Help with boolean algebra simplification I have the following boolean expression:
$$(A \land B) \lor (\lnot A \land C) \lor (B \land C)$$
I know this can be simplified to 
$$(A \land B) \lor (\lnot A \land C)$$
I can see that doing truth tables, drawing a circuit, a venn diagram. I understand it simplifies to that.
What I have trouble with are the actual steps of simplification using the boolean algebra laws. I'm probably missing something really really obvious, because I'm trying to freshen up my simplification skills via exercises and even with complex expressions don't have problems solving them, but this one leaves me stumped.
Could someone please provide a step by step simplification that displays how to simplify it?
Thank you very much for your time.
 A: The Concensus Theorem says:
$XY \lor Y'Z \lor XZ = XY \lor XZ$
So your exercise is really a particular application of this!
But let's assume the Concensus Theorem is not available for you to use.
Then you can do:
$(A \land B) \lor (\neg A \land C) \lor (B \land C) =$ (Adjacency)
$(A \land B) \lor (\neg A \land C) \lor (A \land B \land C) \lor (\neg A \land B \land C)=$ (Absorption x 2)
$(A \land B) \lor (\neg A \land C)$
This assumes:
Adjacency
$P = (P \land Q) \lor (P \land \neg Q)$
Absorption
$P = P \lor (P \land Q)$
A: $$(A \land B) \lor (\lnot A \land C) \lor (B \land C)$$
Insert a tautology:
$$(A \land B) \lor (\lnot A \land C) \lor ((\lnot A \lor A) \land B \land C)$$
Expand:
$$(A \land B) \lor (\lnot A \land C) \lor (\lnot A \land B \land C) \lor (A \land B \land C)$$
Reorder:
$$(A \land B) \lor (A \land B \land C) \lor (\lnot A \land C) \lor (\lnot A \land B \land C)$$
Insert tautology:
$$(A \land B \land true) \lor (A \land B \land C) \lor (\lnot A \land C \land true) \lor (\lnot A \land B \land C)$$
Extract:
$$(A \land B \land (true \lor C)) \lor (\lnot A \land C \land (true \lor B))$$
Absorb:
$$(A \land B \land true) \lor (\lnot A \land C \land true)$$
Simplify:
$$(A \land B) \lor (\lnot A \land C)$$
A: By looking at the similar part, you just want to show $$ C\,\vee \,B \wedge \, C = C$$
This identity is described on wikipedia as the absorption property :https://en.wikipedia.org/wiki/Boolean_algebra#Laws.
Intuitively, you can say that $B$ is absorbed by $C$ (via truth table for example).
Indeed: If $C = 1$, then $C \vee \cdots =1 $. If $C=0$, $B \wedge C = 0$ so the end result will be $0$
