Logical proof using sentential logic $\Big(\big(A \& B \big) \lor \big((\neg{}A \& B) \lor (A \& \neg{}B)\big)\Big)$ I have a premise $(A \lor B)$ 
and need to achieve the conclusion $\Big(\big(A \& B \big) \lor \big((\neg{}A \& B) \lor (A \& \neg{}B)\big)\Big)$
This intuitively makes sense since $(A \lor B)$ means at least one of them is true.
I just can't figure out how the conclusion can be achieved formally. 
I tried assuming the conclusion is false and achieving a falsum, but that didn't get anywhere. 
I also tried using disjunction introduction rule backwards but that didn't work either. 
Any strategies on how to attack this proof?
 A: 
I have a premise $(A \lor B)$ 
  and need to achieve the conclusion $\Big(\big(A \& B \big) \lor \big((\neg{}A \& B) \lor (A \& \neg{}B)\big)\Big)$
This intuitively makes sense since $(A \lor B)$ means at least one of them is true.

Exactly, so use a proof by cases. Here, I'll start it off in the Fitch system.  Add the justifications and fill in the dots.$$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}} \fitch{A\lor B}{\fitch{A}{B\lor\lnot B\\\fitch{B}{A\land B\\(A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B))}\\B\to((A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B)))\\\quad\vdots\\\lnot B\to((A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B)))\\(A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B))}\\A\to ((A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B)))\\\quad\vdots\\B\to((A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B)))\\(A\land B)\lor((\lnot A\land B)\lor(A\land\lnot B))}$$ 
A: You can always check the truth table to see the relation between two logical statements. All the well known / widly used relations are based on this.
Truth table of $A\vee B$ : 
$$\begin{array}{|c|c|c|}
\hline
B\backslash A&1&0\\\hline
1&1&1\\\hline
0&1&0\\\hline
\end{array}$$
Truth table of $\Big(\big(A \& B \big) \lor \big((\neg{}A \& B) \lor (A \& \neg{}B)\big)\Big)$ : 
$$\begin{array}{|c|c|c|}
\hline
B\backslash A&1&0\\\hline
1&1\vee0\vee0=1&0\vee1\vee0=1\\\hline
0&0\vee0\vee1=1&0\vee0\vee0=0\\\hline
\end{array}$$
and you can see the two logical statements are actually equivalent, that is,
$$
A\vee B\Leftrightarrow\Big(\big(A \& B \big) \lor \big((\neg{}A \& B) \lor (A \& \neg{}B)\big)\Big).
$$
