$i,j\geq 1$. Observe that$$\label{equ:H_exp}
H_i(j)\left\{
   \begin{aligned}
   &=H_{i-1}(j),\quad j<f(i);\\
   &=H_{i-1}(j)-1,\quad j=f(i);\\
   &\leq \operatorname{Inc}_{j-1}H_i(j-1),\quad j>f(i).
   \end{aligned}\right.$$ Indeed, the first two equalities directly follow from the definition of $I_i$, and the third inequality is from Theorem \[thm:Macaulay\]. Similarly, we have$$\label{equ:hatH_exp}
\hat{H}_i(j)\left\{
   \begin{aligned}
   &=\hat{H}_{i-1}(j),\quad j<f(i);\\
   &=\hat{H}_{i-1}(j)-1,\quad j=f(i);\\
   &=\operatorname{Inc}_{j-1}\hat{H}_i(j-1),\quad j>f(i).
   \end{aligned}\right.$$ Here, the third one becomes equality since $\hat{I_i}$ is $f(i)$-compressed and $\operatorname{gdeg}(\hat{I_i})=f(i)\leq j-1$ and so the conditions for equality in Theorem \[thm:Macaulay\] is satisfied. On the other hand, we observe that $H_1(f(1))=\hat{H}_1(f(1))=(f(1))_{d-1}-1$. Then it can be inductively proved from equations (\[equ:H\_exp\]) and (\[equ:hatH\_exp\]) that:$$\begin{split}
H_i(j)&=H_{i-1}(j)\leq \hat{H}_{i-1}(j)=\hat{H}_i(j),\quad \text{for}\ j<f(i);\\
H_i(j)&=H_{i-1}(j)-1\leq \hat{H}_{i-1}(j)-1=\hat{H}_i(j),\quad \text{for}\ j=f(i);\\
H_i(j)&\leq \operatorname{Inc}_{j-1}H_i(j-1)\leq\operatorname{Inc}_{j-1}\hat{H}_i(j)=\hat{H}_i(j),\quad \text{for}\ j>f(i).
\end{split}$$ Here, the fact that $r<t\Rightarrow\operatorname{Inc}_j(r)<\operatorname{Inc}_j(t)$ is applied. So we have proved that $H_i(j)\leq\hat{H}_i(j),$ for all $i\geq 1$ and $j\geq 1$. Then $H_N(f(N+1))\leq \hat{H}_N(f(N+1))=0$. We have $m_{N+1}\in\{m|\deg(m)=f(N+1)\}\subseteq\langle \hat{m}_1,...,\hat{m}_n\rangle$, which is contrary to the definition of $\{m_1,\cdots,m_n\}$. $\Box$

We consider to compute the greatest length $N$ using the condition $\hat{H}_N(f(N))=\hat{H}_N(f(N+1))=0$. To this end, we define $\Omega(d-1,f,t)$ to be the number $k$ such that $$\label{equ:Omega}\begin{split}\hat{H}_k(f(k))&=(d\underbrace{-2,d-2,\cdots,d-}_{f(1)-t\ \text{times}}2)_{f(k)}\\
&=(d-2)_{f(k)}+\cdots+(d-2)_{(f(k)-f(1)+t+1)}.\end{split}$$ Then from this definition $N=\Omega(d-1,f,f(1))$. Note that $$\hat{H}_1(f(1))=(f(1))_{d-1}-1=(d\underbrace{-2,d-2,\cdots,d-}_{f(1)\ \text{times}}2)_{f(1)},$$ then $\Omega(d-1,f,0)=1$. Computing $\Omega(d,f,t)$ is presented in the following theorem.

\[thm:Omega\] Given a number $d\in{\mathbb{N}}$, an increasing function $f:{\mathbb{N}}\rightarrow{\mathbb{N}}$, and a number $t\in\{0,1,\cdots,f(1)\}$, $\Omega(d,f,t)$ 