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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)
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The cut at an extremal element is closed under ff

Statement

Let (P,)(P, \le) be a chain-complete poset, f:PPf : P \to P progressive, MM the smallest admissible set, and xMx \in M extremal (Extremal element and its cut (Bourbaki–Witt)). Then the cut MxM_x satisfies f(y)Mxf(y) \in M_x for every yMxy \in M_x.

Facts & Assumptions

Given: A chain-complete poset (P,)(P, \le), a progressive f:PPf : P \to P, the smallest admissible set MM, an extremal xMx \in M, and an element yMxy \in M_x.

[A1]

xx is extremal: for every zMz \in M with z<xz < x, f(z)xf(z) \le x (Extremal element and its cut (Bourbaki–Witt)).

[A2]

Mx={zM:zx or f(x)z}M_x = \{z \in M : z \le x \text{ or } f(x) \le z\} (Extremal element and its cut (Bourbaki–Witt)).

[L1]

MM is admissible, so it is closed under ff and under suprema of its chains (A smallest admissible set exists, Admissible subset (Bourbaki–Witt)).

[L2]

ff is progressive: zf(z)z \le f(z) for every zPz \in P (Chain-complete poset).

[L3]

\le is a partial order: it is reflexive (uuu \le u) and transitive (uvu \le v and vwv \le w imply uwu \le w), and its strict form u<vu < v means uvu \le v together with uvu \ne v (Partial order and partially ordered set).

Proof

technique · cases
1.1

Since yMxy \in M_x we have yMy \in M, and MM is closed under ff, so f(y)Mf(y) \in M.

A2L1
1.2

Membership of MxM_x gives yxy \le x or f(x)yf(x) \le y, and the relation yxy \le x holds exactly when y<xy < x or y=xy = x, by the definition of the strict order.

A2L3
1.3

Suppose y<xy < x.

assume-case below
1.4

Suppose y=xy = x.

assume-case equal
1.5

Suppose f(x)yf(x) \le y.

assume-case above
2.1

In the case y<xy < x, extremality of xx gives f(y)xf(y) \le x, so f(y)f(y) lies in MM and satisfies f(y)xf(y) \le x, hence f(y)Mxf(y) \in M_x.

step 1.3A1step 1.1A2
2.2

In the case y=xy = x, we get f(y)=f(x)f(y) = f(x), and f(x)f(x)f(x) \le f(x) by reflexivity, so f(y)Mf(y) \in M satisfies the second alternative, hence f(y)Mxf(y) \in M_x.

step 1.4step 1.1A2L3
2.3

In the case f(x)yf(x) \le y, progressivity gives yf(y)y \le f(y), so f(x)f(y)f(x) \le f(y) by transitivity, hence f(y)Mxf(y) \in M_x.

step 1.5L2step 1.1A2L3
3.1

The three cases cover every yMxy \in M_x, and each yields f(y)Mxf(y) \in M_x.

step 1.2step 2.1step 2.2step 2.3cases-exhaustive

Remarks

  • The case y=xy = x is the one that explains the shape of the cut. It is precisely why MxM_x is defined with f(x)zf(x) \le z rather than x<zx < z: the image f(x)f(x) must itself land inside MxM_x, and it does so on the upper side.
  • Extremality of xx is used only in the first case, and it is exactly what stops ff from carrying an element from strictly below xx into the forbidden zone strictly between xx and f(x)f(x). That zone is what the cut omits, and keeping it empty of elements of MM is what eventually makes MM a chain (The smallest admissible set is a chain).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 7 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources