Boolean Algebra There seems to be some discrepancy between my answer and the solution's. Can somebody please tell me what I have done wrong? Thanks! 
$$\begin{align}
\left(A \lor B\right) \land \left(B \lor C\right) \land \left(\lnot A \lor C \right) & = \left(A+B \right)\left(B + C\right)\left( !A+C \right) \\
& = (A+B)(B!A+BC+C!A+CC) \\ 
& = (A+B)(B!A+BC+!AC+C) \\
&=(A+B)(B!A+C) \\
&=AB!A+AC+BB!A+BC\\
&=F+AC+B!A+BC \\
&=(A \land C) \lor (B \land \lnot A) \lor (B \land C) \text{ My answer}\\ \\
& \text{But solution says:} \\
&(A\land C)\lor(B \land \lnot A)
\end{align}
$$
 A: If you work out the truth tables, you’ll see that the two answers are equivalent. Here’s a computational reduction:
$$\begin{align*}
B\land C&=B\land C\land T\\
&=B\land C\land(A\lor\lnot A)\\
&=(B\land C\land A)\lor(B\land C\land\lnot A)\;,
\end{align*}$$
so
$$\begin{align*}
(A\land C)\lor(B\land\lnot A)\lor(B\land C)&=(A\land C)\lor(B\land\lnot A)\lor(B\land C\land A)\lor(B\land C\land\lnot A)\\
&=\Big((A\land C)\lor(B\land C\land A)\Big)\lor\Big((B\land\lnot A)\lor(b\land C\land\lnot A)\Big)\\
&=(A\land C)\lor(B\land\lnot A)\;.
\end{align*}$$
The trick is to realize that one of $A$ and $\lnot A$ must be true, so if $B\land C$ is true, then either $B\land C\land A$ is true, or $B\land C\land\lnot A$ is true $-$ but these are both absorbed into the other two disjuncts, $A\land C$ and $B\land\lnot A$, that we already had.
A: Both answers are good, but the $(B \land C)$ term is not necessary. If both $B$ and $C$ are true, then one of $(A \land C)$ or $(B \land \neg A)$ are satisfied too, depending on whether $A$ is true or not. Nevertheless, the result of the whole expressions is the same.
A: The $B \land C$ at the end of your answer is already satisfied by the previous parts, so it is not needed. So it is good, but it can be further simplified.
