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Boolean Algebra
Definition : A Boolean Algebra is a math construct (B,+,. , ‘, 0,1) where B is a non-empty set, + and. are binary operations in B, ‘ is a unary operation in B, 0 and 1 are special elements of B, such that:
a) + and. are communative: for all x and y in B, x+y=y+x, and x.y=y.x b) + and. are associative: for all x, y and z in B, x+(y+z)=(x+y)+z, and x.(y.z)=(x.y).z c) + and. are distributive over one another: x.(y+z)=xy+xz, and x+(y.z)=(x+y).(x+z) d) Identity laws: 1.x=x.1=x and 0+x=x+0=x for all x in B e) Complementation laws: x+x’=1 and x.x’=0 for all x in B
Examples:
(B=set of all propositions, ∨, ∧, ¬ , T, F) (B=2 A, U, ∩, c^ , Φ,A)
Theorem 1 : Let (B,+,. , ‘, 0,1) be a Boolean Algebra. Then the following hold:
a) x+x=x and x.x=x for all x in B b) x+1=1 and 0.x=0 for all x in B c) x+(xy)=x and x.(x+y)=x for all x and y in B
Proof:
a) x = x+0 Identity laws = x+xx’ Complementation laws = (x+x).(x+x’) because + is distributive over. = (x+x).1 Complementation laws = x+x Identity laws x = x.1 Identity laws = x.(x+x’) Complementation laws = x.x +x.x’ because + is distributive over. = x.x+0 Identity laws = x.x b) x+1 =x+(x+x’) Complementation laws = (x+x)+x’ + is associative = x+x’ using (a) = 1 Complementation laws 0.x =(x’.x).x Complementation laws = x’.(x.x). is associative = x’.x using (a) =0 Complementation laws c) x+(xy) = x.1+x.y Identity laws =x.(1+y) because + is distributive over.
=x.1 using (b) =x Identity laws
x.(x+y) = x.x+x.y Distributivity laws =x+x.y by (a) =x Just shown above. Q.E.D.
Definition: An element y in B is called a complement of an element x in B if x+y=1 and xy=
Theorem 2 : For every element x in B, the complement of x exists and is unique. Proof: Existence. Let x be in B. x’ exists because ‘ is a unary operation. X’ is a complement of x because it satisfies the definition of a complement (x+x’=1 and xx’=0). Uniqueness. Let y be a complement of x. We will show that y=x’. Since y is a complement of x, we have x+y=1 and xy=yx=0. y=y.1=y.(x+x’)=yx+yx’=0+yx’=xx’+yx’=(x+y)x’=1.x’=x’ => y=x’. QED
Corollary 1 : (x’)’=x. Proof, since x’+x=1 and x’x=0, it follows that x is a complement of x’. Since the complement of x’ is unique, it follows then that (x’)’ , which is a complement of x’, and x, which is also a complement of x’, must be equal. Thus, (x’)’=x. QED
Theorem 3 (De Morgan’s Laws): a) (x+y)’=x’y’ b) (xy)’=x’+y’ Proof: a) Show that x’y’+(x+y)=1 and (x’y’)(x+y)=0. x’y’+(x+y)=(x’y’+x)+y=(x’+x)(y’+x)+y=1.(y’+x)+y=(y’+x)+y=(x+y’)+y=x+(y’+y)=x+1= (x’y’)(x+y)=(x’y’)x+(x’y’)y=(y’x’)x+x’(y’y)=y’(x’x)+x’0=y’0+0=0+0= b) The proof is similar and left as an exercise.
QED.
Definition : Let (B,+,. , ‘, 0,1) be a Boolean Algebra. Define the following ≤ relation in B:
x ≤ y if xy=x
Theorem 4 : The relation ≤ is a partial order relation.
Proof : We need to prove that ≤ is reflexive, antisymmetric and transitive
Reflexivity: since xx=x (by Theorem 1-a), it follows that x ≤ x Antisymmetry: need to show that x≤y and y≤x => x=y. x≤y and y≤x => xy=x and yx=y =>
Definition : A literal is any Boolean variable x or its complement x’.
