Boolean Algebra cont

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Boolean Algebra contThe digital abstraction
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• Theorem Absorption Law
• For every pair of elements a , b ? B,
• a a b a
• a ( a b ) a

Proof (1)
Identity
Distributivity
Commutativity
Theorem For any a ? B, a 1 1
Identity
(2) duality.
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• Theorem Associative Law
• In a Boolean algebra, each of the binary
  operations ( ) and ( ) is associative. That
  is, for every a , b , c ? B,
• a ( b c ) ( a b ) c
• a ( b c ) ( a b ) c

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Proof (1) Let
Distributivity
Commutativity
Distributivity
Distributivity
Idempotent Law
Absorption Law
Absorption Law
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Commutativity
Distributivity
Distributivity
Idempotent Law
Absorption Law
Commutativity
Absorption Law
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Same transitions
Putting it all together
before
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Also,
(2) Duality
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• Theorem 11 DeMorgans Law
• For every pair of elements a , b ? B,
• ( a b ) a b
• ( a b ) a b

Proof (1) We first prove that (ab) is the
complement of ab. Thus, (ab) ab
By the definition of the complement and its
uniqueness, it suffices to show (i)
(ab)(ab) 1 and (ii) (ab)(ab) 0.
(2) Duality (ab) ab
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Distributivity
Commutativity
Associativity
a and b are the complements of a and b
respectively
Theorem For any a ? B, a 1 1
Idempotent Law
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Commutativity
Distributivity
Commutativity
Associativity
Commutativity
a and b are the complements of a and b
respectively
Theorem For any a ? B, a 0 0
Idempotent Law
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Algebra of Sets
Consider a set S. B all the subsets of S
(denoted by P(S)).
plus ? set-union ? times ? set-intersection n
Additive identity element empty set Ø
Multiplicative identity element the set S.
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Theorem The algebra of sets is a Boolean
algebra.
Proof By satisfying the axioms of Boolean
algebra
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A1. Cummutativity of ( ? ) and ( n ).
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A2. Distributivity of ( ? ) and ( n ).
Let
and
or
or
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This can be conducted in the same manner as ?. We
present an alternative way
Definition of intersection

Also, definition of intersection
definition of union
Similarly,

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Taking () and () we get,
Distributivity of union over intersection can be
conducted in the same manner.
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A3. Existence of additive and multiplicative
identity element.
A4. Existence of the complement.
Algebra of sets is Boolean algebra.
All axioms are satisfied
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Boolean expression - Recursive definition
base 0 , 1 , a ? B expressions. recursion
step Let E1 and E2 be Boolean expressions.
Then, E1 ( E1 E2 ) ( E1
E2 )
Dual transformation - Recursive definition Dual
expressions ? expressions base 0 ? 1 1 ?
0 a ? a , a ? B\0,1 recursion step Let E1
and E2 be Boolean expressions. Then,
E1 ? dual(E1) ( E1 E2 ) ? dual(E1)
dual(E2) ( E1 E2 ) ? dual(E1)
dual(E2)
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Let fd be the dual of a function f ( x1 , x2 ,
, xn )
Lemma In switching algebra, fd f ( x1 ,
x2 , , xn )
Proof Let f ( x1 , x2 , , xn ) be a Boolean
expression. We show that applying the complement
on the whole expression together with replacing
each variable by its complement, yields the dual
transformation definition.
Induction basis 0 , 1 expressions.
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Induction hypothesis Lemma holds for Boolean
expressions E1 and E2 . That is
Induction step show that it is true for E1
( E1 E2 ) ( E1 E2 )
then,
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If
then,
If
then,
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Definition A function f is called self-dual
if f fd
Lemma For any function f and any two-valued
variable A, the function g Af Afd is a
self-dual.
Proof (holds for any Boolean algebra)
Dual definition
Distributivity
Commutativity
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Distributivity
Commutativity
A is the complement of A
Identity
Commutativity
Notice that the above expression has the
form ab ac bc where a A, bf,
c fd.
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We now prove a stronger claim
Identity
a is the complement of a
Distributivity
Commutativity
Commutativity
Distributivity
Theorem For any a ? B, a 1 1
Identity
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For example
self-dual
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Easier proof (1) for switching algebra only
(using dual properties)
Switching algebra
OR
Identity
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Easier proof (2) for switching algebra only
(case analysis)
A 0
0 1
Identity
Commutativity
Theorem For any a ? B, a 0 0
Absorption Law
Identity
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A 1
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Example of a transfer function for an inverter
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true only if
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BUT, this is not always the case. For example
Moreover, in this example it can be proved that
no threshold values exist, which are consistent
with definition 3 from lecture notes.
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Using the assumption
slope lt -1
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slope -1
f (x) x
slope -1
true if
slope lt -1