Boolean algebra

1. Boolean algebra • Last time we talked about Boolean functions, Boolean expressions, and truth tables. • Today we’ll learn how to how use Boolean algebra to simplify Booleans expressions. • Last time, we saw this expression and converted it to a circuit: (x + y’)z + x’ ©2000-2002 Howard Huang

2. Simplifying circuits • The expression and circuit on the previous page are actually equivalent to the simplified ones below: x’ + z • Simpler hardware is always better! • In many cases, the simpler circuit is also faster. • Less hardware means smaller size, which also reduces costs. • Simpler circuits draw less power. • How do we know this circuit is the same as the one on the last page? Boolean algebra

3. Boolean algebra is special • Boolean algebra gives us a way to simplify Boolean functions, much like regular algebra allows us to simplify mathematical functions. • The AND and OR operations are similar to multiplication and addition: • AND is the same as multiplication for the values 0 and 1 • OR is almost the same as addition, except for 1 + 1. • But there are important differences to watch out for, too. • Boolean algebra uses are only two values. • OR is not quite the same as addition, and NOT is a new operation. Boolean algebra

4. Formal definition of Boolean algebra • A Boolean algebra requires: • A set of values B, which contains at least two elements 0 and 1 • Two 2-argument operations + and  • A one-argument operation ' • The values and operations must satisfy the axioms below. Boolean algebra

5. Satisfying the axioms • The operations we chose do satisfy all of the axioms. We can show this by looking at the definitions of our operations. • For example, the axiom x + x’ = 1 is satisfied, because: • We have only two possible values for x: 0 and 1. • The complement of these values are 1 and 0, respectively. • 0 + 1 = 1, and 1 + 0 = 1. Boolean algebra

6. Proofs with truth tables • We can always prove equivalences using truth tables, to explicitly show that two expressions always produce the same results. • Let’s use truth tables to prove DeMorgan’s law, (x + y)’ = x’y’. • The columns on the left in each table are the inputs. We have two inputs, x and y, so there are four possible input combinations. • The columns on the right, in blue, are outputs. • Intermediate columns can be useful in deriving the outputs. • Since both of the columns in blue are the same, this shows that (x + y)’ and x’y’ really are equivalent. Boolean algebra

7. Similarities with regular algebra • Several of the axioms (in blue) look just like regular algebraic rules. • The associative laws show that there is no ambiguity in a term such as xyz or x + y + z, so we can use multiple-input primitive gates: Boolean algebra

8. The complement operation • The magentaaxioms deal with the complement operation, which doesn’t exist in normal algebra. • Most of them almost make sense if you translate them into English: • “It is raining or it is not raining” is always true (x + x’ = 1) • “It is raining and it is not raining” can never be true (x  x’ = 0) • “I am not not handsome” means “I am handsome” ((x’)’ = x) Boolean algebra

9. DeMorgan’s • DeMorgan’s axioms are especially important for circuit design. (x + y)’ = x’y’(xy)’ = x’ + y’ • We’ve already proved one of these axioms, but here are some rough examples in English. • “I am not rich-or-famous” • This is the same as “I am not rich, and I am not famous.” • In other words, I am nothing. • “I am not old-and-bald” • This is equivalent to “I am not old, or I am not bald.” • I could still be one of the other three possibilities: young and bald, old and hairy, or young and hairy. Boolean algebra

10. Other differences with regular algebra • Finally, the red axiomsare completely different from regular algebra. • The first three make sense if you think about it logically: • “Blah or true” is always true, even if “blah” is false (x + 1 = 1) • “I am handsome or I am handsome” is redundant (x + x = x) • “I am handsome and I am handsome” is redundant (x  x = x) • The last one, x + yz = (x + y)(x + z), is the trickiest one of the bunch. Boolean algebra

11. Duality • Look carefully at the left and right columns of axioms. • They are duals, where we’ve: • exchanged all AND operations with OR operations, and • exchanged all 0s with 1s. • The dual of any equation is always true. Boolean algebra

