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For every      n ≥  3 n \geq 3       there exists a finite nonabelian algebra whose congruence lattice has      M n {M_n}       as a      0 , 1 0, 1      -sublattice. This answers a question of R. McKenzie and D. Hobby.

$M\sb n$ as a $0,1$-sublattice of ${\rm Con}\,A$ does not force the term condition