Size of decision tree for f is polynomial in the DNF size of f and CNF size of f I've been having hard time with proving the following claim:
Let $f:\{T,F\}^n\rightarrow \{T,F\}$ be a boolean function. Let $size_{DT}(f)$ denote the number of leaves in the smallest (w.r.t the number of leaves) decision tree for $f$. Also, let $size_{CNF}(f),size_{DNF}(f)$ denote the number of clauses, terms in the minimal $CNF,DNF$ formulas for $f$ respectively.

Prove that $size_{DT}(f)\in poly(n,size_{DNF}(f),size_{CNF}(f))$.
As pointed out by the OP (user13772), this is false.  Jukna et al. constructed explicit Boolean functions $f$ that require deterministic decision trees of size $2^{\Omega(\log^2 N)}$, where $N$ is the number of monomials in a minimal DNF for $f$ and $\lnot f$.  Note that every Boolean function on $n$ inputs can be expressed as a deterministic decision tree of size $2^{O(\log n \log^2 N)}$.


S. Jukna, A. Razborov, P. Savický, I. Wegener, On P versus NP$\cap$co-NP for decision trees and read-once branching programs, Computational Complexity 8 (1999), 357–370.  doi:10.1007/s000370050005  (preprint)