Question 1190973
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We'll need these Boolean Algebra rules
<table border = "1" cellpadding = "5"><tr><td>Name</td><td>AND form</td><td>OR form</td></tr><tr><td>Identity Law</td><td>1*A = A</td><td>A+0 = A</td></tr><tr><td>Null Law</td><td>0*A = 0</td><td>A+1 = 1</td></tr><tr><td>Idempotent Law</td><td>A*A = A</td><td>A+A = A</td></tr><tr><td>Inverse Law</td><td>A*A' = 0</td><td>A+A' = 1</td></tr><tr><td>Commutative Law</td><td>A*B = B*A</td><td>A+B = B+A</td></tr><tr><td>Associative Law</td><td>A*(B*C) = (A*B)*C</td><td>A+(B+C) = (A+B)+C</td></tr><tr><td>Distributive Law</td><td>A+B*C = (A+B)*(A+C)</td><td>A*(B+C) = A*B+A*C</td></tr><tr><td>Absorption Law</td><td>A*(A+B) = A</td><td>A+A*B = A</td></tr><tr><td>De Morgan's Law</td><td>(A*B)' = A' + B'</td><td>(A+B)' = A'*B'</td></tr></table>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: *[Tex \large A' = \overline{A}]
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).



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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<table border = "1" cellpadding = "5"><tr><td>X</td><td>Y</td><td>Z</td><td>X’</td><td>X*Y</td><td>X’*Y</td><td>X*Z</td><td>X*Y+X’*Y+X*Z</td><td>Y + X*Z</td></tr><tr><td>1</td><td>1</td><td>1</td><td>0</td><td>1</td><td>0</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>0</td><td>0</td><td>1</td><td>0</td><td>0</td><td>1</td><td>1</td></tr><tr><td>1</td><td>0</td><td>1</td><td>0</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>1</td><td>1</td><td>1</td><td>0</td><td>1</td><td>0</td><td>1</td><td>1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td><td>1</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>0</td><td>0</td><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr></table>The last two columns are identical, which means X*Y + X'*Y + X*Z is the same as Y + X*Z


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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
<table border = "1" cellpadding = "5"><tr><td>X</td><td>Y</td><td>Z</td><td>X’</td><td>Y’</td><td>Z*X</td><td>X+Y</td><td>X’+Y+Z</td><td>(X+Y)*(X’+Y+Z)</td><td>Y + Z*X</td></tr><tr><td>1</td><td>1</td><td>1</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>0</td><td>1</td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>0</td><td>0</td><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>1</td><td>1</td><td>1</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>0</td><td>0</td><td>1</td><td>0</td><td>0</td></tr><tr><td>0</td><td>0</td><td>0</td><td>1</td><td>1</td><td>0</td><td>0</td><td>1</td><td>0</td><td>0</td></tr></table>



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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.



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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)


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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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