How was the rule of absorption applied to the following logical expression?
$$ (A \land \neg B \land C) \lor (A \lor B \lor \neg C) \lor (\neg A \land B) \equiv (A \lor B \lor \neg C) \lor (\neg A \land B).$$
The step is not really clear to me, and I hope someone can explain it.
On
Given associativity of $\lor$, you can just drop parentheses:
$$ (A \land \neg B \land C) \lor (A \lor B \lor \neg C) \lor (\neg A \land B) $$
$$=(A \land \neg B \land C) \lor A \lor B \lor \neg C \lor (\neg A \land B)$$
So now the $A$ terms absorps the $(A \land \neg B \land C)$ term:
$$=A \lor B \lor \neg C \lor (\neg A \land B) $$
And note that the $B$ term will absorp the $(\neg A \land B) $ term:
$$=A \lor B \lor \neg C $$
When you drop unnecessary parentheses, these kinds of simplifications become a lot easier to see!
Using associativity, $$ (A \wedge \neg B \wedge C) \vee (A \vee B \vee \neg C) \vee (\neg A \wedge B) = ((A \wedge \neg B \wedge C) \vee A) \vee (B \vee \neg C)\vee (\neg A \wedge B). $$ Now, by absorption and associativity, $$(A \wedge \neg B \wedge C) \vee A = (A \wedge (\neg B \wedge C)) \vee A = A,$$ and the expression becomes $$A \vee (B \vee \neg C)\vee (\neg A \wedge B),$$ which is the desired result, using associativity again.