About the height of primes associated to a squarefree monomial ideal $I\subseteq J$ Edit Let $S=K[x_1,\dots,x_n]$ be a polynomial ring in $n$ indeterminates with coefficients in a field $K$. For a monomial ideal $I$ of $S$, $G(I)$ denotes the minimal generating set of $I$. For example, if $I=(x_1x_2,x_2x_3,x_1x_2x_3)$ then $G(I)=\{x_1x_2,x_2x_3\}$, since $x_1x_2x_3$ is not minimal.
Given two squarefree monomial ideals $I,J$ such that $G(I)\subseteq G(J)$, let
\begin{align}
I&=\mathfrak{p}_1\cap\dots\cap\mathfrak{p}_r,\\
J&=\mathfrak{q}_1\cap\dots\cap\mathfrak{q}_s,
\end{align}
be the minimal primary decompositions of $I$ and $J$, respectively.
It is known that each associated prime of $I$ (and $J$) is of the form $(x_{i_1},\dots,x_{i_\ell})$ for some subset $\{i_1,\dots,i_\ell\}\subseteq\{1,\dots,n\}$. Moreover it is known that each associated prime is minimal.
Suppose that all primes $\mathfrak{q}_i$ have the same height: $q=\text{height}(\mathfrak{q}_1)=\ldots=\text{height}(\mathfrak{q}_s)$. Is it true that
$$
\text{height}(\mathfrak{p}_i)\le q,
$$
for all $i=1,\dots,r$?
Many examples seems to suggest this is true. However, I haven't found a way to prove it. I thought that maybe I could use Alexander duality, but didn't progress.
Any help is appreciated, also references to the literature are welcome.

No, this is not true. For a simple example, take $S=K[x_1,x_2,x_3]$, let $J=(x_3)$, and let $I=(x_1,x_2) \cap (x_3)$.  Of course, $I \subseteq J$, but $I$ has a minimal prime of height $2$.
