Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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  • 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.
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A coordinate scaling and a coordinate transposition send the unit cube to a set of measure equal to the absolute value of the determinant

Statement

Let n1 and assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Work with real matrices and identify a matrix with the linear map it defines by (Ax)i=j<naijxj (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0).

  1. Coordinate scaling. Let p<n, let c0 be real and let Dp(c) be the elementary matrix obtained from the identity by multiplying row p by c (Elementary matrices obtained by applying one elementary row operation to an identity matrix). Then Dp(c) sends x to the point whose p-th coordinate is cxp and whose other coordinates are those of x, the image Dp(c)[(0,1]n] is Lebesgue measurable, and λn(Dp(c)[(0,1]n])  =  c  =  detDp(c).
  2. Coordinate transposition. Let n2, let pq be below n and let Epq be the elementary matrix interchanging rows p and q. Then Epq exchanges the p-th and q-th coordinates, Epq[(0,1]n]=(0,1]n, and λn(Epq[(0,1]n])  =  1  =  detEpq.

Facts & Assumptions

Given: A natural number n1, the Axiom of Countable Choice, and the elementary matrices Dp(c) and Epq over R.

[L1]

If aibi are real for i<n, then any box obtained from the coordinate interval product i<n[ai,bi] by independently choosing for each endpoint whether it is included has Lebesgue measure i<n(biai) (A box in Rn with parameters aibi is Lebesgue measurable of measure i<n(biai), whichever of its faces are included). In particular (u,v]n=B(u,v) (Half-open boxes in Rn and their volume).

[F1]

An elementary matrix is a matrix obtained by applying one elementary row operation to the identity matrix In; there are three types: Epq interchanges rows p and q; Dp(c) multiplies row p by c0; and Tpq(c) adds c times row q to the distinct row p (Elementary matrices obtained by applying one elementary row operation to an identity matrix, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes).

[F2]

Let n1 and let AMn(R) be a matrix over a commutative ring; interchanging two rows changes det(A) to det(A), and multiplying one row by any cR changes it to cdet(A) (For every square matrix, including singular ones, a row swap negates the determinant, scaling a row by any scalar scales it, and row addition leaves it unchanged, claims 1 and 2; For n1, the determinant over a commutative ring by the Leibniz formula, and detA for a real matrix).

[F3]

If A is upper or lower triangular over a commutative ring, with n1, then det(A)=i<naii (The determinant of a triangular matrix is the product of its diagonal entries).

[F4]

For every linear L:RmRn there is a unique matrix A such that (Lh)i=j<maijhj (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0).

[F5]

The absolute value satisfies c>0 for c0, c=c for c0 and c=c for c0 (Absolute value in an ordered field, Basic properties of the absolute value).

Proof

technique · direct
1.1

The identity matrix is triangular with every diagonal entry 1, so detIn=1; the row-operation table applied to In then gives detDp(c)=c and detEpq=1, hence detDp(c)=c and detEpq=1.

F1F2F3F5
1.2

Reading off the matrix entries, Dp(c) sends x to the point with p-th coordinate cxp and the other coordinates unchanged, and Epq sends x to the point with p-th coordinate xq, q-th coordinate xp and the others unchanged.

F1F4
2.1

For claim 1, Dp(c)[(0,1]n]={x:0<xi1 for ip, xpc(0,1]}. When c>0 this is the half-open box with p-th side (0,c]; when c<0 it is the box with p-th side [c,0) and all other sides (0,1]. In either case [L1] gives Lebesgue measurability and measure i<n(biai)=c.

step 1.2L1F5
2.2

For claim 2, Epq restricts to a bijection of (0,1]n onto itself, since exchanging two coordinates of a point all of whose coordinates lie in (0,1] again gives such a point and the map is its own inverse; hence the image is (0,1]n, of measure 1.

step 1.2L1
3.1

Steps 1.1, 2.1 and 2.2 are the two claims.

step 1.1step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

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Sources