If $R$ is a noetherian local domain which is catenary, and $a_1,...,a_n$ are elements of the maximal ideal of $R$ with $\operatorname{height}(a_1,...,a_n)=n$, could we conclude that $\operatorname{height}(a_1,...,a_i)=i$ for each $i$ with $1≤i≤n$?
Being a domain is a necessary condition, as could be seen for $$R=K[X,Y,Z]/{(X,Y)∩(Z)}=K[x,y,z],$$ where $\operatorname{height}(x,y+z)=2$ and $\operatorname{height}(x)=0$. Thanks in advance.
Proposition: $\DeclareMathOperator{\ht}{ht}$If $R$ is a catenary local domain, and $I$ is an $R$-ideal, then $\ht I + \dim R/I = \dim R$.
Proposition: If $(R, m)$ is Noetherian local, a sequence of elements $a_1, \ldots, a_d \in m$ ($d = \dim R$) form a system of parameters iff $\dim R/(a_1, \ldots, a_i) = \dim R - i$ for each $i = 1, \ldots, d$.
Thus in a Noetherian catenary local domain $R$, elements $a_1, \ldots a_n \in m$ with $n \le \dim R$ satisfy $\ht (a_1, \ldots, a_i) = i$ for each $i = 1, \ldots, n$ iff $a_1, \ldots, a_n$ form part of a system of parameters.
Edit: I had initially overlooked the assumption that $\ht(a_1, \ldots, a_n) = n$. This does imply that $a_1, \ldots, a_n$ is part of a system of parameters, as follows: by the first proposition above, $\dim R/(a_1, \ldots, a_n) = \dim R - n =: d - n$. Choose a s.o.p. of $R/(a_1, \ldots, a_n)$, necessarily of length $d - n$, say $\overline{b_1}, \ldots, \overline{b_{d-n}}$. Then $(\overline{b_1}, \ldots, \overline{b_{d-n}})$ is $\overline{m}$-primary, so $(a_1, \ldots, a_n, b_1, \ldots, b_{d-n})$ is $m$-primary, so $a_1, \ldots, a_n, b_1, \ldots, b_{d-n}$ is an s.o.p. of $R$.