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CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31
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The diamond M3 and pentagon N5 violate distributivity by explicit joins and meets

Statement refuted

Every finite lattice is distributive.

Facts & Assumptions

Given: The diamond M3={0,1,a,b,c}, where a,b,c are incomparable atoms, and the pentagon N5={0,a,b,c,1}, where 0<a<b<1, 0<c<1, and c is incomparable with a,b.

[F1]

Distributivity requires x∧(y∨z)=(x∧y)∨(x∧z) for all elements (Lattices, distributive lattices, and order ideals).

Counterexample

technique · direct
1.1

In M3, one has b∨c=1, a∧b=0, and a∧c=0. Therefore a∧(b∨c)=a, while (a∧b)∨(a∧c)=0.

givenF1
1.2

In N5, one has a∨c=1, b∧a=a, and b∧c=0. Therefore b∧(a∨c)=b, while (b∧a)∨(b∧c)=a.

givenF1
2.1

Since a≠0 in M3 and a≠b in N5, each lattice violates the distributive identity in [F1]. Both are finite, so either one refutes the Statement.

step 1.1step 1.2F1∎

Remarks


\node[font=\normalsize] at (2.8,3.35) {$N_5$}; \node[elt] (n0) at (2.2,0) {$0$}; \node[elt] (na) at (1.5,.95) {$a$}; \node[elt] (nb) at (1.5,2.05) {$b$}; \node[elt] (nc) at (4.1,1.45) {$c$}; \node[elt] (n1) at (2.2,2.9) {$1$}; \draw[gray!75] (n0)--(na)--(nb)--(n1) (n0)--(nc)--(n1); \end{tikzpicture} ```

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Used by

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Dependency tree · two levels

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Sources