Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Applying the Nullstellensatz by hand to over
Example
Let
Facts & Assumptions
Given: the polynomial and the sets above.
If , the coefficient of is nonzero, and , then is nonzero at some point of (Alon's Combinatorial Nullstellensatz: if , the coefficient of in is nonzero, and , then for some ).
The grid-reduction lemma reduces degrees without changing the values on the grid (Reducing modulo lowers each below , preserves the values on the grid, and preserves any top-degree coefficient whose exponents stay below the grid sizes).
Verification
The total degree of is , and the coefficient of is . Since and , [L1] applies.
Indeed , so the theorem's conclusion is visible directly.
Reducing modulo replaces by , so on the grid the polynomial agrees with , illustrating [L2].
Depends on
- Alon's Combinatorial Nullstellensatz: if $\deg f=\sum_it_i$, the coefficient of $x_1^{t_1}\cdots x_n^{t_n}$ in $f$ is nonzero, and $\lvert S_i\rvert>t_i$, then $f(s_1,\dots,s_n)\ne0$ for some $s_i\in S_i$
- Reducing $f$ modulo $g_i(x_i)=\prod_{s\in S_i}(x_i-s)$ lowers each $\deg_{x_i}$ below $\lvert S_i\rvert$, preserves the values on the grid, and preserves any top-degree coefficient whose exponents stay below the grid sizes
- Monomials, coefficients, degree in each variable and total degree in $F[x_1,\dots,x_n]$
- If $\deg_{x_i}P<\lvert S_i\rvert$ for each $i$ and $P$ vanishes on $S_1\times\cdots\times S_n$, then $P=0$
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
15 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
- N. Alon, Combinatorial Nullstellensatz (standard reference, not scraped)