disjunctive normal form

We want a “true” output when the inputs match the first or fourth rows, and only then. The inputs match the first row precisely when both x and y are true (i.e. when the conjunction x∧y is true), and they match the fourth row precisely when both x is not true and y is not true (i.e. when the conjunction x′∧y′ is true). So take the disjunction of these two conjunctions: p⁡(x,y) =(x∧y)∨(x′∧y′). Procedure 3.2.1: To Produce the Disjunctive Normal Form Polynomial for a Given Boolean Truth Table Given a truth table with nonzero output, we may obtain a Boolean polynomial in disjunctive normal form with that truth table as follows. 1. Identify rows the in truth table for which the desired output is 1 2. For each such row, form the conjunction of all variables, but negate those variables that have input value 0 for that row. 3. Form a polynomial by taking the disjunction of all those conjunctions. The fourth, fifth, seventh, and eigth rows have outcome 1. The corresponding conjunctions are is both in disjunctive normal form and will have the desired truth table. is both in disjunctive normal form and will have the desired truth table. We can get a much simpler expression for q⁡(x,y,z) by not using the procedure (though of course the result will not be in disjunctive normal form). The polynomials in the solutions to the preceding examples are in disjunctive normal form, but the alternative solution to the second example is not. What do you think conjunctive normal form should mean? Can you come up with a procedure which takes a truth table and determines a Boolean polynomial in conjunctive normal form with the desired truth table?