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Notes and Exercises on Boolean Algebra Dr. Holmes June 12, 2006 1 Boolean Expressions from Truth Tables From the truth table for any operation on truth values we can read out an expression for that operation using ∧, ∨, ¬. For example, x y z T T T F T T F T T F T F T F F F F T T T F T F T F F T F F F F T can be expressed as (x ∧ y ∧ ¬z) ∨ (¬x ∧ y ∧ z) ∨ (¬x ∧ y ∧ ¬z) ∨ (¬x ∧ ¬y ∧ ¬z) This expression is a disjunction (or-sentence) of conjunctions (and-sentences) each of which is a conjunction of letters and negations of letters representing a row of the truth table. If there are no rows in the truth table which are true, the expression x ∧ ¬x will represent it. It should be clear that this will work for truth tables for expressions with any number of letters. 1 Exercise: Write out an expression in ∨, ∧, ¬ for the operation on three propositions represented by this truth table: x y z T T T T T T F F T F T T T F F T F T T F F T F T F F T T F F F F Exercise: Construct a truth table for the operation represented by (x → y) ∧ (y → z) then derive an equivalent expression in ∨, ∧, ¬ from the truth table. 2 Algebraic Notation An expression in ∨, ∧, ¬ can be written in an algebraic style. Write x ∧ y as xy, x ∨ y as x + y, and ¬x as x. Negations of complicated expressions like ¬(x + y) will not be written using overlines. Write T as 1 and F as 0 in algebraic notation. A selection of the algebra rules we saw in section 6 can be used to reduce any expression to a form like the ones we derived from truth tables. An expression is said to be in disjunctive normal form just in case it is a disjunction of conjunctions of single letters and negations of single letters. The expressions derived from truth tables have the additional character- istic that every conjunction contains each letter or its negation just once. The following rules will help us to reduce expressions to disjunctive nor- mal form: Distributive laws: x(y + z) = xy + xz and (x + y)z = xz + yz. Use these to eliminate conjunctions involving disjunctions. deMorgan’s laws: ¬(xy) = (x) + y; ¬(x + y) = xy 2 Applying these rules repeatedly will convert any expression into a dis- junction of conjunctions of letters and their negations (or into a conjunction of letters and their negations, or into a single letter or the negation of a single letter). If any conjunction includes a letter x and also includes x, it can be written xxy (where y may be a complicated expression) which will simplify to 0 (false). We have a rule xx = 0; now show 0y = 0 to ﬁnish the demonstration. 0y = yyy = yy (here the rule yy = y is used)= 0. Now 0 can be eliminated as a term in a disjunction because we have a rule 0 + x = x already. Applying these rules can give one more possible expression as a ﬁnal form (which we accept as DNF): 0. Of course any repeated expression in a disjunction or conjunction can be eliminated using idempotent laws xx = x + x = x. Example: convert (x ∨ y ∨ ¬z) ∧ ¬(x ∧ y) to DNF. In algebraic notation, this is (x + y + z)(¬(xy)) = (x + y + z)(x + y) = xx + xy + yx + yy + zx + zy (here I used the distributive law repeatedly just as in ordinary algebra, not writing every step) = 0 + xy + yx + 0 + zx + zy = xy + yx + zx + zy. Exercise: Convert (x ∨ ¬y ∨ z) ∧ ¬(x ∧ y) to DNF. 3 Other Rules We can turn a DNF expression into an expression like those derived from truth tables by making sure that all the conjunctions contain the same letters. To add a letter (say y) to a conjunction (represented by x) use the rule x = xy + xy, justiﬁed by the proof x = x1 = x(y + y) = xy + xy. Exercise: Try converting the expressions from the previous example and exercise into forms from which truth tables can be read. I give the proofs of the Absorption Laws from the class example, though you will not be asked to use these. First Absorption Law: x + xy = x Proof: x = xy + xy = (xy + xy) + xy = xy + (xy + xy) = xy + x = x + xy Exercise: Justify each step in this proof (using algebraic rules from the book or from this document). Second Absorption Law: x + xy = x + y 3 Proof: x + y = (xy + xy) + (xy + xy) = xy + xy) + xy (eliminate repeated xy) = x + xy Exercise: Justify each step in this proof (using algebraic rules from the book or from this document). 4