Answer :
Sure, let's break down and simplify each Boolean expression step-by-step to the indicated number of literals.
### (a) [tex]\(A^{\prime} C^{\prime} + A B C + A C^{\prime}\)[/tex] to three literals
1. Combine similar terms: Notice that [tex]\(A^{\prime} C^{\prime}\)[/tex] and [tex]\(A C^{\prime}\)[/tex] share a common literal [tex]\(C^{\prime}\)[/tex].
2. Apply the Distributive Law:
[tex]\[ A^{\prime} C^{\prime} + A B C + A C^{\prime} = C^{\prime}(A^{\prime} + A) + A B C \][/tex]
3. Apply the Complement Law: [tex]\(A^{\prime} + A = 1\)[/tex],
[tex]\[ C^{\prime}(A^{\prime} + A) + A B C = C^{\prime} \cdot 1 + A B C = C^{\prime} + A B C \][/tex]
So, the expression is now simplified to:
[tex]\[ C^{\prime} + A B C \][/tex]
And this is in three literals as required.
### (b) [tex]\(\left(x^{\prime} y^{\prime} + z\right)^{\prime} + z + x y + w z\)[/tex] to three literals
1. Apply De Morgan's Theorem to [tex]\((x^{\prime} y^{\prime} + z)^{\prime}\)[/tex]:
[tex]\[ (x^{\prime} y^{\prime} + z)^{\prime} = x y \cdot z^{\prime} \][/tex]
So the original expression becomes:
[tex]\[ x y z^{\prime} + z + x y + w z \][/tex]
2. Combine like terms: Notice that [tex]\( x y \)[/tex] appears in the original equation.
[tex]\[ x y z^{\prime} + z + x y + w z = x y (z^{\prime} + 1) + z + w z \][/tex]
3. Simplify using the Complement Law: [tex]\(z^{\prime} + 1 = 1\)[/tex]
[tex]\[ x y (1) + z + w z = x y + z + w z \][/tex]
4. Factor out [tex]\(z\)[/tex] in the last two terms:
[tex]\[ x y + z(1 + w) \][/tex]
5. Simplify using the Identity Law: [tex]\(1 + w = 1\)[/tex]
[tex]\[ x y + z \][/tex]
And this reduces the expression to three literals.
### (c) [tex]\(A^{\prime} B\left(D^{\prime} + C^{\prime} D\right) + B\left(A + A^{\prime} C D\right)\)[/tex] to one literal
1. Distribute terms:
[tex]\[ A^{\prime} B \cdot D^{\prime} + A^{\prime} B \cdot C^{\prime} D + B A + B A^{\prime} C D \][/tex]
2. Group terms involving [tex]\(A^\prime\)[/tex]:
[tex]\[ A^{\prime} B D^{\prime} + A^{\prime} B C^{\prime} D + B A + B A^{\prime} C D \][/tex]
3. Factor [tex]\(A^\prime B\)[/tex] from [tex]\(A^{\prime} B\left(D^{\prime} + C^{\prime} D\right)\)[/tex]:
[tex]\[ A^{\prime} B (D^{\prime} + C^{\prime} D) + B (A + A^{\prime} C D) \][/tex]
4. Use Absorption Law [tex]\(A + A^{\prime} B = A + B\)[/tex]:
[tex]\[ A^\prime + B = B \][/tex]
And this reduces the expression to one literal [tex]\(B\)[/tex].
### (d) [tex]\(\left(A^{\prime} + C\right)\left(A^{\prime} + C^{\prime}\right)\left(A + B + C^{\prime} D\right)\)[/tex] to four literals
1. First factor out the common term:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime})(A + B + C^{\prime} D) \][/tex]
2. Combine the first two terms using the distribution:
First, realize that [tex]\((A^{\prime} + C)(A^{\prime} + C^{\prime}\)[/tex] implies two cases:
If [tex]\(A^\prime\)[/tex] is true:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime}) \sim A^\prime \][/tex]
if [tex]\(A^\prime\)[/tex] is false:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime}) \sim C \cdot C^{\prime} = 0 \][/tex]
Factor:
[tex]\[ A^\prime + 0 = A^\prime, so this is A^\prime \][/tex]
Now [tex]\((A^\prime)(A + B + C^{\prime} D) = A^\prime + B + C^{\prime} D\)[/tex]
And we cannot simplify farther with less than 4 literals:
[tex]\[ A^\prime + B + C^{\prime} + (D) \][/tex]
Thus, we have achieved the four literals needed.
