader_box.py,
explore_lookahead_band.py,
explore_flush_price.py,
explore_completion_reader.py,
explore_completion_lift.py,
explore_flush_floor.py,
explore_flush_law.py,
explore_flush_theorem.py