Number of functions in basis of $M$. Consider all monotonic Boolean functions (for all $a \preceq b , f(a) \le f(b)$)
I read the fact that number of functions in basis of $M$ more than $2$, but less than $5$. It's easy to see that number more than two, because basis should contain at least $0$ and $1$ ( because other function $f \in T_{0} \cap T_{1}$). But I can't see why there is no more than 4 functions in basis.
I tried to make a table of functions with Boolean classes ($T_{0}, T_{1}, M, S, L$, where $S$ - self-dual functions , $L$ - linear functions , $T_{0}$ and $T_{1}$ are saves zero and one functions). And make basis with five functions , but there is no any contradiction. Any help?
Basis with 3 elements : $xy \vee xz \vee yz, 0, 1$. With 4 elements : $xy, x \vee y, 0, 1$.
 A: The general idea here is the same as the one for showing that every basis of $P_2$ contains at most $4$ functions. We use Post's completeness criterion (connected to five maximal classes) to get five functions generating $P_2$ and then show that from these functions we can always produce the generating family consisting of at most four functions.
To employ this idea here we need to introduce additional closed classes contained in $M$. Namely, let $D = [0, 1, x \vee y], K = [0, 1, xy]$. Now, assume $M = [F]$, for some set  $F$ of boolean functions. As you have already noted $0, 1 \in F$. 
Since both $K$ and $D$ are strictly contained in $M$, there must be some functions $f_D \in F \setminus D$ and $f_K \in F \setminus K$, because otherwise, e.g., if $F \subseteq D$, we would have $M = [F] \subseteq [D] = D$, a contradiction. 
Now, $D$ consists of two constants $0, 1$ and all the functions of the form $x_{i_1} \vee x_{i_2} \vee \dots \vee x_{i_n}$. Without loss of generality assume that $f_D(x_1, \dots, x_n)$ essentially depends on all of its variables (note that $n \geqslant 2$). Since $f_D \notin D$ there must be a vector $\alpha \in B^n$ with exactly one $1$ such that $f_D(\alpha) = 0$, because if $f_D(1, 0, \dots, 0) = f_D(0, 1, 0, \dots, 0) = \dots = f_D(0, 0, \dots, 0, 1) = 1$, then by monotonicity $f_D(\alpha) = 1$ for all $\alpha \in B^n \setminus \{(0, 0, \dots, 0)\}$, and hence either $f_D = 1$ or $f_D = x_1 \vee x_2 \vee \dots \vee x_n$, a contradiction. Again, w.l.o.g. assume that $f_D(1, 0, \dots, 0) = 0$. Since $f_D \in M$ and essentially depends on $x_1$, there must be some $a_2, \dots, a_n \in B$ such that $f_D(x_1, a_2, \dots, a_n) = x_1$. Note that $(a_2, \dots, a_n) \neq (0, \dots, 0)$ because $f_D(1, 0, \dots, 0) = 0$. Now, assume that $a_2 = \dots = a_t = 1, a_{t+1} = \dots = a_n = 0$. Substituting $x_1$ for $x$, all the variables $x_2, \dots, x_t$ for $y$ and $x_{t+1}, \dots, x_n$ for $0$ we obtain $f_D(x, y, \dots, y, 0, \dots, 0) = xy.$ Generally, if $a_i = 1$ for $i \in I \subseteq \{2, \dots, n\}$ and $a_i = 0$ otherwise, then substitute $x_i$ for $y$ if $i \in I$ and for $0$ otherwise. This shows that $xy \in [0, f_D]$. Dually you can show that $x \vee y \in [1, f_K]$.
Therefore we have $0, 1, xy, x \vee y \in [0, 1, f_D, f_K]$. So we need at most $4$ functions $0, 1, f_D, f_K$ from $F$ to obtain a generating family $\{0, 1, xy, x\vee y\}$ of $M$.
Moreover, now it isn't difficult to show that $D, K, M_0, M_1$ (where $M_0 = M \cap T_0, M_1 = M \cap T_1$) are the only maximal classes in $M$. The proof is very similar to the proof of the fact that $L, M, S, T_0, T_1$ are the only maximal classes in $P_2$. Indeed, it is easy to check that none of these classes is contained in the remaining ones. Let $N \in \{D, K, M_0, M_1\}$ and $f \in M \setminus N$, then $N \cup \{f\}$ is not contained in $D, K, M_0, M_1$, so there are $0, 1, f_D, f_K \in N \cup \{f\}$ and hence, as we have shown, $M = [0, 1, xy, x \vee y] \subseteq [0, 1, f_D, f_K] \subseteq [N, f]$, so $[N, f] = M$ and $N$ is maximal in $M$. Now assume $F \subseteq M$ is maximal, then it must be contained in (and hence be equal to) one of the classes $D, K, M_0, M_1$, because otherwise there are $0, 1, f_D, f_K \in F$ and so $[F] = M$, a contradiction.
Here you can check the Post's lattice to find those classes (in this Hasse diagram they are denoted by $\Lambda, \text{V}, MP_0, MP_1$ respectively) and to make sure that they are indeed the only maximal classes in $M$.
