Boolean algebra, logic expression minimization

Question

$$(A\land \neg A)\lor((A \land B) \lor(A\land B\land \neg C ))$$ I'm trying to minimize this Boolean expression with boolean algebra but i cant minimize it completly can i get some help?

Answer 1

You can simplify like the following: $$P = (A\land \neg A)\lor((A \land B) \lor(A\land B\land \neg C ))$$ $$A\land \neg A = \bot \Rightarrow P = (A \land B) \lor(A\land B\land \neg C ) = D \lor (D\land \neg C), D = A\land B \Rightarrow P = D = A\land B$$

In the above $R =D\lor(D\land \neg C) = D$ because if $D$ is true, so the $R$ is true, and if $D$ is false, so $R$ is false. Therefore, we can conclude that $R = D$.

Answer 2

It can be simplified as follows:

\begin{array}{l} & (A \land \neg A) \lor ((A \land B) \lor (A \land B \land \neg C)) & \text{ Given }\\ & F \lor ((A \land B) \lor (A \land B \land \neg C)) & \text{ Complement }\\ & (A \land B) \lor (A \land B \land \neg C) & \text{ Identity }\\ & (A \land B) \land (T \lor \neg C) & \text{ Distributive }\\ & (A \land B) \land T & \text{ Identity }\\ & A \land B & \text{ Identity }\\ \end{array}