Law of Excluded Middle in Logic Proof I'm having some difficulty doing a proof for the following:
$$\neg A \vee \neg(\neg B \wedge (\neg A \vee B))$$
It is said that you could use the law of excluded middles.
Any help or guidance would be appreciated. Thanks in advance!
 A: Consider this:
$$
\begin{align*}
\neg A\lor\neg(\neg B\land(\neg A\lor B)) &\equiv \neg A\lor(B\lor\neg(\neg A\lor B))\\
&\equiv \neg A\lor(B\lor (A\land\neg B)) \\
&\equiv \neg A\lor((B\lor A)\land(B\lor\neg B)) \\
&\equiv \neg A\lor((B\lor A)\land \top) \\
&\equiv \neg A\lor(B\lor A)
\end{align*}
$$
where $B\lor\neg B\equiv\top$ by the law of excluded middle. Applying it again should show the original expression is a tautology, which I believe is what you want to prove.
A: Using distributivity,
$\neg A \bigvee \neg((\neg B \bigwedge \neg A) \bigvee (\neg B \bigwedge B))$
$\equiv \neg A \bigvee \neg (\neg B \bigwedge \neg A)$
$\equiv \neg A \bigvee (B \bigvee A)$
$\equiv \neg A \bigvee A$
as required. 
A: If you want to use LEM explicitly, then: 
Case 1, B is true:
$$\begin{array} {rcl}
\lnot A \lor \lnot (\lnot B \land (\lnot A \lor B)) 
&\triangleright&
\lnot A \lor \lnot (\lnot \text{T} \land (\lnot A \lor \text{T})) \\
&\triangleright&
\lnot A \lor \lnot (\text{F} \land \text{T}) \\
&\triangleright&
\lnot A \lor \text{T} \\
&\triangleright&
\text{T} \\
\end{array}$$
Case 2, B is false:
$$\begin{array} {rcl}
\lnot A \lor \lnot (\lnot B \land (\lnot A \lor B)) 
&\triangleright&
\lnot A \lor \lnot (\lnot \text{F} \land (\lnot A \lor \text{F})) \\
&\triangleright&
\lnot A \lor \lnot (\text{T} \land \lnot A) \\
&\triangleright&
\lnot A \lor \lnot \lnot A \\
&\triangleright&
\text{T} \\
\end{array}$$
You can be even more explicit if you split it into 4 cases, for the values of $A$ and $B$.
A: The following proof uses the law of the excluded middle (LEM) in a Fitch-style proof checker to prove the tautology:


Although not likely what the exercise was intending, another way of looking at this problem is to use material implication and De Morgan's laws to convert the original statement to the following:
$$A \to (\lnot B \to (A \land \lnot B))$$
Here is a proof of that:

This proof checker does not have the material implication rule, so I would not be able to show that approach directly using it.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf
