Let A, B and C be digital inputs. Prove that the following boolean equation holds true for any given values for inputs.
(A AND B) OR (A AND (NOT B) AND C) OR (B AND (NOT C)) = (A AND C) OR (B AND (NOT C))
This is a question with no solution in my circuit design book, I've googled and tried simplifying the equation to no avail.
Let us simplify the LHS, by taking $A$ out of the first two monomials.
$$\begin{align} (A\wedge B)\vee(A\wedge \neg B\wedge C)\vee(B\wedge\neg C) &= [A\land (B \lor (\neg B\land C))] \lor (B\land \neg C)\\ &= [A\land (B\lor C)]\lor (B\land\neg C)\\ &= (A\land B) \lor (A\land C)\lor (B\land \neg C). \end{align}$$
Note that this is not exactly the answer you were supposed to have. For the last step, suppose that $A\land B$ is true. If $C$ is true, then we have $A\land C$, and if $C$ is false, we have $B\land \neg C$. Hence we may rewrite the last line as $(A\land C)\lor (B\land\neg C)$. This reasoning could be done "internally", only by computation, by writing $A\land B$ as $A\land B\land(C\lor \neg C)$, as in the following:
$$\begin{align} (A\wedge B)\vee(A\wedge \neg B\wedge C)\vee(B\wedge\neg C) &=(A\land B) \lor (A\land C)\lor (B\land \neg C) \quad\text{from above}\\ &= (A\land B\land(C\lor\neg C))\lor (A\land C)\lor (B\land\neg C)\\ &= (A\land B\land C)\lor (A\land B\land\neg C)\lor (A\land C)\lor (B\land\neg C)\\ &= (A\land C)\lor (B\land \neg C) . \end{align}$$