# Grobner basis of the toric ideal $I_{A_P}$ with respect to $<_{rev}$ consists of those binomials $t_αt_β − t_{α\cap β} t_{α\cup β}$

I try to understand the proof of the Theorem. 10.1.3.(page 185) from ''Monomial Ideals'' by Herzog & Hibi.

The reduced Grobner basis of the toric ideal $I_{A_P}$ with respect to $<_{rev}$ consists of those binomials $$t_αt_β − t_{α\cap β} t_{α\cup β}$$ with neither α ⊂ β nor β ⊂ α.

Proof.

If α and β are poset ideals of $P$, then each of α∩β and α∪β is a poset ideal of P.

It is clear that the binomial $t_αt_β − t_{α∩β}t_{α∪β}$ belongs to $I_{A_P}$ and, in the case of neither α ⊂ β nor β ⊂ α, its initial monomial is $t_αt_β$.

Once we show that the set of those binomials $t_αt_β −t_{α∩β}t_{α∪β}$ with neither α ⊂ β nor β ⊂ α is a Grobner basis of $I_{A_P}$ with respect to $<_{rev}$, it follows immediately that such a set of binomials is the reduced Grobner basis of $I_{A_P}$ with respect to $<_{rev}$.

Let $G$ denote the reduced Grobner basis of $I_{A_P}$ with respect to $<_{rev}$. Let $$\begin{equation*} \prod_{i=1}^{q} t_{\alpha_i} - \prod_{i=1}^{q} t_{\beta_{i}} \end{equation*}$$ be a binomial belonging to $G$ with $\begin{equation*} \prod_{i=1}^{q} t_{\alpha_i} \end{equation*}$ its initial monomial.

What we must prove is that there are $k$ and $l$ such that one has neither $\alpha_k ⊂ α_l$ nor $α_l ⊂ α_k$.

Why is necessary and sufficiently to prove the above sentence?

I posted this question on MSE also.