Simplify Implication Expression (Predicate/Prop Logic) I'm trying to do some past paper questions for revision and find myself perplexed on some of the expressions that need normalized/simplified which involves an implies.
For example:
(A ∧ ¬B) → B ∨ C ∨ ¬ (A ∧ ¬C) 
Now I know that A → B can be normalized to ¬A or B, but I can't seem to find an example when it comes to multiple things implying something to learn from.  I'd appreciate if someone could explain to me how I would simplify such an expression.
 A: I assume that $\to$ is the main connective. First, eliminate $\to$ in favor of $\lor$ and $\neg$, then use De Morgan to eliminate $\land$, and then remove repeated terms:
$$\begin{align}
(A\land \neg B) \to B\lor C\lor \neg(A\land \neg C) &\iff \neg(A\land \neg B) \lor B\lor C\lor \neg(A\land \neg C) \\
&\iff \neg A\lor B \lor B\lor C\lor \neg A\lor C \\
&\iff \neg A\lor B \lor C.\\
\end{align}$$
If you wish, you can now reintroduce $\to$, yielding the equivalent formula:
$$
A\to (B \lor C).
$$
A: The rule that you cite:
$$A \rightarrow B = \neg A \vee B$$
also works when $A$ is a compound expression. (All of these rules do). In your example, the simplification would go like this:
$$
\begin{aligned}
(A \wedge \neg B) &\rightarrow B \vee C \vee \neg(A \wedge \neg C)\\
\neg(A \wedge \neg B) &\vee (B \vee C \vee \neg(A \wedge \neg C))\\
\neg A \vee \neg \neg B &\vee B \vee C \vee \neg(A \wedge \neg C)\\
\neg A \vee B &\vee B \vee C \vee \neg A \vee \neg \neg C\\
\neg A &\vee B \vee C\\
\end{aligned}
$$
Here I've used DeMorgan's laws twice.
