Help with boolean algebra simplification and equivalent I have the following boolean expression:
See number 5 and 7
What I have trouble with are the actual steps of simplification using the boolean algebra laws for number 5 and 7 . I'm probably missing something really really obvious, because I'm trying to freshen up my simplification skills via exercises and even with complex expressions don't have problems solving them, but these one leaves me stumped.
Could someone please provide a step by step simplification that displays how to simplify it?
Thank you very much for your time.
 A: Here is my attempt at 5:
\begin{align*}
((A \wedge (\neg B \vee C)) \wedge (A \rightarrow (B \rightarrow \neg C)) \rightarrow A &= \neg ((A \wedge (\neg B \vee C)) \wedge (A \rightarrow (B \rightarrow \neg C))) \vee A\\
&= \neg (A \wedge (\neg B \vee C)) \vee \neg (A \rightarrow (B \rightarrow \neg C))) \vee A\\
&= \neg A \vee \neg(\neg B \vee C) \vee \neg (A \rightarrow (B \rightarrow \neg C)) \vee A\\
&= (\neg A \vee A )\vee \neg(\neg B \vee C)) \vee \neg (A \rightarrow (B \rightarrow \neg C)))\\
&= 1 \vee (\neg(\neg B \vee C) \vee \neg (A \rightarrow (B \rightarrow \neg C))\\
&= 1
\end{align*}
We can stop there because $1 \vee O$ where $O$ is anything is just $1$. 
Now for 7:
\begin{align*}
(\neg A \rightarrow \neg B) \rightarrow ((A \rightarrow B) \rightarrow \neg A) &= (\neg A \rightarrow \neg B) \rightarrow ((A \rightarrow B) \rightarrow \neg A)\\
&=\neg (\neg A \rightarrow \neg B) \vee ((A \rightarrow B) \rightarrow \neg A)\\
&=\neg (\neg A \rightarrow \neg B) \vee (\neg (\neg A \vee B) \vee \neg A)\\
&=\neg (A \vee \neg B) \vee (\neg (\neg A \vee B) \vee \neg A)\\
&=\neg (A \vee \neg B) \vee ((A \wedge \neg B) \vee \neg A)\\
&=(\neg A \wedge B) \vee ((A  \vee \neg A) \wedge (\neg B  \vee \neg A))\\
&=(\neg A \wedge B) \vee (1 \wedge (\neg B  \vee \neg A))\\
&=(\neg A \wedge B) \vee (\neg B  \vee \neg A)\\
&= ((\neg A \wedge B) \vee \neg A )\vee \neg B  \\
&= \neg A \vee \neg B  \\
\end{align*}
