Exam 12: A: Boolean Algebra

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mark each statement TRUE or FALSE. - xxx+xy+yy=xyx x x + x y + y y = x y

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determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. - x+y+z=xyzx + y + z = x y z

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Use the Quine-McCluskey method to simplify the Boolean expression xˉyz+xˉyzˉ+xˉyˉz+xˉyˉzˉ+xyz\bar { x } y z + \bar { x } y \bar { z } + \bar { x } \bar { y } z + \bar { x } \bar { y } \bar { z } + x y z

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determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. - x(x+y)=x+yxx ( x + y ) = x + y x

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determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. - xˉz+xz=z\bar { x } z + x z = z

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mark each statement TRUE or FALSE. - x+y=xˉyˉx + y = \bar { x } \quad \bar { y }

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 Let F(x,y,z)=(yˉz)(x+xˉy)\text { Let } F ( x , y , z ) = ( \bar { y } z ) ( x + \bar { x } y ) \text {. } Show that F can be simplified to give y+xzˉy + x \bar { z }

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mark each statement TRUE or FALSE. - xy=xy\overline { x \mid y } = x \downarrow y

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mark each statement TRUE or FALSE. -When written as a sum of minterms in the variables x and y,1=xy+xyˉ+xˉy+xˉyˉy , 1 = x y + x \bar { y } + \bar { x } y + \bar { x } \bar { y }

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Using only the five properties associative laws, commutative laws, distributive laws, identity laws, and com- plement laws, prove that xx=xx x = x is true in all Boolean algebras.

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mark each statement TRUE or FALSE. -{+, ·} is a functionally complete set of operators.

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fill in the blanks. -There are ____ Boolean functions with 3 variables.

(Short Answer)
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If f(w,x,y,z)f ( w , x , y , z ) = (xˉ+yzˉ)\overline{( \bar { x } + y \bar { z } )} + (wˉx)( \bar { w } x ) , find f(0,0,0,0)f ( 0,0,0,0 )

(Short Answer)
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fill in the blanks. -When written as a product of maxterms (in the variables x and y), (x+y)z=( x + y ) z = ____.

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Show that the Boolean function F given by F(x,y,z)=xˉ+y+xy+x+y simplifies to x+yˉ, by using F \text { given by } F ( x , y , z ) = \overline { \bar { x } + y } + x y + \overline { x + y } \text { simplifies to } x + \bar { y } \text {, by using } only the definition of a Boolean algebra.

(Essay)
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Give a reason for each step in the proof that x + 1 = x is true in Boolean algebras. Your reasons should come from the following: associative laws for addition and multiplication, commutative laws for addition and multiplication, distributive law for multiplication over addition and distributive law for addition over multiplication, identity laws, unit property, and zero property. 1=x+xˉ=x+xˉ1=(x+xˉ)(x+1)=1(x+1)=x+11 = x + \bar { x } = x + \bar { x } \cdot 1 = ( x + \bar { x } ) ( x + 1 ) = 1 \cdot ( x + 1 ) = x + 1

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mark each statement TRUE or FALSE. - xˉyˉ=((xx)(yy))((xx)(yy))\bar { x } \bar { y } = ( ( x \mid x ) \mid ( y \mid y ) ) \mid ( ( x \mid x ) \mid ( y \mid y ) )

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Write xy+xyzˉx y + x y ^ { \bar { z } } as a sum-of-products in the variables x, y, and z.

(Short Answer)
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Give a reason for each step in the proof that x + xy = x is true in Boolean algebras. Your reasons should come from the following: associative laws for addition and multiplication, commutative laws for addition and multiplication, distributive law for multiplication over addition and distributive law for addition over multiplication, identity laws, unit property, zero property, and idempotent laws. x+xy=x\cdot1+xy=x(y+)+xy=(xy+x)+xy=xy+(xy+x)=(xy+xy)+x=xy+x= x(y+)=x\cdot1=x

(Essay)
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If f(w,x,y,z)f ( w , x , y , z ) = (xˉ+yzˉ)\left( \bar { x } + y \bar { z } \right) + (wˉx)( \bar { w } x ) , find ff (1,1,1,1) .

(Short Answer)
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