All non-distributive lattices of 6 elements I'm struggling to find all non-distributive lattices of 6 elements. I looked for those, that have at least 1 sub-lattice isomorphic to M3 or N5 and found some, but I don't know how to guarantee that i found all of them.
Thanks in advance, any help appreciated.
 A: (Apologies: I previously misremembered the names and characterization of nondistributive lattices. I made corrections.)
Being nondistributive, the lattice must contain either $N_5$ or $M_3$ as sublattices.
Case 1: contains a copy of $M_3$
If the added point is the greatest element, your picture is just $M_3$ with a new point above the top of the diamond. Similarly if you are adding the smallest element, it's $M_3$ with a point below the diamond.
Otherwise, the top and bottom of the diamond must be the greatest and least elements. The added point must be between them, somehow.
You could add it so that it becomes like the three existing middle points. Otherwise, it would have to be above or below one of the three existing middle points.
It cannot be above two existing middle points at once: that would spoil the fact that the join of those two elements is the greatest element. Similarly it cannot be below two of the middle elements.
It also cannot be both above one middle element and below another, since that would spoil the join of those two elements.
So the new point can be between a middle point and the greatest element, or between the least point and an element, and if you think about this, either option produces the same lattice (a pentagon within $M_3$.) So up to this point, we've only discovered four nonisomorphic lattices.
Case 2: contains $N_5$ but not $M_3$
Refer to this diagram:
.
Again, you could add a new greatest element or a new least element.
Adding a point which is between $0$ and $1$ (and has no relations to the other points) creates a graph with $M_3$ in it, so we already counted it.
Putting the point between $a$ and $1$ or between $a$ and $0$ produces (up to isomorphism) a single new type of graph we haven't seen.
Putting a new point anywhere beween $1, b,c,0$ produces a single graph (which we haven't seen.)
Again, a point cannot be added so that is simultaneously above $a$ and one of $b$ or $c$ because that invalidates our declaration that $1$ is the least upper bound of the two, and similarly it can't be simultaneously below $a$ and one of $b$ or $c$ because that invalidates the fact that $0$ is the greatest lower bound of $a$ and $b$ or $c$.
So in this path, we've discovered four new graphs.
In total, we have $8$ distinct possibilities.
