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Boolean Theorems
Investigating the various Boolean theorems (rules) can help us to
simplify logic expressions and logic circuits.
Multivariable Theorems
The theorems presented below involve more than
one variable:
(9)
(10)
x + y = y + x (commutative law)
x * y = y * x (commutative law)
(11)
(12)
(13a)
(13b)
(14)
(15)
x+ (y+z) = (x+y) +z = x+y+z (associative law)
x (yz) = (xy) z = xyz (associative law)
x (y+z) = xy + xz
(w+x)(y+z) = wy + xy + wz + xz
x + xy = x [proof see below]
x + x'y = x + y
Proof of (14)
x + xy
= x (1+y)
= x * 1 [using theorem (6)]
= x [using theorem (2)]
DeMorgan's Theorem
DeMorgan's theorems are extremely useful in simplifying
expressions in which a product or sum of variables is inverted.
The two theorems are:
(16) (x+y)' = x' * y'
(17) (x*y)' = x' + y'
Theorem (16) says that when the OR sum of two variables is
inverted, this is the same as inverting each variable individually
and then ANDing these inverted variables.
Theorem (17) says that when the AND product of two variables is
inverted, this is the same as inverting each variable individually
and then ORing them.
Example
X
= [(A'+C) * (B+D')]'
= (A'+C)' + (B+D')' [by theorem (17)]
= (A''*C') + (B'+D'') [by theorem (16)]
= AC' + B'D
Three Variables DeMorgan's Theorem
(18) (x+y+z)' = x' * y' * z'
(19) (xyz)' = x' + y' + z'
Implications of DeMorgan's Theorem
For (16): (x+y)' = x' * y'
For (17): (x*y)' = x' + y'
Boolean algebraic properties
Another type of mathematical identity, called a "property" or a "law," describes how differing
variables relate to each other in a system of numbers. One of these properties is known as the
commutative property, and it applies equally to addition and multiplication. In essence, the
commutative property tells us we can reverse the order of variables that are either added
together or multiplied together without changing the truth of the expression:
Along with the commutative properties of addition and multiplication, we have the associative
property, again applying equally well to addition and multiplication. This property tells us we
can associate groups of added or multiplied variables together with parentheses without
altering the truth of the equations.
Lastly, we have the distributive property, illustrating how to expand a Boolean expression
formed by the product of a sum, and in reverse shows us how terms may be factored out of
Boolean sums-of-products:
To summarize, here are the three basic properties: commutative, associative, and distributive.
Boolean Functions
NOT.
So
is the complement of A, defined by the truth table
A
0 1
1 0
A truth table defines the boolean function by specifying the boolean output value
associated with each boolean input value. The circuit symbol for the NOT function is the
inverter, Figure 148. The bubble denotes inversion (complement).
Figure 148: NOT gate.
Interestingly, in Quantum Computing there is a
gate, but that is another story.
OR.
Here + denotes OR, defined by:
A B X=A+B
0 0
0
0 1
1
1 0
1
1 1
1
OR is like two switches A, B is parallel; one or both open lets current flow. The circuit
symbol is shown in Figure 149.
Figure 149: OR gate.
AND.
Here denotes AND, defined by:
A B
0 0
0
0 1
0
1 0
0
1 1
1
AND is like two switches A, B is series; both must be open for current to flow. The
circuit symbol is shown in Figure 150.
Figure 150: AND gate.
These three are the most basic logic functions, and define the Boolean algebra on the set
. Also important are the following functions.
NAND.
NAND is defined by:
A B
0 0
1
0 1
1
1 0
1
1 1
0
NAND is the complement of AND. The circuit symbol is shown in Figure 151. (We
encountered the physical NAND gate in section 10.3.)
Figure 151: NAND gate.
NOR.
NOR is defined by:
A B
0 0
1
0 1
0
1 0
0
1 1
0
NOR is the complement of OR. The circuit symbol is shown in Figure 152.
Figure 152: NOR gate.
NAND and NOR are very important building blocks.
XOR.
XOR is defined by:
A B
0 0
0
0 1
1
1 0
1
1 1
0
XOR is true if one and only one of A, B is true; hence the term exclusive OR. . The
circuit symbol is shown in Figure 153.
Figure 153: XOR gate.
XOR is useful in



arithmetic circuits
controlled inverter
parity generation and checking
EQV.
EQV is defined by:
A B
0 0
1
0 1
0
1 0
0
1 1
1
EQV is true only if both A and B are the same, and so is the complement of XOR. It is
called equivalence or exclusive NOR. The circuit symbol is shown in Figure 154.
Figure 154: EQV gate.
EQV is useful for detecting equality.
