Answer :
Sure, let's simplify the given expressions step-by-step using De Morgan's Theorems.
### Part (a)
Given expression: [tex]\(\overline{A+B+C \cdot(\overline{A B C})}\)[/tex]
1. Applying De Morgan's Theorem:
- The complement of a sum (OR operation) of variables is equal to the product (AND operation) of the complements of the variables.
[tex]\[ \overline{A + B + C \cdot (\overline{A B C})} = \overline{A} \cdot \overline{B} \cdot \overline{(C \cdot (\overline{A B C}))} \][/tex]
2. Simplify the inner expression:
- Simplify [tex]\(\overline{(C \cdot (\overline{A B C}))}\)[/tex]:
[tex]\[ \overline{(C \cdot (\overline{A B C}))} = \overline{C} + A + B \][/tex]
- This follows from De Morgan's Theorem and the properties of Boolean algebra.
3. Combine the results:
- Substitute back into the expression:
[tex]\[ \overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B) \][/tex]
So, the simplified form of [tex]\(\overline{A+B+C \cdot(\overline{A B C})}\)[/tex] is:
[tex]\[ \boxed{\overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B)} \][/tex]
### Part (b)
Given expression: [tex]\(\overline{x+\bar{y}+\bar{y} z}\)[/tex]
1. Applying De Morgan's Theorem:
[tex]\[ \overline{x + \overline{y} + \overline{y} z} = \overline{x} \cdot y \cdot \overline{(\overline{y} z)} \][/tex]
2. Simplify the inner expression:
- Simplify [tex]\(\overline{(\overline{y} z)}\)[/tex]:
[tex]\[ \overline{(\overline{y} z)} = y + \overline{z} \][/tex]
- Again, this follows from De Morgan's Theorem and Boolean algebra.
3. Combine the results:
- Substitute back into the expression:
[tex]\[ \overline{x} \cdot y \cdot (y + \overline{z}) \][/tex]
So, the simplified form of [tex]\(\overline{x+\bar{y}+\bar{y} z}\)[/tex] is:
[tex]\[ \boxed{\overline{x} \cdot y \cdot (y + \overline{z})} \][/tex]
In conclusion, we've simplified the expressions using De Morgan's Theorems:
- For part (a): [tex]\(\boxed{\overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B)}\)[/tex]
- For part (b): [tex]\(\boxed{\overline{x} \cdot y \cdot (y + \overline{z})}\)[/tex]
### Part (a)
Given expression: [tex]\(\overline{A+B+C \cdot(\overline{A B C})}\)[/tex]
1. Applying De Morgan's Theorem:
- The complement of a sum (OR operation) of variables is equal to the product (AND operation) of the complements of the variables.
[tex]\[ \overline{A + B + C \cdot (\overline{A B C})} = \overline{A} \cdot \overline{B} \cdot \overline{(C \cdot (\overline{A B C}))} \][/tex]
2. Simplify the inner expression:
- Simplify [tex]\(\overline{(C \cdot (\overline{A B C}))}\)[/tex]:
[tex]\[ \overline{(C \cdot (\overline{A B C}))} = \overline{C} + A + B \][/tex]
- This follows from De Morgan's Theorem and the properties of Boolean algebra.
3. Combine the results:
- Substitute back into the expression:
[tex]\[ \overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B) \][/tex]
So, the simplified form of [tex]\(\overline{A+B+C \cdot(\overline{A B C})}\)[/tex] is:
[tex]\[ \boxed{\overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B)} \][/tex]
### Part (b)
Given expression: [tex]\(\overline{x+\bar{y}+\bar{y} z}\)[/tex]
1. Applying De Morgan's Theorem:
[tex]\[ \overline{x + \overline{y} + \overline{y} z} = \overline{x} \cdot y \cdot \overline{(\overline{y} z)} \][/tex]
2. Simplify the inner expression:
- Simplify [tex]\(\overline{(\overline{y} z)}\)[/tex]:
[tex]\[ \overline{(\overline{y} z)} = y + \overline{z} \][/tex]
- Again, this follows from De Morgan's Theorem and Boolean algebra.
3. Combine the results:
- Substitute back into the expression:
[tex]\[ \overline{x} \cdot y \cdot (y + \overline{z}) \][/tex]
So, the simplified form of [tex]\(\overline{x+\bar{y}+\bar{y} z}\)[/tex] is:
[tex]\[ \boxed{\overline{x} \cdot y \cdot (y + \overline{z})} \][/tex]
In conclusion, we've simplified the expressions using De Morgan's Theorems:
- For part (a): [tex]\(\boxed{\overline{A} \cdot \overline{B} \cdot (\overline{C} + A + B)}\)[/tex]
- For part (b): [tex]\(\boxed{\overline{x} \cdot y \cdot (y + \overline{z})}\)[/tex]
