(not ((b or not c) and (not a or not c))) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))


Let's simplify the Boolean expression step by step:

### Original Expression:

\[
\neg ((b \lor \neg c) \land (\neg a \lor \neg c)) \lor \neg (c \lor \neg (b \land c)) \lor (a \land \neg c) \land (\neg a \lor (a \land b \land c) \lor (a \land ((b \land \neg c) \lor (\neg b))))
\]

#### Step 1: Simplifying \(\neg ((b \lor \neg c) \land (\neg a \lor \neg c))\)

Using De Morgan’s law, \(\neg (A \land B) = \neg A \lor \neg B\):

\[
\neg ((b \lor \neg c) \land (\neg a \lor \neg c)) = \neg (b \lor \neg c) \lor \neg (\neg a \lor \neg c)
\]
\[
= (\neg b \land c) \lor (a \land c)
\]

#### Step 2: Simplifying \(\neg (c \lor \neg (b \land c))\)

Again using De Morgan’s law:

\[
\neg (c \lor \neg (b \land c)) = \neg c \land (b \land c)
\]

Since \(\neg c \land (b \land c)\) can never be true (because \(c\) and \(\neg c\) can't be true at the same time), this simplifies to:

\[
\neg c \land (b \land c) = \text{false}
\]

#### Step 3: Simplifying \((a \land \neg c) \land (\neg a \lor (a \land b \land c) \lor (a \land ((b \land \neg c) \lor \neg b)))\)

Let’s first simplify the second part:

\[
\neg a \lor (a \land b \land c) \lor (a \land ((b \land \neg c) \lor \neg b))
\]

This simplifies as follows:

- \(\neg a \lor (a \land \ldots)\) is always true because if \(a\) is false, \(\neg a\) is true, and if \(a\) is true, \(a \land \ldots\) can still be true.

So this part simplifies to:

\[
\neg a \lor (a \land \ldots) = \text{true}
\]

Now, returning to the whole term:

\[
(a \land \neg c) \land \text{true} = a \land \neg c
\]

#### Step 4: Combining everything

Now, let's combine the simplified terms:

\[
(\neg b \land c) \lor (a \land c) \lor \text{false} \lor (a \land \neg c)
\]

\[
= (\neg b \land c) \lor (a \land c) \lor (a \land \neg c)
\]

We now have two cases for \(a\):

- If \(a = 1\), the expression simplifies to \(c \lor \neg c = 1\).
- If \(a = 0\), the expression simplifies to \(\neg b \land c\).

Thus, the final simplified form is:

\[
a \lor (\neg b \land c)
\]

### Final Answer:

\[
a \lor (\neg b \land c)
\]

---

Would you like any further details or explanations on any step?

Here are five related questions you might want to explore:

1. How do De Morgan's laws work in Boolean algebra?
2. What are the rules for simplifying Boolean expressions?
3. How can truth tables help in simplifying expressions?
4. How can Boolean algebra be applied in circuit design?
5. What are common techniques for reducing complex logic expressions?

**Tip:** Always simplify Boolean expressions by applying De Morgan's laws and other logical identities like distributive, associative, and commutative properties.

not ((b or not c) and (not a or not c)) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))

This solution simplifies the complex Boolean expression step by step using key Boolean algebra concepts such as De Morgan's laws, distributive, commutative, and identity properties. Suitable for college-level students or those studying logic circuits.

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