SOLUTION: Simplify the following functions using Boolean Algebra a. F(X, Y, Z) = X Y + X’ Y + X Z b. F(X, Y, Z) = (X + Y) (X’ + Y + Z) c. F(X, Y, Z) = X Y’ Z + X Y Z + Y’ Z d.

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Question 1190973: Simplify the following functions using Boolean Algebra
a. F(X, Y, Z) = X Y + X’ Y + X Z
b. F(X, Y, Z) = (X + Y) (X’ + Y + Z)
c. F(X, Y, Z) = X Y’ Z + X Y Z + Y’ Z
d. F(X, Y, Z) = XY + X’YZ
e. F(X, Y, Z) = X’Y + XYZ’
f. F(X, Y, Z) =(XY) + (X + Y +Z)’X + YZ
g. F(X, Y, Z) = (XY)’ + (X+Y +Z)’
h. F(X,Y,Z) = YZ + (X+Y)’ + (XYZ)’
i. F(X,Y,Z) = (X+Y +Z)’ (X+Y)

Could you explain me this second homework question step-by-step please? Thank you!
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

We'll need these Boolean Algebra rules
NameAND formOR form
Identity Law1*A = AA+0 = A
Null Law0*A = 0A+1 = 1
Idempotent LawA*A = AA+A = A
Inverse LawA*A' = 0A+A' = 1
Commutative LawA*B = B*AA+B = B+A
Associative LawA*(B*C) = (A*B)*CA+(B+C) = (A+B)+C
Distributive LawA+B*C = (A+B)*(A+C)A*(B+C) = A*B+A*C
Absorption LawA*(A+B) = AA+A*B = A
De Morgan's Law(A*B)' = A' + B'(A+B)' = A'*B'
The star or asterisk symbol is sometimes omitted.
This means something like A*B is the same as AB

In some Boolean algebra textbooks, the tickmark is replaced with a horizontal bar overhead.
Eg:
I'll use the tickmark notation here.


Since you posted quite a lot of problems, I'll do the first 5 to get you started.
I'll go from part (a) to part (e).


========================================================

Part (a)

X*Y + X'*Y + X*Z
(X+X')*Y + X*Z ... Distributive Law
(1)*Y + X*Z ... Inverse Law
Y + X*Z ... Identity Law

Answer: Y + X*Z

Here's the verification table
XYZX’X*YX’*YX*ZX*Y+X’*Y+X*ZY + X*Z
111010111
110010011
101000111
100000000
011101011
010101011
001100000
000100000
The last two columns are identical, which means X*Y + X'*Y + X*Z is the same as Y + X*Z

========================================================
Part (b)

(X+Y)*(X' + Y + Z)
W*(X' + Y + Z) .... let W = X+Y
W*X' + W*Y + W*Z .... Distributive Law
X'*W + Y*W + Z*W .... Commutative Law
X'*(X+Y) + Y*(X+Y) + Z*(X+Y) .... Replace every W with X+Y
X'*X+X'*Y + Y*X+Y*Y + Z*X+Z*Y .... Distributive Law
0+X'*Y + Y*X+Y*Y + Z*X+Z*Y .... Inverse Law
X'*Y + Y*X+Y*Y + Z*X+Z*Y .... Identity Law
X'*Y + Y*X+Y + Z*X+Z*Y .... Idempotent Law
X'*Y + Y*X+Y+Z*Y + Z*X .... Commutative Law
(X'+ X+1+Z)*Y + Z*X .... Distributive Law
(1+1+Z)*Y + Z*X .... Inverse Law
(1+Z)*Y + Z*X .... Idempotent Law
1*Y + Z*Y + Z*X .... Distributive Law
Y + Z*Y + Z*X .... Identity Law
Y + Z*X .... Absorption Law


Answer: Y + Z*X

Side note: This is equivalent to the result of part (a)

Verification Table
XYZX’Y’Z*XX+YX’+Y+Z(X+Y)*(X’+Y+Z)Y + Z*X
1110011111
1100001111
1010111111
1000101000
0111001111
0101001111
0011100100
0001100100



========================================================
Part (c)

X*Y'*Z + X*Y*Z + Y'*Z
X*Z*(Y' + Y) + Y'*Z .... Distributive Law
X*Z*(1) + Y'*Z .... Inverse Law
X*Z + Y'*Z .... Identity Law
(X + Y')*Z ... Distributive Law


Answer: (X + Y')*Z

I'll let you create the verification table.


========================================================
Part (d)

X*Y + X'*Y*Z
(X*Y + X'*Y)*(X*Y + Z) ... Distributive Law (use the "AND" form)
( (X + X')*Y )*(X*Y + Z) ... Distributive Law
( (1)*Y )*(X*Y + Z) ... Inverse Law
Y*(X*Y + Z) ... Identity Law
Y*X*Y + Y*Z ... Distributive Law
X*Y*Y + Y*Z ... Commutative Law
X*Y + Y*Z ... Idempotent Law
Y*(X+Z) ... Distributive Law


Answer: Y*(X+Z)

========================================================
Part (e)

X'*Y + X*Y*Z'
(X'*Y + X*Y)*(X'Y + Z') ... Distributive Law (use the "AND" form)
( (X' + X)*Y)*(X'Y + Z') ... Distributive Law
( (1)*Y)*(X'Y + Z') ... Inverse Law
Y*(X'Y + Z') ... Identity Law
Y*X'Y + Y*Z' ... Distributive Law
X'*Y*Y + Y*Z' ... Commutative Law
X'*Y + Y*Z' ... Idempotent Law
Y*(X'+Z') ... Distributive Law
Y*(X*Z)' ... De Morgan's Law

Answer: Y*(X*Z)'
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