We have learned that,
\[\begin{eqnarray} \bar{A} &=& A \uparrow A ~, \nonumber \\ AB &=& (A \uparrow B) \uparrow (A \uparrow B) ~, \nonumber \\ A + B &=& (A \uparrow A) \uparrow (B \uparrow B) ~, \nonumber \end{eqnarray}\]Now, using the above expressions, show that,
\[(A \uparrow A) \uparrow ( ((A \uparrow B) \uparrow (A \uparrow B)) \uparrow ((A \uparrow B) \uparrow (A \uparrow B)) ) = A ~, \nonumber\]