CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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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.

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The diamond $M_3$ and pentagon $N_5$ violate distributivity by explicit joins and meets

Statement refuted

Every finite lattice is distributive.

Facts & Assumptions

Given: The diamond $M_3=\{0,1,a,b,c\}$ , where $a,b,c$ are incomparable atoms, and the pentagon $N_5=\{0,a,b,c,1\}$ , where $0<a<b<1$ , $0<c<1$ , and $c$ is incomparable with $a,b$ .

[F1]

Distributivity requires $x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$ for all elements (Lattices, distributive lattices, and order ideals).

Counterexample

technique · direct

1.1

In $M_3$ , one has $b\vee c=1$ , $a\wedge b=0$ , and $a\wedge c=0$ . Therefore $a\wedge(b\vee c)=a$ , while $(a\wedge b)\vee(a\wedge c)=0$ .

givenF1

1.2

In $N_5$ , one has $a\vee c=1$ , $b\wedge a=a$ , and $b\wedge c=0$ . Therefore $b\wedge(a\vee c)=b$ , while $(b\wedge a)\vee(b\wedge c)=a$ .

givenF1

2.1

Since $a\ne0$ in $M_3$ and $a\ne b$ in $N_5$ , each lattice violates the distributive identity in [F1]. Both are finite, so either one refutes the Statement.

step 1.1step 1.2F1∎

Remarks

Depends on

Lattices, distributive lattices, and order ideals

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 2 results over 2 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

MIT OpenCourseWare 18.212, Lecture 16: Distributive lattices (standard reference, not scraped)

The diamond M3M_3M3​ and pentagon N5N_5N5​ violate distributivity by explicit joins and meets