$$\neg(A \Rightarrow B)\lor \neg B$$
Does this this expression simplify to:?
$$\neg A\Rightarrow\neg B\lor \neg B$$
Which further simplifies to:
$$\neg A\Rightarrow\neg B$$
$$\neg(A \Rightarrow B)\lor \neg B$$
Does this this expression simplify to:?
$$\neg A\Rightarrow\neg B\lor \neg B$$
Which further simplifies to:
$$\neg A\Rightarrow\neg B$$
On
No. The negation of an implication is not the implication using the negations. $$\neg (X\implies Y) \not\equiv (\neg X \implies \neg Y)$$
It goes thus:
$$\begin{align} (A \implies B)\implies B & \equiv \neg (A\implies B) \lor B & \ni X\implies Y \equiv \neg X \lor Y \\ &\equiv \neg(\neg A \lor B) \lor B & \text{again} \\ & \equiv (A \land \neg B) \lor B & \text{DeMorgan's Laws} \\ & \equiv (A \lor B) \land (\neg B\lor B) & \text{association} \\ & \equiv (A \lor B) & \text{tautology} \end{align}$$
On
The negation of an implication always gives problems :)
Maybe the easiest way to understand it is that
$$ \lnot ( A \to B) $$
Means: "it is definitly not the case that A implies B"
and then think,
I can only know that for sure that A does not imply B when A is true and B is false
therefore
$$ \lnot ( A \to B) \to (A \land \lnot B) $$
I guess you probaby imagined that $$ \lnot ( A \to B) $$ means something like
In technical terms this using is "not as failure" but then you are working with a very strange logic (closed world logic) (see http://en.wikipedia.org/wiki/Closed-world_assumption ) and you better stay away from that.
Good luck
Let us first look at the conditions under which $\lnot(A\implies B)\lor \lnot B$ is true. Intuition is often better for $\land$ and $\lor$ than it is for $\implies$, so we eliminate the $\implies$.
The first term is equivalent to $\lnot(\lnot A\lor B)$, which is equivalent to $A\land \lnot B$.
And $(A\land \lnot B)\lor \lnot B$ is equivalent to $\lnot B$.
The second "formula" in the post is not a formula, since crucial parentheses are missing. But if we give precedence to $\implies$, it is not equivalent to $\lnot B$.
The formula $\lnot A\implies \lnot B$ is not equivalent to $\lnot B$, so it is not equivalent to $(A\land \lnot B)\lor \lnot B$.