Proposition 1: Let $R$ be a commutative, Noetherian ring and $\ \mathfrak{p} \in \operatorname{Spec}(R)$. If $\operatorname{ht}(\mathfrak{p})=h$, then there exist $y_1, \ldots, y_h \in R$ such that $\ \mathfrak{p} \in \operatorname{Min}((y_1, \ldots, y_h))$.
I followed this proof (Proposition 12, pg 5). The construction of the $y_i$ involves picking elements in $\mathfrak{p} \setminus \cup Q$, where $Q$ are the minimal primes of $R$. As a corollary, we get:
Proposition 2: Let $(R, \mathfrak{m})$ be a commutative, Noetherian, local ring of Krull dimension $d$. Then,
$$\dim R = \min \{c \mid \exists x_1, \ldots, x_c \in \mathfrak{m}, \ \sqrt{(x_1, \ldots, x_c)} = \mathfrak{m} \}.$$
Definition: Any $x_1, \ldots, x_d$ that verify $\sqrt{(x_1, \ldots, x_d)} = \mathfrak{m}$ are called a system of parameters (sop). An element $x$ is called a parameter if it belongs to a system of parameters. Proposition 1 guarantees the existence of a sop.
Thanks to the proof of Proposition 1, we know that every element of $\mathfrak{m} \setminus \cup Q$, where $Q$ is a minimal prime of $R$, is a parameter.
Question: Is every parameter also in $\mathfrak{m} \setminus \cup Q$ ? I get the feeling it is true but I'm not sure.
I've tried proving it straight away, picking a sop $x_1, \ldots, x_d$ such that $x_1 \in Q$, with $Q$ a minimal prime of $R$, and then using the sop given to us by Proposition 1, $y_1, \ldots, y_d$, such that no parameter is in the union of the minimal primes, but didn't go far.