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CounterexampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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The step set {(1,1),(1,−2)} breaks the reflection argument

Statement refuted

The reflection argument on the A page depends on the step set {U,D} with height changes ±1. It does not extend unchanged to arbitrary step sets.

Facts & Assumptions

Given: the step set S={(1,1),(1,−2)} and the level 0.

[L1]

A diagonal path with h(0)≤c≤h(n) or h(n)≤c≤h(0) meets the level c somewhere (A diagonal path with h(0)≤c≤h(n) or h(n)≤c≤h(0) satisfies h(i)=c for some i).

[L2]

For diagonal paths with steps (1,1) and (1,−1) whose endpoints lie above the level c and which first visit c at some index τ, the initial segment up to τ may be reflected across the line y=c to obtain the bijection of Reflecting the initial segment at the first visit to level c.

[L3]

For diagonal paths with steps (1,1) and (1,−1) and endpoints strictly above a level, the reflection principle identifies paths touching that level with paths from the reflected starting height, and subtracts their count from the total (The reflection principle: paths from (0,a) to (n,b) staying strictly above level c are counted by a difference of two binomial coefficients).

Counterexample

technique · direct
1.1

The one-step path from (0,1) to (1,−1) with step (1,−2) starts above the level 0 and ends below it, but its heights are only 1 and −1, so it never has height 0. This path is outside [L1]'s diagonal-step hypothesis and shows that the conclusion of [L1] fails if that hypothesis is dropped.

L1given
2.1

Because the path of step 1.1 never visits the level 0, the first-visit reflection of [L2] is undefined on it. So the bijection on which the reflection count rests is absent.

L2step 1.1
3.1

The naive analogue of the count fails too. For these same steps there is no path from (0,1) to (2,1), so the total count is 0 and the count of paths staying strictly above 0 is also 0; but there is one path from (0,−1) to (2,1), namely UU. Illegally extending the subtraction pattern of [L3] would therefore give 0−1=−1, which is not a count of paths.

L3step 2.1∎

Remarks

  • The broken step is exactly the one hidden in the ordinary proof: when the height jump can skip over the forbidden level, "changes side" no longer means "meets the level first."

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources