How do these additional cases fit into this Theorem about the optimal substructure of a longest common subsequence? Theorem 15.1 (Optimal Substructure of an LCS)

Theorem Let the $X=(x_1,x_2,\dots,x_m)$ and $Y=(y_1,y_2,\dots,y_n)$ be sequences, and let $Z =(z_1,z_2,\dots,z_k)$ be any LCS.

If $x_m = y_n$, then $z_k = x_m = y_n$ and $Z_{k-1}$ is an LCS of $X_{m-1}$ and $Y_{n-1}$.

If $x_m \neq y_n$, then $z_k \neq x_m$  implies that Z is an LCS of $X_{m-1}$ and Y.

If $x_m \neq y_n$, then $z_k \neq y_n$ implies that Z is an LCS of X and $Y_{n-1}$.



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The above cases are intuitive and easily provable.
I do not understand additional cases not explicitly mentioned in the theorem:
What happens here:
$x_m \neq y_n$ and $z_k \neq x_m \neq y_n$
The above statement combines cases 2 and 3. Does that mean Z is an LCS of both LCS of $X_{m-1}$ and Y AND X and $Y_{n-1}$? That doesn't make sense to me.
If $x_m = y_n$ then any longest common sequence must end with $x_m = y_n$. This is what the first item covers.
If $x_m \neq y_n$, then there are three types of longest common sequences:

Longest common sequences ending in $x_m$.
Longest common sequences ending in $y_n$.
Longest common sequences ending in neither.

These are all the cases. There is no other case.