### (a) [tex]\(A^{\prime} C^{\prime} + A B C + A C^{\prime}\)[/tex] to three literals
1. Combine similar terms: Notice that [tex]\(A^{\prime} C^{\prime}\)[/tex] and [tex]\(A C^{\prime}\)[/tex] share a common literal [tex]\(C^{\prime}\)[/tex].
2. Apply the Distributive Law:
[tex]\[ A^{\prime} C^{\prime} + A B C + A C^{\prime} = C^{\prime}(A^{\prime} + A) + A B C \][/tex]
3. Apply the Complement Law: [tex]\(A^{\prime} + A = 1\)[/tex],
[tex]\[ C^{\prime}(A^{\prime} + A) + A B C = C^{\prime} \cdot 1 + A B C = C^{\prime} + A B C \][/tex]
So, the expression is now simplified to:
[tex]\[ C^{\prime} + A B C \][/tex]
And this is in three literals as required.
### (b) [tex]\(\left(x^{\prime} y^{\prime} + z\right)^{\prime} + z + x y + w z\)[/tex] to three literals
1. Apply De Morgan's Theorem to [tex]\((x^{\prime} y^{\prime} + z)^{\prime}\)[/tex]:
[tex]\[ (x^{\prime} y^{\prime} + z)^{\prime} = x y \cdot z^{\prime} \][/tex]
So the original expression becomes:
[tex]\[ x y z^{\prime} + z + x y + w z \][/tex]
2. Combine like terms: Notice that [tex]\( x y \)[/tex] appears in the original equation.
[tex]\[ x y z^{\prime} + z + x y + w z = x y (z^{\prime} + 1) + z + w z \][/tex]
3. Simplify using the Complement Law: [tex]\(z^{\prime} + 1 = 1\)[/tex]
[tex]\[ x y (1) + z + w z = x y + z + w z \][/tex]
4. Factor out [tex]\(z\)[/tex] in the last two terms:
[tex]\[ x y + z(1 + w) \][/tex]
5. Simplify using the Identity Law: [tex]\(1 + w = 1\)[/tex]
[tex]\[ x y + z \][/tex]
And this reduces the expression to three literals.
### (c) [tex]\(A^{\prime} B\left(D^{\prime} + C^{\prime} D\right) + B\left(A + A^{\prime} C D\right)\)[/tex] to one literal
1. Distribute terms:
[tex]\[ A^{\prime} B \cdot D^{\prime} + A^{\prime} B \cdot C^{\prime} D + B A + B A^{\prime} C D \][/tex]
2. Group terms involving [tex]\(A^\prime\)[/tex]:
[tex]\[ A^{\prime} B D^{\prime} + A^{\prime} B C^{\prime} D + B A + B A^{\prime} C D \][/tex]
3. Factor [tex]\(A^\prime B\)[/tex] from [tex]\(A^{\prime} B\left(D^{\prime} + C^{\prime} D\right)\)[/tex]:
[tex]\[ A^{\prime} B (D^{\prime} + C^{\prime} D) + B (A + A^{\prime} C D) \][/tex]
4. Use Absorption Law [tex]\(A + A^{\prime} B = A + B\)[/tex]:
[tex]\[ A^\prime + B = B \][/tex]
And this reduces the expression to one literal [tex]\(B\)[/tex].
### (d) [tex]\(\left(A^{\prime} + C\right)\left(A^{\prime} + C^{\prime}\right)\left(A + B + C^{\prime} D\right)\)[/tex] to four literals
1. First factor out the common term:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime})(A + B + C^{\prime} D) \][/tex]
2. Combine the first two terms using the distribution:
First, realize that [tex]\((A^{\prime} + C)(A^{\prime} + C^{\prime}\)[/tex] implies two cases:
If [tex]\(A^\prime\)[/tex] is true:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime}) \sim A^\prime \][/tex]
if [tex]\(A^\prime\)[/tex] is false:
[tex]\[ (A^{\prime} + C)(A^{\prime} + C^{\prime}) \sim C \cdot C^{\prime} = 0 \][/tex]
Factor:
[tex]\[ A^\prime + 0 = A^\prime, so this is A^\prime \][/tex]
Now [tex]\((A^\prime)(A + B + C^{\prime} D) = A^\prime + B + C^{\prime} D\)[/tex]
And we cannot simplify farther with less than 4 literals:
[tex]\[ A^\prime + B + C^{\prime} + (D) \][/tex]
Thus, we have achieved the four literals needed.
