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Consider Boolean function $f \in P_{2}$ with n variables. It's easy to show that any function could be represented as sum of disjunctive terms. Consider $L(f)$ is number of variables uses in minimal (with minimal quantity of variables in formula) DNF that describes function $f$.

Actually I'm interested in $\max_{f\in P_{2}} L(f)$. I guess $L_{max}(f) \le n2^{n-1}$. I build some forms , when $n = 2,3$: $x_{1}\bar{x_{2}} \vee \bar{x_{1}}x_{2}$ and so on. Actually I think that bound maximum could be proved by induction. But I don't know connection between $n$ and $n+1$. Any hints.

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Hint: Let us introduce additional function $l(n) = \max_{f \in P^{(n)}_2} l(f)$, where $l(f)$ is the length of $f$, that is, the number of disjunctive terms in the shortest DNF of $f$. It is easy to see that $L(f) \leqslant n \cdot l(f)$, since any disjunctive term contains at most $n$ variables.

Show that $l(n) \leqslant 2^{n-1}$ (and hence $L(n) \leqslant n2^{n-1}$) using induction on the number of variables $n$. The connection between $n+1$ and $n$ is established via the following decomposition $$ f(x_1, \dots, x_n, x_{n+1}) = \bar{x}_{n+1}f(x_1, \dots, x_n, 0) \vee x_{n+1}f(x_1, \dots, x_n, 1). $$

It is actually known that $l(n) = 2^{n-1}$ and $L(n) = n2^{n-1}$. The lower bound is given by the linear function $f(x_1, \dots, x_n) = x_1 \oplus x_2 \dots \oplus x_n$.

Its perfect DNF is the only DNF of this function (the set of ones $N_f$ of this function is located in the layers of the boolean cube $B^n$ with odd numbers, and so no pair of neighbour points, i.e., $\alpha, \beta \in B^n$ with $\|\beta\| = \|\alpha\| + 1$, belongs to $N_f$) and hence $l(f) = 2^{n-1}$ and $L(f) = n2^{n-1}$.