1. 6 - 1 Simplification Theorems Useful for simplification of expressions & therefore simplification of the logic network which results. XY + XY' = ( X + Y ) ( X + Y' ) = X + XY = X ( X + Y ) = ( X + Y' ) Y = e.g.: ( X + Y ) ( X + Y' ) = Can also use truth tables. e.g.: simplify: Z = ( A+B'C ) ( A + C' + BD + B'C ) - much simpler, with reduction in number of gates. X X XY X X XY’ + Y = X + Y Example proof: XX + XY’ + YX + YY’ X 0 = X(1+Y+Y’) = X X (A+B’C + C’+BD ) X Y  X(X+Y) = X  Z = A+B’C From 6 to 2 gates

2. 6 - 2 Laws and Theorems of Boolean Algebra Operations with 0 and 1: 1. X + 0 = X1D. X · 1 = X0' = 1 2. X + 1 = 12D. X · 0 = 01' = 0 Idempotent laws: 3. X + X = X3D. X · X = X Involution law: 4. (X')' = X Laws of complementarity: 5. X + X' = 15D. X · X' = 0 Commutative laws: 6. X + Y = Y + X6D. XY = YX Associative laws: 7. ( X + Y ) + Z = X + ( Y + Z )7D. ( XY ) Z = X ( YZ ) = XYZ = X + Y + Z

3. 6 - 3 Laws and Theorems of Boolean Algebra(continued) Distributive Laws: 8. X(Y+Z) = XY + XZ8D. X + YZ = (X + Y)(X + Z) Simplification theorems: 9. XY + XY' = X9D. ( X + Y ) ( X + Y' ) = X 10. X + XY = X10D. X ( X + Y ) = X 11. ( X + Y' ) Y = XY11D. XY' + Y = X + Y DeMorgan's laws: 12. ( X + Y + Z +... )' = X'Y'Z'...12D. ( XYZ... )' = X' + Y' + Z' +... 13. [ f (X 1, X 2,..., X n, 0, 1, +, ·) ]' = f ( X 1 ', X 2 ',..., X n ', 1, 0, ·, +) Duality: 14. ( X + Y + Z +...)D = XYZ...14D. ( XYZ...)D = X + Y + Z +... 15. [ f (X 1, X 2,..., X n, 0, 1, +, · ) ] D = f ( X 1, X 2,..., X n, 1, 0, ·, + ) Theorem for multiplying out and factoring: 16. ( X + Y ) ( X' + Z ) = XZ + X'Y16D. XY + X'Z = ( X + Z ) ( X' + Y ) Consensus theorem: 17. XY + YZ + X'Z = XY + X'Z17D. ( X + Y ) ( Y + Z ) ( X' + Z ) = ( X + Y ) ( X' + Z )

4. 6 - 4 Multiplying Out & Factoring - Yields convenient forms for Boolean expressions. - An expression is said to be in sum-of-products (SOP) form when all products are products of single variables only. e.g.: ABC + B'CDE' + A'B' BCDE + AB'E + HI' + C + D However: ( A + B ) C ABC + B'C ( D + E' ) Logic network realization of a SOP expression is always one or more AND gates feeding a single OR gate. e.g.: A B C D E B’ A’ E’ ABC B’CDE’ A’B’ ABC+B’CDE’+A’B’ AND ABC + B'CDE' + A'B' Not SOP But easy to make

5. 6 - 5 An expression is said to be in the product-of-sums (POS) form when all sums are the sums of single variables. e.g.:( A + B + C ) ( A + D' ) ( B' + C ) AB' ( C + D ) E but ( A + B + C ) ( AB + D ) ( AB' + C ) ( D' + E ) Realization of a POS expression is one or more OR gates feeding into and AND gate. e.g.: POS Not POS A B C A D' C B' ( A + B + C ) ( A + D' ) ( B' + C )

6. 6 - 6 A Boolean expression can always be put into the SOP form by: - use distributive laws: 1. X ( Y + Z ) = XY + XZ 2. ( X + Y ) ( X + Z ) = X + YZ - apply 2nd law 1st when possible ( = less work ) e.g.: ( A + B + C ) ( A + D + E )(This is a POS expression) If use 1st law only, redundant terms must be eliminated: e.g. (above):= AA + AD + AE + AB + BD + BE + AC + CD + CE = as above, but longer process & more room for error. X Y X Z A + (B+C)(D+E) = A+BD+BE+CD+CE SOP Form = A(1+D+E+B+C)+BD+BE+CD+CE = A+BD+BE+CD+CE (as above)

7. 6 - 7 Similarly a Boolean expression can be put into a POS form by: FACTORING - again by using distributive laws - apply 1st law first, when possible e.g.: AB + AC + DE + DF e.g.: AB + CD = = A(B+C) +D(E+F)1 X YZ (AB+C)(AB+D) y z x = (A+C)(B+C)(A+D)(B+D) POS form X YZ = (A(B+C)+D)(A(B+C)+E+F) 2 Y Z X y z x = (A+D)(B+C+D)(A+E+F)(B+C+E+F) POS form

8. 6 - 8 Inversion (De Morgan's Laws) ( X + Y )' = ( XY )’ = In general: ( X 1 + X 2 +... + X n )' = ( X 1 X 2 ··· X n )' = Can be proved by induction from 2-variable laws. e.g.:F = ( A'BC' + D ) ( E + F' ) F' = X’Y’ X’ + Y’ X 1 ’X 2 ’X 3 ’…..X n ’ X 1 ’ + X 2 ’+ ……+X n ’ X Y (XY)’ = X’ +Y’ = (A’BC’+D)’ + (E+F’)’ = (A’BC’)’ D’ + E’F = (A+B’+C)D’ + E’F

9. 6 - 9 Both laws can be combined into a single general law for complementation: - to form the complement of a Boolean expression, replace : each variable with its complement, each with e.g. (above): F = ( A'BC' + D ) ( E + F' ) F' = Parentheses must be used to keep the proper hierarchy of operations. i.e., operations take place in the complement in the same order as in the original expression. Symbolically, DeMorgan's law can be written: [ F( X 1, X 2, ···, X n, 0, 1, +, · )]' = 0101 1010 + + ((A+B’+C)D’)+(E’F) F(X 1 ’,X 2 ’,….X n ’,1,0,,+)