Basic Boolean Operations
Basic Operations are AND, OR, and Complement
(inverse) Truth Tables list all combinations of inputs and
the corresponding outputs for combinational circuits.
An AND gate produces a 1 iff all inputs are 1
An OR gate produces a 1 iff one or more inputs are
1
A NOT gate (inverter) produces an output opposite
to the input
X Y | XY | X+Y | X'
-----+----+-----+--0 0| 0| 0 |1
0 1| 0| 1 |1
(provide symbols)
1 0| 0| 1 |0
1 1| 1| 1 |0
Combining AND with NOT and OR with NOT, we
have NAND & NOR
X Y | (XY)' | (X+Y)'
-----+-------+------0 0| 1 | 1
0 1| 1 | 0
(provide symbols)
1 0| 1 | 0
1 1| 0 | 0
Show circuit diagram and truth table for F = XY +
X'Y'Z
Draw circuit for F = (AB + C)' and ask class for
Boolean expression
Ask class to draw circuit for F = (A + C)(B' + D)
Real logic circuits behavior is a function of time.
Logic values cannot change instantaneously. Also, there is
a time delay associated with every logic gate.
Axioms
Axioms: minimal set of definitions assumed to be true
X = 0 if X <> 1
X = 1 if X <> 0
If X = 0, X' = 1
If X = 1, X' = 0
0.0=0
1+1=1
1.1=1
0+0=0
0.1=1.0=0
1+0=0+1=1
The above five axioms completely define switching
algebra.All other facts about the system can be proved
starting with the axioms.
Single-Variable Theorems:
X+0=X
X+1=1
X+X=X
(X')' = X
X + X' = 1
X(1) = X -- Identities
X(0) = 0 -- Null elements
XX = X
-- Idempotency
-- Involution
XX' = 0 -- Complements
These theorems are easily proved by showing that
they are valid for both possible values of X.Any boolean
expression can be substituted for the variable X in the
above theorems.
Two- and Three-Variable Theorems:
XY = YX
-- Commutativity
X(YZ) = (XY)Z
Y) + Z -- Associativity
X+Y=Y+X
X + (Y + Z) = (X +
From the above we see that any number of boolean
variables ANDed together will have a value of '1' iff every
variable has a value of '1'.
Likewise, any number of boolean variables ORed
together
will have a value of '0' iff every variable has a value
of '0'.
(D1): X(Y + Z) = XY + XZ (D2): X + YZ = (X +
Y)(X + Z) -- Distributivity
X + XY = X
Covering
X(X + Y) = X
XY + XY' = X
-- Combining
(X + Y)(X + Y') = X
(X + Y')Y = XY
--
XY' + Y = X
+Y
Any boolean expression can be substituted for X
and Y in the above theorems.
Example:
Simplify AB'(C + D) + (C + D)'
AB'(C + D) + (C + D)' = AB' + (C + D)'
by XY' + Y = X + Y, where X = AB' and
Y = (C + D)'
Multiplying Out and Factoring:
SUM OF PRODUCTS: An expression is said to be
in sum-of-products form when all products are the
products of single variables only.
Example:
AB + AC + CD' + E is a sum of
products
A(B+C) + CD + E' is not a
sum of products
The distributive laws are used to
multiply out an
expression to obtain a sum-ofproducts form. Applying D2 first, when applicable, can
reduce the amount of work required to arrive at a simplified
sum-of-products form.
Multiplying Out Example:
(A + B)(B + C)(B + D')(ACD' + E)
= (A + B)(B + CD')(ACD' + E)
by D2
= (B + ACD')(ACD' + E)
by D2
& Comm. law
= ACD' + BE
by D2 &
Comm. law
(A + B')(A + D' + C)(A + D' + B)
= (A + B')(A + D' + BC)
D2
= A + B'(D' + BC)
D2
= A + B'D'
D1
PRODUCT OF SUMS: An expression is said to be
in product-of-sums form when all sums are sums of single
variables.
Example:
(A + B')CD
is a product of
sums
(AB + C)(B + C)
is not a
product of sums
The distributive laws can be used to factor an
expression repeatedly until the expression is in a productof-sums form. Applying D1 first can reduce the amount of
work required to fully
factor an expression.
Factoring Example:
ABC + A'B'C + CD'
= C(AB + A'B') + CD'
by D1
= C(AB + A'B' + D')
by D1
= C((AB + A')(AB + B') + D')
by D2
= C((A + A')(B + A')(A + B')(B + B') + D')
by D2
= C((B + A')(A + B') + D')
= C(B + A'+ D')(A + B'+ D')
by D2
ABC + A'B'C
= C(AB + A'B')
D1
= C((AB + A')(AB + B'))
D2
= C((A' + A)(A' + B)(B' + A)(B' + A))
D2
= C(A' + B)(B' + A)