ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generated

How 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.

Empty and singleton RSK boundaries

Example

For $n = 0$ : the only word in $X_{0}$ is the empty word, it corresponds to the pair $(\emptyset, \emptyset)$ of empty tableaux, and the hook length formula reads $f^{\emptyset} = 0! / 1 = 1$ . For $n = 1$ : the only word is $(1)$ ; row insertion gives $P = [1]$ and the recording tableau $Q = [1]$ , so $X_{1}$ corresponds to the single pair $([1], [1])$ , and $f^{(1)} = 1! / 1 = 1$ . At $n = 0$ the removal recursion is not asserted, its index set $Rem (\emptyset) = \emptyset$ being empty, and $f^{\emptyset} = 1$ is the convention; at $n = 1$ it reads $f^{(1)} = f^{\emptyset}$ .

Facts & Assumptions

Given: The sets $X_{0}$ and $X_{1}$ of words, the empty tableau $\emptyset$ , and the hook products $P (\emptyset) = 1$ and $P ((1)) = 1$ .

[L1]

For $n \geq 0$ , $X_{n}$ is the set of words $(w_{1}, \dots, w_{n})$ of pairwise distinct real numbers with ${w_{1}, \dots, w_{n}} = {1, \dots, n}$ ; the empty word is the unique element of $X_{0}$ , and $X_{1} = {(1)}$ ; the Robinson-Schensted map is a bijection from $X_{n}$ onto the pairs of standard tableaux of common shape $λ ⊢ n$ (The Robinson-Schensted correspondence).

[L2]

The empty word inserts to the empty tableau; row-inserting the single letter $1$ into $\emptyset$ appends it in the only box, and the recording tableau carries the label $1$ in that box (Row insertion and the bumping route, The Robinson-Schensted correspondence).

[L3]

The hook product $P (λ) = \prod_{x \in [λ]} h (x)$ is the empty product $1$ for $λ = \emptyset$ , and $P ((1)) = h (1, 1) = 1$ ; the hook length formula reads $f^{λ} = n! / P (λ)$ for $λ ⊢ n \geq 0$ , so $f^{\emptyset} = 0! / 1$ and $f^{(1)} = 1! / 1$ (Hook, arm, leg, and hook length of a box, The hook length formula).

[L4]

For $λ ⊢ n$ with $n \geq 1$ , $f^{λ} = \sum_{x \in Rem (λ)} f^{λ - x}$ ; at $n = 0$ the index set $Rem (\emptyset) = \emptyset$ is empty and the recursion is not asserted, the value $f^{\emptyset} = 1$ being the convention for the unique empty tableau, while $Rem ((1)) = {(1, 1)}$ and $(1) - (1, 1) = \emptyset$ (The removal recursion for standard tableaux, Hook, arm, leg, and hook length of a box).

Verification

technique · direct

1.1L1L2given

( $n = 0$ pair.) The empty word inserts no letters, so $P (\emptyset) = \emptyset$ ; no box is ever added, so the recording tableau is $\emptyset$ as well; hence the unique element of $X_{0}$ corresponds under [L1] to the pair $(\emptyset, \emptyset)$ of standard tableaux of the common shape $\emptyset ⊢ 0$ .

1.2L1L2given

( $n = 1$ pair.) The set $X_{1}$ has the single word $(1)$ ; inserting $1$ into the empty tableau appends it in the only box, so $P = [1]$ , and the recording tableau carries $1$ in that box, so $Q = [1]$ ; hence $X_{1}$ corresponds to the single pair $([1], [1])$ of standard tableaux of shape $(1)$ .

2.1L3step 1.1algebra

( $n = 0$ hook formula.) $λ = \emptyset$ has no boxes, so its hook product is the empty product $P (\emptyset) = 1$ , and [L3] gives $f^{\emptyset} = 0! / 1 = 1$ , the number of standard $\emptyset$ -tableaux (the empty tableau alone), in agreement with the single pair of step 1.1.

2.2L3step 1.2algebra

( $n = 1$ hook formula.) For $λ = (1)$ the unique hook length is $h (1, 1) = 1 - 1 + 1 - 1 + 1 = 1$ , so $P ((1)) = 1$ and [L3] gives $f^{(1)} = 1! / 1 = 1$ , in agreement with the single pair of step 1.2.

3.1L4L3step 2.1

( $n = 0$ removal recursion.) The empty partition has no removable node, so $Rem (\emptyset) = \emptyset$ and the recursion of [L4] is not asserted at $n = 0$ ; the convention $f^{\emptyset} = 1$ of [L3] is consistent with the count of one empty tableau.

4.1L4step 2.2step 2.1∎

( $n = 1$ removal recursion.) $Rem ((1)) = {(1, 1)}$ and $(1) - (1, 1) = \emptyset$ , so the recursion of [L4] reads $f^{(1)} = f^{\emptyset} = 1$ , which matches steps 1.2 and 2.2.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

David A. Craven, Groups, Geometries and Representation Theory (Spring Term 2013 lecture notes, 42 pp.) (standard reference, not scraped)
C. Schensted, Longest Increasing and Decreasing Subsequences, Canadian Journal of Mathematics 13 (1961), 179-191 (13 pp.) (standard reference, not scraped)
Charlotte Chan, Representation Theory of Symmetric Groups (Oxford Hilary Term 2011 lecture notes, 40 PDF pp.) (standard reference, not scraped)