polynomial grade Hamilton-Marley in the paper "Non-Noetherian Cohen–Macaulay rings" have   
 
I can't understand highlighted part. my attempt is:
$$\text{p-grade} ((x')R',R')=\text{p-grade} ((x')R',R/(x')R)=\text{p-grade} ((x')R',R)-\ell +1.$$ So I should prove now $\text{p-grade} ((x')R',R)=\ell$.  
Thank you.

Here are definition and 2.6(a):  


I think this theorem in Northcott's book Finite free resolutions (Chapter 5, Theorem 19) might be of help:
Let $f \colon R \to S$ be a ring homomorphism, $I$ an ideal of $R$ and $M$ an $S$-module. Then $\text{p.grade}_{R}(I,M) = \text{p.grade}_{S}(IS,M)$.
Now let $I = (\textbf{x})R$ and $M = R'$. Then observe that $IR' = (x_{\ell})R'$, so that by the above theorem, we have that $\text{p.grade}_{R}((\textbf{x})R, R') = \text{p.grade}_{R'}( (x_{\ell})R', R')$. 
With the same notation, since $\textbf{x}'$ is a regular sequence of length $\ell - 1$ contained in $I = (\textbf{x})R$, by Prop. 2.6(a) we have $\ell = \text{p.grade}_{R}((\textbf{x})R,R) = \text{p.grade}_{R}((\textbf{x})R, R') + \ell - 1 = \text{p.grade}_{R'}((x_{\ell})R',R') + \ell - 1$. 
Hence we see that $\text{p.grade}_{R'}((x_{\ell})R',R') = 1$. 