12. Simplification with axioms • We can now start doing simplifications using these axioms: x’y’ + xyz + x’y = x’(y’ + y) + xyz [ Distributive: x’y’ + x’y = x’(y’ + y) ] = (x’  1) + xyz [ Axiom 7: y’ + y = 1 ] = x’ + xyz [ Axiom 2: x’  1 = x’ ] = (x’ + x)(x’ + yz) [ Distributive ] = 1  (x’ + yz) [ Axiom 7: x’ + x = 1 ] = x’ + yz [ Axiom 2 ] Boolean algebra

13. Simpler expressions yield simpler hardware • Here are the circuits resulting from the original and simplified expressions on the previous page. Boolean algebra

14. Some more laws • Here are some more useful equations. • These are usually called laws, because they can all be proven from the more fundamental axioms. • You can also prove them using truth tables. • Notice that each law also has a dual. • Feel free to use these laws in homeworks and exams, if you like. Boolean algebra

15. The complement of a function • The complement of a function always outputs 0 where the original function output 1, and 1 where the original produced 0. • In a truth table, we can just exchange 0s and 1s in the output column. • On the left is a truth table for f(x,y,z) = x(y’z’ + yz) • On the right is the table for the complement, denoted f’(x,y,z). Boolean algebra

16. Complementing a function algebraically • When you complement an expression algebraically, you can use DeMorgan’s axiom to keep “pushing” the complement operators inwards: • Another clever method to complement an expression is to take the dual of the expression, and then complement each literal: • If f(x,y,z) =x(y’z’ + yz)… • …the dual of f is x + (y’ + z’)(y + z)… • …then complementing each literal gives x’ + (y + z)(y’ + z’)… • …so f’(x,y,z) = x’ + (y + z)(y’ + z’) f(x,y,z) = x(y’z’ + yz) f’(x,y,z) = ( x(y’z’ + yz) )’ [ complement both sides ] = x’ + (y’z’ + yz)’ [ because (xy)’ = x’ + y’ ] = x’ + (y’z’)’ (yz)’ [ because (x + y)’ = x’ y’ ] = x’ + (y + z)(y’ + z’) [ because (xy)’ = x’ + y’, twice] Boolean algebra

17. Standard forms of expressions • We can write expressions in many ways, but some ways are more useful than others. • A sum of products (SOP) expression is characterized by: • There are only OR (sum) operations at the “outermost” level. • Each term in the sum must be a product of literals. • For example: f(x,y,z) = y’ + x’yz’ + xz • The advantage is that a sum of products expression can be implemented using a fairly simple two-level circuit: • Literals are at the “0th” level. • ANDgates are at the first level. • A single OR gate is at the second level. Boolean algebra

18. Hey! This looks like a truth table! Minterms • A minterm is a special product term, where each input variable appears exactly once. • A function with n variables has up to 2n minterms. For example, a three-variable function like f(x,y,z) has up to 8 minterms: • Each minterm is true for exactly one combination of inputs: x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz Minterm Is true when… Shorthand x’y’z’ xyz = 000 m0 x’y’z xyz = 001 m1 x’yz’ xyz = 010 m2 x’yz xyz = 011 m3 xy’z’ xyz = 100 m4 xy’z xyz = 101 m5 xyz’ xyz = 110 m6 xyz xyz = 110 m7 Boolean algebra

19. Sum of minterms form • Every function can be written as a sum of minterms, which is a special kind of sum of products form. • The sum of minterms form for any function is unique. • If you have a truth table for a function, you can write a sum of minterms expression just by picking out the rows of the table where the function output is 1. f = m0 + m1 + m2 + m3 + m6 = m(0,1,2,3,6) = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’ f’ = m4 + m5 + m7 = m(4,5,7) = xy’z’ + xy’z + xyz f’ contains all the minterms not in f. Boolean algebra

20. Summary • We can build complex functions from just the basic Boolean values “true” and “false,” and the operations AND, OR and NOT. • Any Boolean expression can be implemented with a circuit, which uses primitive logic gates to compute products, sums and complements. • We saw two ways to prove equivalence of expressions: • Truth tables show that all possible inputs yield the same results. • Boolean algebra is especially useful for simplifying expressions, and hence circuits. • Expressions can be written in many different ways, so it’s sometimes useful to use standard representations like sums of products or sums of minterms. • Next time we’ll study an alternative, graphical simplification method. Then we’ll start using all this stuff to build and analyze bigger, more useful, circuits. Boolean algebra