1. --- outputs logical functions of inputs --- new outputs appear shortly after changed inputs (propagation delay) --- no feedback loops --- no clock Sequential logic --- outputs logical functions of inputs and previous history of circuit (memory) --- after changed inputs, new outputs appear in the next clock cycle --- frequent feedback loops Combinational logic

2. Fundamentals of Boolean algebra Named after George Boole He presented an algebraic formulation of the process of “logical thought and reason” This formulation come to be known as Boolean Algebra

3. Postulates of Boolean algebra 1) Definition A Boolean algebra is a closed algebraic system containing a set K of two or more elements and the two operators ‘’ or ‘/\’ or ‘  ’, called AND, and ‘+’ or ‘\/’ or ‘  ’, called OR; Closed system: for every a and b in set K, ab belongs to K and a+b belongs to K.

4. Postulates of Boolean algebra 2) Existence of 1 and 0 There exist unique elements 1 (one) and 0 (zero) in set A" such that for every a in K a) a + 0 = a, b) a 1 = a, where 0 is the identity element for the + operation and 1 is the identity element for the operation.

5. Postulates of Boolean algebra 3) Commutativity of the + and operations For every a and b in K a) a + b = b + a. b) a b = b a 4) Associativity of the + and operations For every a, b, and c in K a) a + (b + c) = (a + b) + c. b) a (b c) = (a b) c.

6. Postulates of Boolean algebra 5) Distributivity of + over and over + For every a, b, and c in K a) a + (b c) = (a + b) (a + c), b) a (b + c) = (a b) + (a c). 6) Existence of the complement For every a in K there exists a unique element called ā (complement of a) in K such that a) a + ā = 1. b) a ā = 0.

7. Venn diagrams for the postulates Operations on sets Sets  closed regions Sets correspond to elements Intersection  corresponds to Union  corresponds to +

8. Venn diagrams for the postulates

9. Venn diagrams Examples of Venn diagrams

10. Venn diagrams a + b c = (a + b) (a + c)

11. Venn diagrams a + b c = (a + b) (a + c)

12. Boolean algebra Duality –If an expression f(x 1, x 2, … x n, +,, 0, 1) is valid, then f(x 1, x 2, … x n,, +, 1, 0) obtained by interchanging + and, 0 and 1 is also valid a (b + c) = (a b) + (a c) a + (b c) = (a + b) (a + c) Postulates 2 – 6 are stated in dual form

13. Fundamental theorems of Boolean algebra –Prove part (b) by exchanging + with, and use the dual form of the postulates

14. Fundamental theorems of Boolean algebra

15. a ā = 0[P6(b)] a + ā = 1[P6(a)] Therefore, ā is the complement of a, and also a is the complement of ā. Because the complement of ā is unique, it must be equal to a.

16. Fundamental theorems of Boolean algebra

17. Why a + ab = a Fundamental theorems of Boolean algebra aabb

18. Fundamental theorems of Boolean algebra

23. Example using DeMorgan’s theorem Fundamental theorems of Boolean algebra

25. Boolean algebra postulates and theorems

26. Theorems Proofs by perfect induction Proofs by exhaustion: Let variables assume all possible values and show validity of result in all cases

27. Example: Show X + 0 = X (a) Keep axioms handy (b) Elaborate cases: if X = 0, have X + 0 = 0 + 0 = 0 = X if X = 1, have X + 0 = 1 + 0 = 1 = X

28. More Theorems Can prove by exhaustion....but have more cases For distributive laws, T8 looks like ordinary algebra T8’ also true (swap operators, factor, swap back) T9, T10 for logic minimization - drop irrelevant terms

29. T9, T10, T11 for logic minimization - drop superfluous terms Proof: X + X  Y = X  1 + X  Y = X  (1+Y) = X  1 = X X  (X+Y) = (X+0)  (X+Y) = X+(0  Y) = X+0 = X T10 (Combining): X  Y + X  Y’ = X and (X + Y)  (X + Y’) = X Proof: X  Y + X  Y’ = X  (Y + Y’) = X  1 = X (X + Y)  (X + Y’) = X + (Y  Y’) = X + 0 = X T11 (Consensus): X  Y+X’  Z+Y  Z = X  Y+X’  Z and (X+Y)  (X’+Z)  (Y+Z)= (X+Y)  (X’+Z) Proof: If Y  Z = 0 X  Y+X’  Z+Y  Z = X  Y+X’  Z+ 0 = X  Y+X’  Z else Y = Z = 1 left side: X  Y+X’  Z+YZ = something + YZ = something + 1 =1 right side: X  Y+X’  Z = X + X’ = 1 So, in either case, X  Y+X’  Z+YZ = X  Y+X’  Z If Y+Z = 1 (X+Y)  (X’+Z)  (Y+Z)= (X+Y)  (X’+Z)  1= (X+Y)  (X’+Z) else Y = Z = 0 left side: (X+Y)  (X’+Z)  (Y+Z)= something  (Y + Z) = something  0 = 0 right side: (X+Y)  (X’+Z) = (X+0)  (X’+0) = X  X’ = 0 So, in either case, (X+Y)  (X’+Z)  (Y+Z)= (X+Y)  (X’+Z) T9 (Covering): X + X  Y = X and X  (X+Y)=X

30. Duality De Morgan’s Theorems: (X + Y)’ = X’  Y’ (X  Y)’ = X’ + Y’ Dual: Swap 0 & 1, AND & OR, but leave variables unchanged –Result: Theorems still true Why? –f(X, Y) = g(X, Y) –complement[f(X, Y)] = complement[g(X, Y)] –dual[f(X’, Y’)] = dual[g(X’, Y’)] –but X’, Y’ just dummy variables, replace with originals Counterexample? X + X  Y = X (T9) X  X + Y = X (dual) X + Y = X (T3) !! error ? X + (X  Y) = X (T9) X  (X + Y) = X (dual) (X  X) + (X  Y) = X (T8) X + (X  Y) = X (T3) parentheses, operator precedence

31. N-variable Theorems Prove via induction Most important: DeMorgan theorems

32. DeMorgan Symbol Equivalence Bubble-pushing...

33. Likewise for OR