chapter 10.1 and 10.2: boolean algebra
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Chapter 10.1 and 10.2: Boolean Algebra. Based on Slides from Discrete Mathematical Structures: Theory and Applications. Learning Objectives. Learn about Boolean expressions Become aware of the basic properties of Boolean algebra. Two-Element Boolean Algebra. Let B = {0, 1}. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 10.1 and 10.2: Boolean Algebra
Based on Slides fromDiscrete Mathematical Structures: Theory and Applications
Discrete Mathematical Structures: Theory and Applications 2
Learning Objectives
Learn about Boolean expressions
Become aware of the basic properties of Boolean algebra
Discrete Mathematical Structures: Theory and Applications 3
Two-Element Boolean AlgebraLet B = {0, 1}.
Discrete Mathematical Structures: Theory and Applications 4
Two-Element Boolean Algebra
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Two-Element Boolean Algebra
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Two-Element Boolean Algebra
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Boolean Algebra
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Boolean Algebra
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Find a minterm that equals 1 ifx1 = x3 = 0 and x2 = x4 = x5 =1,and equals 0 otherwise.
x’1x2x’3x4x5
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Therefore, the set of operators {. , +, ‘} is functionally complete.
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Sum of products expression
Example 3, p. 710
Find the sum of products expansion of
F(x,y,z) = (x + y) z’
Two approaches:
1) Use Boolean identifies
2) Use table of F values for all possible 1/0 assignments of variables x,y,z
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F(x,y,z) = (x + y) z’
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F(x,y,z) = (x + y) z’
F(x,y,z) = (x + y) z’= xyz’ + xy’z’ + x’yz’
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Functional Completeness
This means that the set of operators {. , +, '} is functionally complete.
Summery:A function f: Bn B, where B={0,1}, is a Boolean function.
For every Boolean function, there exists a Boolean expression with the same truth values, which can be expressed as Boolean sum of minterms.Each minterm is a product of Boolean variables or their complements.
Thus, every Boolean function can be represented with Boolean operators ·,+,'
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Functional Completeness
0100111
The question is:Can we find a smaller functionally complete set?Yes, {. , '}, since x + y = (x' . y')'Can we find a set with just one operator?Yes, {NAND}, {NOR} are functionally complete:
NAND: 1|1 = 0 and 1|0 = 0|1 = 0|0 = 1
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{NAND} is functionally complete, since {. , '} is so andx' = x|xxy = (x|y)|(x|y)
NOR: