Basic logical math properties So I have a question in logical math.
I want to know how to simplify the next expression and maybe understand the rule.
So I have those 3 atomic formulas $A,B,C$  (if there is a way)
$$ \neg(\neg(A\lor \neg(A\lor B))\lor \neg(B\lor C))$$
I was told that is actually  $ ((A\land \neg B)\lor (B\land C))$
Thanks in advanced !!
 A: $$\neg(\neg(A\lor \neg(A\lor B))\lor \neg(B\lor C))$$
Use DeMorgan's !!
By DeMorgan's, we know that 


*

*$\lnot (P \lor Q) \equiv (\lnot P \land \lnot Q)$.

*$\lnot (P\land Q) \equiv (\lnot P \lor \lnot Q)$.


Try to use this multiple times...and see what this gives you.
$$\neg(\neg(A\lor \neg(A\lor B))\lor \neg(B\lor C)) $$
$$\equiv \lnot \lnot (A \lor (\lnot A\land  \lnot B)] \land (B \lor C) \tag{(2) DM I}$$
$$\equiv [A \lor (\lnot A\land \lnot B)] \land (B\lor C)\tag{(3) DN}$$
*Note: In the first line, I used DeMorgan's twice. In the second line,} I applied double negation: $$\lnot \lnot P \equiv P\tag{Double Negation, DN}$$
Now, we can apply the distributive "laws": $$P \land (Q \lor R) \equiv (P \land Q) \lor (P\land R)\tag{DL I}$$
$$P \lor (Q \land R) \equiv (P\lor Q) \land (P \lor R)\tag{DL II}$$

$$[A \lor (\lnot A\land \lnot B)] \land (B\lor C)) \equiv ((A\lor \lnot A)\land (A \lor \lnot B))\land (B\lor C))$$
Knowing that $A \lor \lnot A \equiv T$, we can omit $A \lor \lnot A$ without losing any information whatsoever. That leaves us with
$$ ((A\lor \lnot A)\land (A \lor \lnot B))\land (B\lor C))\equiv \bf (A \lor \neg B)\land (B\lor C)$$
A: One (maybe not elegant, but definitely leading to success) way is to use truth tables and compare. Here you'd need to check all $8$ combinations of truth and falsehood among the three atomics.
A: You have
$$ \neg(\neg(A\lor \neg(A\lor B))\lor \neg(B\lor C))$$
By DeMorgan's Law, this becomes:
$$ (\neg\neg(A\lor \neg(A\lor B))\land \neg\neg(B\lor C))$$
By the double negation law,
$$ ((A\lor \neg(A\lor B))\land (B\lor C))$$
By DeMorgan's Law,
$$ ((A\lor (\neg A\land \neg B))\land (B\lor C))$$
By the distributive law,
$$ ((A\lor \neg A)\land (A \lor \neg B))\land (B\lor C))$$
Using the Tautology law,
$$  (A \lor \neg B)\land (B\lor C)$$
Which is what I got, though it seems to be "backwards" to what you have
