How to determine if given lattice is distributive or not? I am trying to understand how to determine whether given lattice is distributive. I came across following:

A lattice is distributive if and only if none of its sublattices is isomorphic to $M_3$ or $N_5$; a sublattice is a subset that is closed under the meet and join operations of the original lattice. Note that this is not the same as being a subset that is a lattice under the original order (but possibly with different join and meet operations).


               L1                                                    L2
Above are distributive lattices which contains $N_5$ (solid lines, left) and $M_3$ (right) as subset, but not as sublattice, respectively
  On left, $b∧e=f∉N_5$
  On right, $d∧e=b∉M_3$

Doubts: 


*

*I was able to understand that $f∉N_5$, but I dont think its correct to say $b∉M_3$. So does $M_3$ indeed a sub lattice of L2 and hence its not distributive?

*Also have doubt about definition of sub lattice. This gives definition of lattice as follows:

A sublattice of a lattice L is a nonempty subset of L that is a lattice with the same meet and join operations as L.
That is if L is a lattice and M≠∅ is a subset of L such that for every pair of elements a, b in M both a ∧ b and a ∨ b are in M, then M is a sublattice of L

What does it mean by "same meet nad join operations"? Does the logic / nature of operation should be the same or for each pair of elements in M, their meet and join should be the same as that in L and should also be present in M? Because in L2, $d\wedge e=b$ but in corresponding $M_3$, $d\wedge e=c$. Is this why $M_3$ not sublattice of L2?
 A: The diagrams of $L_1$ and $L_2$ are misleading representations of lattices since they are not Hasse diagrams.   
This is because in $L_1$ you have one line to much: the one from $c$ to $b$.
It is redundant because $c < f < b$, whence $c < b$ follows from transitivity.
Erase that line and it becomes obvious that the lattice is distributive (up to isomorphism, it is the lattice $\mathbf 3 \times \mathbf 2$).
Likewise in $L_2$ you have three lines to much: between $c$ and $d$; between $c$ and $e$; and between $b$ and $a$.
Again, erase those and everything becomes clear.
In the sub-lattice definition, the meaning of "same meet and join operations" is that if $a$ and $b$ are in the subset which we are testing if it is a sub-lattice, then for that set to be a sub-lattice, $a \wedge b$ and $a \vee b$ must belong to that set, where these operations are calculated in the main lattice.
So you can have a subset which is a lattice on its own, but not a sub-lattice (for example, in the lattice $L_2$ above, erase $b$, and the resulting subset is a lattice, but it is not a sub-lattice of $L_2$ because $d \wedge e = b$, which doesn't belong to the resulting subset; it is still a lattice where $d \wedge e = c$).
