Proving $\neg A\vee(A\wedge \neg B)= \neg A \vee \neg B$. How do I prove using boolean algebra that $\neg A\vee(A\wedge \neg B)= \neg A \vee \neg B$?  I can see it in the logic table and it is logical, but I can't prove it mathematically.
 A: Distribute so you get $(\neg A\vee A)\wedge(\neg A\vee \neg B)$ The first is obviously a tautology so you just get $\neg A\vee \neg B$
A: Here is essentially the answer from ruler501, but written down in a different notation, and with some more detail.
To prove this equivalence, the simplest approach is to start at the most complex side and try to simplify.  Therefore we calculate:
\begin{align}
& \lnot A \lor (A \land \lnot B) \\
\equiv & \qquad \text{"$\;\lor\;$ distributes over $\;\land\;$ -- really the only thing we can do"} \\
& (\lnot A \lor A) \land (\lnot A \lor \lnot B) \\
\equiv & \qquad \text{"left hand part: law of the excluded middle, i.e., $\;P \lor \lnot P \equiv \text{true}\;$"} \\
& \text{true} \land (\lnot A \lor \lnot B) \\
\equiv & \qquad \text{"simplify using $\;P \land \text{true} \equiv P\;$"} \\
& \lnot A \lor \lnot B \\
\end{align}
This proves the equivalence.
A: I like to do these via cases because it is a little quicker than truth tables a lot of the time. Suppose $A = T$ then 
$$\begin{align}\neg A\vee (A\wedge \neg B) &= F\vee (T\wedge \neg B) \\ &= F\vee \neg B \\ &= \neg A\vee \neg B.\end{align}$$
This follows since $F = \neg A$. Then if $A=F$, we have
$$\begin{align}\neg A\vee(A\wedge \neg B) &= T\vee (F\wedge\neg B) \\ &= T \\ &= \neg A\vee \neg B. \end{align}$$
This follows since $\neg A = T.$ This holds regardless of the truth or falsity of $B$ so they are the same.
