the telescoping product. let the record move through R₀ = F, R₁, …, Rₙ by successive breaks. a balance present throughout survives each break in the proportion Rₖ₋₁ / Rₖ, since the seal spends (g − R) / g and retains R / g with g becoming the new record.
every intermediate term cancels against its neighbour, so the result depends on the first and last records alone. the path taken between them — how many breaks, how large, in what order — leaves nothing behind in the answer.
this is why the contract can store no history without losing anything. the record is a sufficient statistic for the whole burn schedule, and a machine that remembers only its deepest silence remembers everything the arithmetic needs.
every halving of the reserve costs a doubling of the silence.
fig. 1 establishes that retention is geometric in depth: the vertical axis falls by a factor of two for every rung crossed, so the curve is a straight line under a log scale.
the blocks of silence required at each rung, F · 2ⁿ, on a log axis.
fig. 2 establishes the arithmetic of the ladder itself: each rung demands twice the silence of the one below it, and the demand outruns any plausible chain lifetime well before the sixteenth.
the fraction is fixed at a half. the amount is not.
fig. 3 establishes the shape of the reserve over time: it climbs while the machine is ignored and falls vertically at each seal, and the drops grow while the climbs lengthen.
the ladder drawn to scale against the chain's own age.
fig. 4 establishes the scale of the requirement against the chain's own age, with the rung spans bracketed against the axis so a rung can be read as a length rather than a number.
beating the record by a little costs a little; doubling it costs half.
fig. 5 establishes that the cost of a break is scale-free in g / R, rising from nothing at g = R toward an asymptote of one, and passing exactly one half at g = 2R.
the fraction never changes; the amount doubles forever.
fig. 6 establishes the separation between share and amount: the overlaid constant line is the fraction, fixed at a half, while the bars beneath it double at every rung.
a single silence is twice as efficient as a climb, and destroys exactly the same amount.
fig. 7 establishes path independence in its sharpest form: climbing rung by rung costs the same as one uninterrupted silence and takes very nearly twice as long.
holding the record down accumulates the reserve that the first missed window destroys.
fig. 8 establishes the cost of suppression: calls at the floor interval hold the record down while the reserve accumulates, and the first missed window destroys the accumulation.
1 · the record is non-decreasing.
the only assignment to R is R := g, executed exactly when g > R, so R is monotone. depth D = log₂(R / F) is monotone in R, and cumulative burn is a sum of non-negative terms. none of the three can fall. ∎
2 · a deposit is never wholly destroyed.
a balance present at depth Da retains F/R at each stage, so by the telescoping product its surviving fraction at depth Dn is 2^−(Dn − Da), which is strictly positive for every finite n. the destroyed fraction is 1 − 2^−(Dn − Da) < 1. ∎
3 · the reserve is never emptied.
the spend is balance × (g − R) / g with g > R > 0, so (g − R) / g < 1 for all finite g and the spend is strictly less than the balance. what remains is balance × R / g > 0. ∎
4 · the total burned depends only on F and the maximum gap.
by proposition 2 the surviving fraction after any sequence of breaks ending at record R is F / R, independent of the intermediate records. the burned share is therefore 1 − F / R, a function of F and max g alone. ∎
5 · seal transfers nothing to its caller.
the only outbound transfers in the seal path are the purchase on the market and the transfer of the purchased tokens to the dead address. no branch pays msg.sender, so the caller's net position is the gas spent. ∎
6 · integer division keeps the retained share plus dust.
the spend is computed as balance × (g − R) / g under integer division, which floors. the floored spend is at most the exact spend, so the residual is at least balance × R / g, and the machine keeps the retained share plus the truncation dust. ∎