Question 1191035: Simplify following functions using Boolean Algebra
a. F(X, Y, Z) =(XY) + ( X + Y +Z)’X + YZ
b. F(X, Y, Z) = (XY)’ + (X+Y +Z)’
c. F(X,Y,Z) = YZ + (X+Y)’ + (XYZ)’
d. F(X,Y,Z) = (X+Y +Z)’ ( X+Y)
Can you please explain me this homework question step-by-step? Thank you very much!
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
We'll need these Boolean Algebra rules
| Name | AND form | OR form | | Identity Law | 1*A = A | A+0 = A | | Null Law | 0*A = 0 | A+1 = 1 | | Idempotent Law | A*A = A | A+A = A | | Inverse Law | A*A' = 0 | A+A' = 1 | | Commutative Law | A*B = B*A | A+B = B+A | | Associative Law | A*(B*C) = (A*B)*C | A+(B+C) = (A+B)+C | | Distributive Law | A+B*C = (A+B)*(A+C) | A*(B+C) = A*B+A*C | | Absorption Law | A*(A+B) = A | A+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.
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Part (a)
X*Y + (X+Y+Z)'*X + Y*Z
X*Y + ((X+Y)'*Z')*X + Y*Z ... De Morgan's Law
X*Y + (X'*Y'*Z')*X + Y*Z ... De Morgan's Law
X*Y + X'*X*Y'*Z' + Y*Z ... Commutative Law
X*Y + 0*Y'*Z' + Y*Z ... Inverse Law
X*Y + 0 + Y*Z ... Null Law
X*Y + Y*Z ... Identity Law
Y*X + Y*Z ... Commutative Law
Y*(X+Z) ... Distributive Law
========================================================
Part (b)
(X*Y)' + (X+Y+Z)'
X'+Y' + X'*Y'*Z' ... De Morgan's Law
X'+Y' + Y'*X'*Z' ... Commutative Law
X'+Y' ... Absorption Law
(X*Y)' ... De Morgan's Law
========================================================
Part (c)
Y*Z + (X+Y)' + (X*Y*Z)'
Y*Z + X'*Y' + X'+(Y*Z)' ... De Morgan's Law
Y*Z+(Y*Z)' + X'*Y' + X' ... Commutative Law
1 + X'*Y' + X' ... Inverse Law
1 + X' + X'*Y' ... Commutative Law
1 + X' ... Absorption Law
1 .... Null Law
This means the entire original expression is equivalent to 1 regardless of what X,Y, and Z are.
We can say that Y*Z + (X+Y)' + (X*Y*Z)' = 1 is an identity.
More specifically, the original expression is a tautology since it always results in true (ie the value 1).
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Part (d)
(X+Y+Z)'*(X+Y)
(W+Z)'*(W) .... Let W = X+Y
W'*Z'*W .... De Morgan's Law
W'*W*Z' .... Commutative Law
0*Z' ... Inverse Law
0 .... Null Law
We can say that (X+Y+Z)'*(X+Y) = 0 is an identity.
The original expression is always equal to 0 regardless of what X,Y, and Z are.
The original expression is a contradiction since it always results in false (ie the value 0).
========================================================
The verification table for each part can be found here
https://docs.google.com/spreadsheets/d/1PQIkJG5gGlyMGEN8upSCosFSfEtkYjzca5wnvuWfNhY/edit?usp=sharing
You don't need to have a Google account to be able to see the spreadsheet in the link.
You can use excel (or similar) to make the same set of tables.
Click on any cell to see what formula I used for it.
For example, when computing the first item in the X+Y column for part d, I typed in =OR(B3,C3)*1 for cell E3.
Don't forget about the equal sign up front or else the formula won't evaluate.
The *1 at the end converts the result from TRUE/FALSE to their numeric counterparts (1 and 0 respectively).
Navigate to any part a through d by clicking at the bottom of the page.
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