Zassenhaus Lemma states that, if $A^* \lhd A$ and $B^* \lhd B$, then $$\frac{A^* (A\cap B)}{A^* (A\cap B^* )}\cong \frac{B^* (A\cap B)}{B^* (A^*\cap B)}.$$
I generalized this lemma, and I want to see whether my proof for the generalization is correct or not.
2nd Isomorphism Theorem: If $H \lhd G$ and $K <G$ then $HK/H \cong K/K \cap H$.
Let's generalize this theorem a little bit:
Lemma 1: If $X < N(A)$, where $N(A)$ is the normalizer of $A$, then $AX/A \cong X/(A \cap X)$.
The proof of this is similar to the 2nd Iso. Th.
Lemma 2: Suppose G is a group and $X$, $A$, and $H$ are subgroups of $G$. $X < N(A)$ implies $$\frac{A(X \cap H)}{A} \cong \frac{(A\cap X)(X \cap H)}{(A\cap X)}.$$
Proof:
$$X < N(A) \Rightarrow X \cap H < N(A) \Rightarrow \frac{A(X \cap H)}{A} \cong \frac{X\cap H}{A \cap X \cap H}, \text{by Lemma 1}$$ Also $$X < N(A \cap X) \Rightarrow X \cap H < N(A \cap X) \Rightarrow \frac{(A \cap X)(X \cap H)}{A\cap X} \cong \frac{X\cap H}{A \cap X \cap H}, \text{by Lemma 1}$$
Hence, the result is followed.
Lemma 3 (Edited): Suppose G is a group and $X$, $A$, and $H$ are subgroups of $G$. $X < N(A) \cap N(H)$ implies $$\frac{AX}{A(X\cap H)} \cong \frac{X}{(A\cap X)(X \cap H)}.$$
Proof: Lemma 1 implies $\frac{AX}{A} \cong \frac{X}{A\cap X}$. This and Lemma 2 imply $$\frac{AX/A}{A(X\cap H)/A} \cong \frac{X/(A\cap X)}{(A \cap X)(X \cap H)/(A \cap X)}.$$
Moreover, $X < N(A) \cap N(H)$ implies $X\cap H \lhd X < N(A)$, which, in turn, implies $A(X \cap H) \lhd AX$. In addition, $H \cap X \lhd X$ and $A \cap X \lhd X$ imply $(A \cap X)(H \cap X) \lhd X$.
Now, the result is followed by the 3rd isomorphism theorem.
Generalized Zassenhaus: If $X < N(A) \cap N(B)$, then $$\frac{AX}{A(X\cap B)} \cong \frac{BX}{B(X \cap A)}.$$
Proof: We use Lemma 3 by considering $X < N(A)$ and $H=B$. We have $\frac{AX}{A(X\cap B)} \cong \frac{X}{(A\cap X)(X \cap B)}$. Now, we use Lemma 3 by considering $X < N(B)$ and $H=A$. We have $\frac{BX}{B(X\cap A)} \cong \frac{X}{(B\cap X)(X \cap A)}$. Hence, the result is followed.
The Zassenhaus lemma is derived with the generalized Zassenhaus by considering $X = A \cap B$, $A=A^* $, and $B=B^* $.
Is the proof correct? If it is, have you seen this generalization before? Does this generalization have a value in group theory?