Basic Laws, theorems, and postulates of Boolean Algebra EEE301 – Digital Electronics Rachaen M. Huq Dept. of EEE, BRACU Rachaen M. Huq

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Associative Law In traditional algebra : (xy)z= x(yz) How would you prove this? Think of something simple. Rachaen M. Huq

Associative Law we would show that for any value of x,y and z, the identity would be true. In Boolean Algebra, (x.y).z = x.(y.z) How can we be sure? What would be true? ? Rachaen M. Huq

Associative Law The truth table x y z xy (xy)z 1 x y z yz X(yz) 1 1 x y z yz X(yz) 1 Rachaen M. Huq

Other laws x + x' = 1 x . x' = 0 x + y = y + x x . y = y . x Rachaen M. Huq

Other laws (x + y) + z = x + (y + z) = x + y + z 0 + x = x + 0 = x 1 . x = x . 1 = x Rachaen M. Huq

Other Laws x . (y + z) = (x . y) + (x . z) You can prove all these using .... ? Rachaen M. Huq

Duality – an interesting property of Boolean algebra every valid Boolean expression (equality) remains valid if the operators and identity elements are interchanged, as follows: +  . 1  0 Example: Given the expression a + (b.c) = (a+b).(a+c) The following would be true as well. a . (b+c) = (a.b) + (a.c)

Duality (x.y.z)' = x'+y'+z’ Duality gives free theorems If (x+y+z)' = x'.y.'z' is valid, then its dual is also valid: (x.y.z)' = x'+y'+z’ If x + 1 = 1 is valid, then its dual is also valid: x . 0 = 0

Important! Understand the difference between identities and equations (a+b)^2 = a^2 + 2ab + b^2 is an identity (can be compared with laws, theorems, postulates) X^2 -2x+1 = 0 ….. Is an equation, i.e for only a (limited) number of real values of x would satisfy both sides of the equal sign. Rachaen M. Huq

Basic Theorems/postulates of Boolean Algebra x+0=x ------------ x.1=x x+x’=1 ------------ x.x’=0 x+x=x --------------x.x=x x+1=1 -------------- x.0=0 (x’)’=x x+y=y+x -----------xy=yx x(yz)=(xy)z --------x+(y+z)=(x+y)+z x(y+z)=xy+xz ----- x+yz=(x+y)(x+z) (x+y)’=x’y’ -------- (xy)’=x’+y’ x+xy=x ------------- x(x+y)=x Table 2.1 Morris Mano Super important !

Why Boolean Algebra is interesting? Two different expressions carry the same meaning. If a Boolean Function is expressed by an expression, which has a shorter version (e.g. x+xy=x) then it is certainly interesting for us. Why? Because length of the expression is related to the size of chips (number of gates), in other words MONEY! Rachaen M. Huq

How Boolean Algebra reduce cost! Lets take an example of two Booleans functions: F1 = x+ xy (express in words what they mean) F2 = x (express in words what they mean) Draw the logic diagrams See the difference. Boolean algebra says that there was NO gate necessary at all, maybe just one buffer. Rachaen M. Huq

Basic Theorems of Boolean Algebra Theorems can be proved using the truth table method. (Exercise: Prove De-Morgan’s theorem using the truth table.) They can also be proved by algebraic manipulation using axioms/postulates or other basic theorems.

Basic Theorems of Boolean Algebra Theorem 6a (absorption) can be proved by: x + x.y = x.1 + x.y (identity) = x.(1 + y) (distributivity) = x.(y + 1) (commutativity) = x.1 (Theorem 2a) = x (identity) By duality, theorem 6b: x.(x+y) = x Try prove this by algebraic manipulation.

Boolean Functions (Solve ?) Examples: F1= xyz' F2= x + y'z F3=(x'y'z)+(x'yz)+(xy') F4=xy'+x'z From the truth table, F3=F4. Can you also prove by algebraic manipulation that F3=F4?

Switch to blackboard for more examples.

Drawing Logic Circuit – Trivial- See at home When a Boolean expression is provided, we can easily draw the logic circuit. Examples: (i) F1 = xz' (note the use of a 2-input AND gate) x z F1 z'

Drawing Logic Circuit (ii) F2 = x + y'z (can assume that variables and their complements are available) x y' z F2 y'z (iii) F3 = xy' + x'z x' z F3 x'z xy' x y'

Solve it yourself Q1. Draw a logic circuit for BD + BE + D’F A’BC + B’CD + BC’D + ABD’

Analysing Logic Circuit When a logic circuit is provided, we can analyse the circuit to obtain the logic expression. Example: What is the Boolean expression of F4? A'B' A' B' C F4 A'B'+C (A'B'+C)' F4 = (A'B'+C)' = (A+B).C'

Solve it yourself What is Boolean expression of F5? z F5 x y

There are always shortcuts (Format approaches) Shortcuts to derive Boolean expressions from the truth table directly and vice versa No intermediate columns (e.g. columns for x’ ,or xy’, or z’) are necessary. Using minterms (mostly used) / maxterms Can be derived by looking closely at the truth tables. Rachaen M. Huq

Canonical Forms Shortest way to express a Boolean Function. Gives the idea about what the function logically does to inputs. But no idea about how many gates required to implement it. Rachaen M. Huq