Theorem: Product of Quaternion Conjugates

Let $\mathbf x, \mathbf y \in \mathbb H$ be quaternions.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
Then:
:$\overline {\mathbf x \times \mathbf y} = \overline {\mathbf y} \times \overline {\mathbf x}$
but in general:
:$\overline {\mathbf x \times \mathbf y} \ne \overline {\mathbf x} \times \overline {\mathbf y}$

Proof:
Consider the matrix form of $\mathbf x$ and $\mathbf y$:
{{begin-eqn}}
{{eqn | l = \mathbf x
      | r = \begin {bmatrix} a & b \\ -\overline b & \overline a \end {bmatrix}
}}
{{eqn | l = \mathbf y
      | r = \begin {bmatrix} c & d \\ -\overline d & \overline c \end {bmatrix}
}}
{{end-eqn}}
where $a, b, c, d \in \C$.
{{begin-eqn}}
{{eqn | l = \overline {\mathbf x \times \mathbf y}
      | r = \overline {\begin {bmatrix} a & b \\ -\overline b & \overline a \end {bmatrix} \begin {bmatrix} c & d \\ -\overline d & \overline c \end {bmatrix} }
      | c = 
}}
{{eqn | r = \overline {\begin {bmatrix} a c - b \overline d & a d + b \overline c \\ - \overline b c - \overline a \overline d & - \overline b d + \overline a \overline c \end {bmatrix} }
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \begin {bmatrix} \overline {a c - b \overline d} & -\paren {a d + b \overline c} \\ \overline b c + \overline a \overline d & \overline {-\overline b d + \overline a \overline c} \end {bmatrix}
      | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}}
}}
{{eqn | r = \begin {bmatrix} \overline a \overline c - \overline b d & - a d - b \overline c \\ \overline {a d} + \overline b c & a c - b \overline d \end {bmatrix}
      | c = Complex Conjugation is Field Automorphism of Complex Numbers, Complex Conjugation is Involution
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \overline {\mathbf y} \times \overline {\mathbf x}
      | r = \overline {\begin {bmatrix} c & d \\ -\overline d & \overline c \end {bmatrix} } \ \overline {\begin{bmatrix} a & b \\ -\overline b & \overline a \end {bmatrix} } 
      | c = 
}}
{{eqn | r = \begin {bmatrix} \overline c & - d \\ \overline d & c \end {bmatrix} \begin {bmatrix} \overline a & - b \\ \overline b & a \end {bmatrix}
      | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}}
}}
{{eqn | r = \begin {bmatrix} \overline a \overline c - \overline b d & - a d - b \overline c \\ \overline {a d} + \overline b c & a c - b \overline d \end {bmatrix}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \overline {\mathbf x \times \mathbf y}
      | c = from above
}}
{{end-eqn}}
but:
{{begin-eqn}}
{{eqn | l = \overline {\mathbf x} \times \overline {\mathbf y}
      | r = \overline {\begin {bmatrix} a & b \\ -\overline b & \overline a \end {bmatrix} } \ \overline {\begin {bmatrix} c & d \\ -\overline d & \overline c \end {bmatrix} }
      | c = 
}}
{{eqn | r = \begin {bmatrix} \overline a & - b \\ \overline b & a \end {bmatrix} \times \begin {bmatrix} \overline c & - d \\ \overline d & c \end {bmatrix}
      | c = {{Defof|Conjugate Quaternion|subdef = Matrix Form}}
}}
{{eqn | r = \begin {bmatrix} \overline a \overline c - b \overline d & - \overline a d - b c \\ a \overline d + \overline {b c} & a c - \overline b d \end {bmatrix}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | o = \ne
      | r = \overline {\mathbf x \times \mathbf y}
      | c = from above
}}
{{end-eqn}}
{{qed}}