Course MATH 0017 Boolean algebra a Slide 1
Course MATH 0017 Boolean algebra (a) Slide 1 Axioms of Boolean algebra We have already met the three primitive operations of Boolean algebra: AND, inclusive OR and NOT (the fourth, exclusive or, is made up from the other three). We originally talked about the values of the operands and results of these functions as TRUE and FALSE values, or, equivalently, as 1 and 0, respectively. We can also think of these as the identity and the null elements, respectively, of the Boolean algebra. Let us now recall the operations of Boolean algebra and then summarise the basic Boolean equalities. Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (b) Slide 1 Basic Boolean functions: Operands AND OR NOT A B A+B A’ 0 0 1 1 Copyright © Dr N M Brooke 0 1 0 0 0 1 1 1 0 1
Course MATH 0017 Boolean algebra (c) Slide 1 Boolean equalities Name AND rule OR rule NOT rule A. 0 = 0 A+0 = A 0’ = 1 A. 1 = A A+1 = 1 1’ = 0 A. A = A A+A = A (A’)’ = A A. A’ = 0 A+A’ = 1 Commutative. A. B = B. A A+B = B+A Associative A. (B. C) = A+(B+C) = (A. B). C (A+B)+C Distributive A. (B+C) = A+(B. C) = A. B+A. C (A+B). (A+C) De Morgan (A. B)’ = (A+B)’ = A’+B’ A’. B’ Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (d) Slide 1 Proving Boolean identities (a) by algebra To show that A. (A + B) = A we can apply the Boolean equalities: A. (A + B) = A. A + A. B = A. 1 + A. B = A. (1 + B) = A. 1 =A Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (e) Slide 1 Proving Boolean identities (b) by truth table A 0 0 1 1 B 0 1 (A+B) 0 1 1 1 A. (A+B) 0 0 1 1 We observe that the value of the function is just the same as the value of A, for all possible combinations of the two Boolean variables A and B. Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (f) Slide 1 And one more, just to be sure: Let us show that A + A. B = A We can proceed as follows: A + A. B = A. 1 + A. B = A. (1 + B) = A. 1 =A The truth table is also easy to draw. Try it. Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (g) Slide 1 Simplifying rules We can prove a small set of simplifying rules. They are summarised below: AND rules OR rules A. (A+B) = A A+A. B = A (A+B). (A+B’) = A A. B+A. B’ = A A. (A’+B) = A. B A+A’. B = A+B Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (h) Slide 1 Armed with the basic manipulative skills, it is possible to prove or simplify quite complex Boolean expressions. For example, let us simplify by applying the Boolean equalities: (A+B+C). (A+B’) = A. (A+B’)+B. (A+B’)+C. (A+B’) = A. A +A. B’ +A. B +B. B’ +A. C +B’. C = A +A. B’ +A. B +A. C +B. B’ +B’. C = A +A. (B’ +B) +A. C +B. B’ +B’. C = A +A. 1 +A. C +B. B’ +B’. C = A. (1 +C) +B. B’ +B’C Copyright © Dr N M Brooke
Course MATH 0017 Boolean algebra (i) Slide 1 We continue the simplification: (A+B+C). (A+B’) = A. (1 +C) +B. B’ +B’C = A. 1 +B. B’ +B’C = A + 0 +B’C = A + B’C We now have to ask whethere is a way to show that this is indeed the simplest form of the original Boolean function (A+B+C). (A+B’). To do this we need to develop the notion of a standard form of expression of a Boolean function. Copyright © Dr N M Brooke
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