Krull dimension of local Noetherian ring (2) Let $(A,\mathfrak{m})$ be a local Noetherian ring and $x \in \mathfrak{m}$.

Prove that $\dim(A/xA) \geq \dim(A)-1$, with equality if $x$ is $A$-regular (i.e. multiplication with $x,$ as a map $A\rightarrow A$ is injective). 

The dimensions are Krull dimensions.
It may have something to do with this question Dimension inequality for homomorphisms between noetherian local rings, but I simply can't figure it out.
Thank you.

Prologue
For any commutative ring $R$ and any ideal $I\subsetneq R$ we have $$ \dim(R/I)+ht(I)\leq \dim (R) \quad (*)$$  This  does not assume $R$ noetherian, nor local, nor... but just follows from the definitions.  
Inequality
Suppose now that $(A,\mathfrak m)$ is local noetherian.
The trick is to use that $\dim(A)$ is the smallest number  of elements in $\mathfrak m$ generating an $\mathfrak m$-primary ideal (cf. Atiyah-Macdonald Theorem 11.14).  Let's do that for $A/xA$:
If $\bar x_1,...,\bar x_k\in \mathfrak m/xA$ generate an $\mathfrak m/xA$-primary ideal, then $x, x_1,..., x_k$ generate an $\mathfrak m$-primary ideal and this immediately yields the required inequality $$\dim(A)\leq \dim(A/xA)+1 \quad (**)$$
Equality
The Prologue implies that equality in $(**)$ will hold if $ht(xA)=1$.
The  principal ideal theorem says that we always have $ht(xA)\leq 1$.
Now to say that $ht(xA)=0$ means that $x\in \mathfrak p$ for some minimal ideal $\mathfrak p$.
But it is well known that minimal ideals consist of zero divisors (= non-regular elements).
Hence if $x$ is regular we have $ht(xA)=1$ (since we don't have $ht(xA)=0$ !) and the required equality follows $$ \dim(A)= \dim(A/xA)+1 \quad (***)$$