Truth Tables of Boolean functions :
Much like the truth tables for logical propositions If f(x,y,z, …) is an n-variable Boolean function, a truth table for f is a table of n+1 columns (one column per variable, and one column for f itself), where the rows represent all the 2 n combinations of 0-1 values of the n variables, and the corresponding value of f for each combination. Examples:
f(x,y)=xy+x’y’; x y f 1 1 1 1 0 0 0 1 0 0 0 1
g(x,y,z) = xy’z’+x’y’z+x’yz’; x y z g 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 0 0 0 1
h(x,y,z,w) = x’y’w’+xyzw+xz’; x y z w h 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0
u(x,y,z,w)= 1 if the string xyzw has an odd number of 1’s; otherwise, it is 0. x y z w u 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0
Definitions of Minterms and Maxterms :
Suppose we’re dealing with n Boolean variables. A minterm is any product of n literals where each of the n variable appears once in the product. o Example, where n=3 and the variables are x, y and z: Then, xyz, xy’z, xy’z’ are all miterms. xy is not a minterm because z is missing. Also, xyzy’ is not a minterm because y appears multiple times (once as y, and another time as y’). o For n=2 where the variables are x and y, there are 4 minterms in total: xy, xy’, x’y, x’y’. A maxterm is any sum of n literals where each of the n variable appears once in the sum. o Example, where n=3 and the variables are x, y and z: x+y+z, x+y’+z’ are both maxterms (of 3 variables). x+y’ is not a maxterm because z is missing.
Definition (Disjunctive Normal Form): A Boolean function/expression is in Disjunctive Normal Form (DNF), also called minterm canonical form , if the function/expression is a sum of minterms.
Examples:
f(x,y,z)= xyz+xy’z+x’yz’+x’y’z is in DNF g(x,y)=xy+x’y’ is in DNF But h(x,y,z)=xy+x’y’z is not in DNF because xy is not a minterm of size 3.
Definition (Conjunctive Normal Form): A Boolean function/expression is in Conjunctive Normal Form (CNF), also called maxterm canonical form, if the function/expression is a product of maxterms.
Examples:
f(x,y,z)= (x+y+z)(x+y+z’)(x’+y+z’)(x’+y’+z) is in CNF g(x,y)=(x+y)(x’+y’) is in CNF But h(x,y,z)=(x+y)(x’+y’+z) is not in CNF because x+y is not a maxterm of size 3.
Observation: Thanks to De Morgan’s Laws, if f is in DNF, then f’ derived from the DNF using De Morgan’s Laws (that is, changing every literal to its complement, and every “.” to “+”, and every “+” to “.”) is in CNF, and vice versa.
Method of Putting a Function in DNF, using Truth Tables :
- Create the truth table of the given Boolean function f
Finally, applying De Morgan’s, we get the CNF of f: f=(f’)’= (x’+y’+z’)(x+y’+z’)(x+y’+z)(x+y+z).
Optimization of Boolean functions using Karnaugh Maps :
x’y’z + xz + xy’z’
y y’ x 1 1 1 x’ 1 z z’ z Minimized form: xz + xy’ + y’z
xz + yz’ + y’z’ y y’ x 1 1 1 1 x’ 1 1 z z’ z
Minimized form: x+z’
xyz’ + xy’z’w’ + x’y’zw + x’yw + y’z’w y y’ x 1 1 w 1 1 w’ x’ 1 1 1 1 w z z’ z
Minimized form: xz’+x’w
x’zw’ + yz’w’ + x’y’z’ + y’z’w’ + x’yz’ y y’ x w 1 1 w’ x’ 1 1 1 1 1 1 w z z’ z
Minimized form: x’w’+z’w’+x’z’
General procedure for Karnaugh-map-based minimization of Boolean functions :
- The Karnaugh map is a table of squares (2 n^ squares when you have n variables)
- Divide the map into regions so that each variable “owns” half of the squares, and its complement owns the other half. Each square will end up being owned by n literals (making up a minterm).
For example, for 3 variables, the map (of 8 sqaures) is
y y’ x x’ z z’ z The variable x owns the top 4 yellow squares. Its complement x’ owns the bottom yellow row of squares. The variable y owns the left half of the squares, and y’ owns the right half. The variavle z owns the left and right column of yellow squares, and its complement z’owns the two middle columns of yellow squares.
For 4 variables, the map (of 16 squares) is:
y y’ x w w’ x’ w z z’ z
o The variable x owns the top two rows of yellow squares. Its complement x’ owns the bottom two yellow rows of squares. o The variable y owns the left half of the squares, and y’ owns the right half. o The variavle z owns the left and right column of yellow squares, and its complement z’owns the two middle columns of yellow squares. o Finally, the variable w owns the top and bottom row of yellow squares, and its complement owns the two middle rows of the yellow squares.
- Fill the map for a given function f: for each minterm in f, put “1” inside the square corresponding to (or owned by) that minterm.
- Grouping the filled squares: Group the 1’s into rectangles of totally filled squares such that o the length and width of each rectangle are powers of 2 o no filled square remains ungrouped o rectangles can overlap o each rectangle must exclusively own at least one filled square.
- Convert each rectangle to a product of literals: For each rectangle, identify the literals where each literal owns the entire rectangle, then multiply those literals.
- The minimized form is the sum of the products derived in the previous steps.
