regular sequence $\iff$ complete intersection

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Let $k$ be a field and $X \subseteq \mathbb{P}^n$ a closed subscheme of dimension $n-r$. We say $X$ is complete intersection if $X = V(I)$, where $I$ is a homogeneous ideal which is generated by $r$ elements.

Next, for a closed immersion of locally noetherian schemes $f : X \to Y$, we say that $f$ is a regular immersion of codimension $r$ if at every point $x \in X$, the kernel of $f^\#_x : \mathscr{O}_{Y, f(x)} \to \mathscr{O}_{X,x}$ is generated by a regular sequence of $r$ elements.

Now for $Y = \mathbb{P}^n$ and $i: X \to Y$, are these two conditions equivalent? In example 3.5. of Liu's "Algebraic geometry and arithmetic curves", the author says that a complete intersection is a regular immersion in this situation.

To show this, I think it's sufficient to show that:

Let $A$ be a noetherian local integral domain of dimension $n$, $I$ a proper ideal and suppose that $\dim A/I = n-r$. Then if $I$ is generated by $r$ elements, then $I$ is generated by a regular sequence of $r$ elements.

For our purpose we can assume that $A$ is regular and is finitely generated over a field.

I showed that for a generator $f_1, \dots, f_r$, $\dim A/(f_1, \dots, f_i) = n-i$. So $f_{i+1}$ in $A/(f_1, \dots, f_i)$ is not contained in any minimal prime ideal. Therefore if these $A/(f_1, \dots, f_i)$ are reduced, this sequence is a regular sequence. And by this, it seems that we can assume these rings reduced.

Thank you very much.