Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Everything in MM is comparable to an extremal element

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 Mx=MM_x = M; that is, for every yMy \in M, either yxy \le x or f(x)yf(x) \le y.

Facts & Assumptions

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

[A1]

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

[L1]

MxM_x is closed under ff (The cut at an extremal element is closed under ff).

[L2]

MxM_x is closed under suprema of its chains (The cut at an extremal element is closed under chain suprema).

[L3]

MM is contained in every admissible subset of PP (A smallest admissible set exists).

[L4]

A subset is admissible when it is closed under ff and under suprema of its chains (Admissible subset (Bourbaki–Witt)).

Proof

technique · direct
1.1

MxM_x is closed under ff.

L1
1.2

MxM_x is closed under suprema of its chains.

L2
2.1

So MxM_x is an admissible subset of PP.

step 1.1step 1.2L4
3.1

By minimality of MM, every admissible subset contains MM, so MMxM \subseteq M_x.

step 2.1L3
4.1

Together with MxMM_x \subseteq M this gives Mx=MM_x = M.

step 3.1A1
5.1

Unfolding the definition of MxM_x, every yMy \in M satisfies yxy \le x or f(x)yf(x) \le y.

step 4.1A1

Remarks

  • This is the first payoff of minimality, and the pattern is worth naming: to prove that everything in MM has a property, collect the elements that have it, show the collection is admissible, and let minimality do the rest. The same move proves Every element of MM is extremal.
  • The conclusion is a comparability statement with a gap. It says nothing about elements strictly between xx and f(x)f(x), and indeed the content of the lemma is that MM has none: an element of MM lies at or below xx, or at or above f(x)f(x), never inside.
  • The hypothesis that xx is extremal is doing real work and cannot be dropped. It is what Every element of MM is extremal later supplies for every element, which is what turns this one-sided statement into total comparability (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: 9 results over 7 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