Lattice breadth $k$ same as having Boolean sublattice of $2^k$ elements?

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The breadth of a lattice is the largest integer $n$ such that any join of elements $X=\{x_1,x_2,\ldots,x_{n+1}\}$ is join of a proper subset of $X$.

Birkhoff's classical book has an exercise: "Show that the smallest lattice with breadth $k$ is the Boolean lattice of $2^k$ elements."

I think that a little more can be said: A lattice with breadth $k$ contains a sublattice isomorphic to Boolean lattice of $2^k$ elements, but not of $2^{k+1}$ elements. But is this false, true and already said somewhere, or true and not already explicitly said?

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The answer is yes and already said. Varieties generated by lattices of breadth two by Jörg Stephan (Order, June 1993) says this in page two.