Simplify the following boolean expressions:
1. F= ((A+BC') + D(E+F))
2. F = ((A+B+C)D)'
3. F = AB'+ A(B+C)' + B(B+C)'
4. F = [AB'(C+BD)A'B'|C
5. F= A'BC + AB'C' + A'B'C' + AB'C +
АВС

Let's simplify the given Boolean expressions step by step:

---

### **1.** \( F = (A + BC') + D(E + F) \)

#### **Solution:**

- \( A + BC' \) is already in its simplest form.
- \( D(E + F) \) expands to \( DE + DF \).
- So, \( F = (A + BC') + DE + DF \).
  
Thus, the simplified form is:

\[
F = A + BC' + DE + DF
\]

---

### **2.** \( F = ((A + B + C)D)' \)

#### **Solution:**

- Apply De Morgan's Law:
  
  \[
  F = (A + B + C)' + D'
  \]
  
- Apply De Morgan’s Law on \( (A + B + C)' \):
  
  \[
  F = A'B'C' + D'
  \]
  
Thus, the simplified form is:

\[
F = A'B'C' + D'
\]

---

### **3.** \( F = AB' + A(B + C)' + B(B + C)' \)

#### **Solution:**

- \( A(B + C)' \) can be expanded by De Morgan's Law as \( A(B'C') \).
- \( B(B + C)' \) can be expanded similarly as \( B(B'C') = 0 \) (since \( BB' = 0 \)).
  
So, the expression simplifies to:

\[
F = AB' + AB'C'
\]

Finally, factoring out \( AB' \), we get:

\[
F = AB'(1 + C') = AB'
\]

---

### **4.** \( F = AB'(C + BD) + A'B' + C \)

#### **Solution:**

- First expand \( AB'(C + BD) \):
  
  \[
  AB'C + AB'BD = AB'C + AB'BD = AB'C + AB'D
  \]
  
- Now the expression becomes:
  
  \[
  F = AB'C + AB'D + A'B' + C
  \]
  
- Combine like terms:
  
  \[
  F = C + AB'D + A'B'
  \]

Thus, the simplified form is:

\[
F = C + AB'D + A'B'
\]

---

### **5.** \( F = A'BC + AB'C' + A'B'C' + AB'C + ABC \)

#### **Solution:**

Group terms and look for common factors:

- Group \( A'BC \) and \( A'B'C' \) together, and \( AB'C' \), \( AB'C \), and \( ABC \) together:
  
  \[
  F = A'(BC + B'C') + AB'(C + C') + ABC
  \]

- Simplify:
  
  - \( BC + B'C' = (B \oplus C)' \)
  - \( C + C' = 1 \)
  
So, the expression simplifies to:

\[
F = A'(B \oplus C)' + AB' + ABC
\]

Thus, the final simplified form is:

\[
F = A'(B \oplus C)' + AB' + ABC
\]

---

### Summary of Simplified Expressions:

1. \( F = A + BC' + DE + DF \)
2. \( F = A'B'C' + D' \)
3. \( F = AB' \)
4. \( F = C + AB'D + A'B' \)
5. \( F = A'(B \oplus C)' + AB' + ABC \)

---

Would you like more details on any of these steps or further clarification? Here are 5 related questions:

1. How does De Morgan's law help in simplifying Boolean expressions?
2. Can Boolean expressions be simplified further after applying De Morgan’s law?
3. What are the key properties of Boolean algebra used for simplification?
4. How is factoring useful in simplifying Boolean expressions?
5. How do we approach simplifying expressions with multiple terms like \( A'BC + AB'C' + A'B'C' \)?

**Tip:** Using Karnaugh maps (K-maps) is another helpful tool to simplify Boolean expressions visually.

Simplify the following boolean expressions:
1. F= ((A+BC') + D(E+F))
2. F = ((A+B+C)D)'
3. F = AB'+ A(B+C)' + B(B+C)'
4. F = [AB'(C+BD)A'B'|C
5. F= A'BC + AB'C' + A'B'C' + AB'C + АВС

Explore how to simplify complex Boolean expressions like F = ((A+BC') + D(E+F)) using key concepts of Boolean algebra, including De Morgan's Law. This guide offers step-by-step solutions suitable for students in Grades 11-12 and introductory university-level courses.

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