Switching Algebra
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pp. 194-206
Logic
What is a logical argument?
an argument that is true by virtue of its form; i.e. if
premises true conclusion must be true
Positive example
P1) All men are mortal
P2) Harry is a man
----------------C) Therefore, Harry is mortal
Negative examples
P1) If I study hard in 200, I will get an A.
P2) I got an A in 200.
-----------------C) Therefore, I studied hard.
P1) Katz believes that 200 is easy.
P2) Katz is a dufus.
------------------C) Therefore, 200 is difficult.
In the mid-19th century, Boole formalized logic.
We will be studying a subset of formal logic known as
the propositional calculus.
Switching Algebra
Axioms of the propositional calculus
(1) All variables must take one of the values 0 or 1
(2) Rules for not
X
X’
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pp. 194-206
----------0
1
1
0
(3) Rules for and
X
Y
X·Y
---------------------0
0
0
0
1
0
1
0
0
1
1
1
(4) Rules for or
X
Y
X+Y
-----------------------0
0
0
0
1
1
1
0
1
1
1
1
Switching Algebra
W
pp. 194-206
Theorems of the propositional calculus
(T1)
(T2)
(T3)
(T4)
(T5)
(T6)
(T7)
(T8)
(T9)
theorem
X+0=X
X+1=1
X+X=X
(X’)’ = X
X + X’ = 1
X+Y=Y+X
(X + Y) + Z = X + (Y + Z)
X · Y + X · Z = X · (Y + Z)
(X · Y)’ = X’ + Y’
Proof by truth table
dual
X ·1=X
X ·0=0
X ·X=X
(X’)’ = X
X · X’ = 0
X·Y=Y·X
(X + Y) + Z = X + (Y + Z)
(X + Y) · (X + Z) = X + Y · Z
(X + Y)’ = X’ · Y’
name
identity
null elements
idempotency
involution
complement
commutativity
associativity
distributivity
DeMorgan’s
Fact
All theorems of the pc can be proven with a truth table, i.e., by
systematically examining every case and showing that the
theorem hold for these cases (this method also is known as
perfect induction).
Examples
Prove (T5):
X
0
1
X’
1
0
X + X’
1
1
Prove the dual of (T9):
X
Y
X+Y
(X + Y)’
X’
Y’
X’ · Y’
0
0
1
1
0
0
1
1
0
1
0
1
1
0
0
0
1
0
0
1
1
1
1
1
Switching Algebra
0
0
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pp. 194-206
Proof by simplification
Method
Successive application of previously proven theorems to produce
a shorter expression
Example
Prove T9:
X+X·Y=X
X+X·Y
=X·1+X·Y
= X · (1 + Y)
=X·1
=X
(T1 dual)
(T8)
(T2)
(T1 dual)
Advantages and disadvantages
advantage - no need to provide destination
disadvantage - need to choose among existing theorems
0
0
Duality
Principle
Any theorem or identity remains true if 0 and 1 are swapped and
· and + are swapped throughout.
Generalized DeMorgan
[F(X1, X2, .... Xn )]’ = FD(X1’, X2’, .... Xn’ )
Switching Algebra
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pp. 194-206
Minterm and Maxterm representations of a function
row#
X
Y
Z
minterm maxterm
0
0
0
0
X’Y’Z’
X+Y+Z
1
0
0
1
X’Y’Z
X+Y+Z’
2
0
1
0
X’YZ’
X+Y’+Z
3
0
1
1
X’YZ
X+Y’+Z’
4
1
0
0
XY’Z’
X’+Y+Z
5
1
0
1
X’YZ’
X+Y’+Z
6
1
1
0
XYZ’
X’+Y’+Z
7
1
1
1
XYZ
X’+Y’+Z’
Principle
A function can be represented as
a) the sum of the minterms where the function is true, OR
b) the product of the maxterms where the function is false
Example
F = X’ + YZ
row#
0
1
2
3
4
5
X
0
0
0
0
1
1
Y
0
0
1
1
0
0
Z
0
1
0
1
0
1
minterm
X’Y’Z’
X’Y’Z
X’YZ’
X’YZ
XY’Z’
X’YZ’
maxterm
X+Y+Z
X+Y+Z’
X+Y’+Z
X+Y’+Z’
X’+Y+Z
X+Y’+Z
F
1
1
1
1
0
0
6
7
1
1
1
1
0
1
XYZ’
XYZ
X’+Y’+Z
X’+Y’+Z’
0
1
minterm representation
X,Y,Z(0,1,2,3,7) = X’Y’Z’ + X’Y’Z + X’YZ’ + X’YZ + XYZ
maxterm representation
X,Y,Z(4,5,6)
= (X’+Y+Z) (X+Y’+Z) (X’+Y’+Z)
Switching Algebra
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pp. 194-206
Why are minterm and maxterm representations equivalent?
a) [X,Y,Z(i,j...n)]’ = X,Y,Z(i,j...n)
b) but, X,Y,Z(i,j...n) is just the inverse of X,Y,Z(p,q,..r),
where p,q,..r are the complement of i,j,..n
c) therefore, by the principle of double negation,
X,Y,Z(i,j...n) = X,Y,Z(p,q,..r)
Terminology review
There are two equivalent representations of a function:


i)
X,Y,Z
sum of minterms
sum of products (SOP)


ii) X,Y,Z
product of maxterms
product of sums (POS)
Combinational-Circuit Analysis
Basic gates
W
pp. 209-214
And gate
Or gate
Not gate
Combinational-Circuit
Analysis
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pp. 209-214
Truth Table from Circuit
Examples
easy
harder