# Boolean Algebra. P1. The OR operation is closed for all x, y B x + y B

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1 Boolean Algebra A Boolean Algebra is a mathematical system consisting of a set of elements B, two binary operations OR (+) and AND ( ), a unary operation NOT ('), an equality sign (=) to indicate equivalence of expressions, and parenthesis to indicate the ordering of the operations, which preserves the following postulates: P1. The OR operation is closed for all x, y B x + y B P2. The OR operation has an identity (denoted by 0) for all x B x + 0 = 0 + x = x P3. The OR operation is commutative for all x, y B x + y = y + x P4. The OR operation distributes over the AND operation for all x, y, z B x + (y z) = (x + y) (x + z) P5. The AND operation is closed for all x, y B x y B P6. The AND operation has an identity (denoted by 1) for all x B x 1 = 1 x = x P7. The AND operation is commutative for all x, y B x y = y x P8. The AND operation distributes over the OR operation for all x, y, z B x (y + z) = (x y) + (x z) P9. Complement for all x B there exists an element x' B, called the complement of x, such that (a) x + x' = 1 (b) x x' = 0 P10. There exist at least two elements x, y B such that x y

2 Theorem 1 The complement of x is unique Proof : Assume x 1 ' and x 2 ' are both complements of x. Then by P9 x + x 1 ' = 1, x x 1 ' = 0, x + x 2 ' = 1, x x 2 ' = 0 x 1 ' = x 1 ' 1 = x 1 ' (x + x 2 ') substitution, x + x 2 ' = 1 = (x 1 ' x) + (x 1 ' x 2 ') AND distributes over OR (P8) = (x x 1 ') + (x 1 ' x 2 ') AND is commutative (P7) = 0 + (x 1 ' x 2 ') substitution, x x 1 ' = 0 = (x x 2 ') + (x 1 ' x 2 ') substitution, x x 2 ' = 0 = (x 2 ' x) + (x 2 ' x 1 ') AND is commutative (P7), twice = x 2 ' (x + x 1 ') AND distributes over OR (P8) = x 2 ' 1 substitution, x + x 1 ' = 1 = x 2 ' Thus, any two elements that are the complement of x are equal. This implies that x' is unique Theorem 2-1 x + 1 = 1 x + 1 = 1 (x + 1) = (x + x') (x + 1) Complement, x + x' = 1 (P9a) = x + (x' 1) OR distributes over AND (P4) = x + x' = 1 Complement, x + x' = 1 (P9a) Theorem 2-2 x 0 = 0

3 Theorem 3-1 AND's identity is the complement of OR's identity 0' = 1 0' = 0 + 0' 0 is the identity for OR (P2) = 1 Complement, x + x' = 1 (P9a) Theorem 3-2 OR's identity is the complement of AND's identity 1' = 0 Theorem 4-1 Idempotent x + x = x x + x = (x + x) 1 = (x + x) (x + x') Complement, x + x' = 1 (P9a) = x + (x x') OR distributes over AND (P4) = x + 0 Complement, x x' = 0 (P9b) = x 0 is the identity for OR (P2) Theorem 4-2 Idempotent x x = x

4 Theorem 5 Involution (x')' = x Let x' be the complement of x and (x')' be the complement of x'. Then by P9, x + x' = 1, xx' = 0, x' + (x')' = 1, and x'(x')' = 0 (x')' = (x')' is the identity for OR (P2) = (x')' + xx' Substitution, xx' = 0 = [(x')' + x][(x')' + x'] OR distributes over AND (P4) = [x + (x')'][x' + (x')'] OR is commutative (P3), twice = [x + (x')'] 1 Substitution, x' + (x')' = 1 = [x + (x')'][x + x'] Substitution, x + x' = 1 = x + [(x')' x'] OR distributes over AND (P4) = x + [x' (x')'] AND is commutative (P7) = x + 0 Substitution, x'(x')' = 0 = x 0 is the identity for OR (P2) Theorem 6-1 Absorption x + xy = x x + xy = (x 1) + xy = x(1 + y) AND distributes over OR (P8) = x(y + 1) OR is commutative (P3) = x 1 x + 1 = 1 (Thm 2-1) = x Theorem 6-2 Absorption x(x + y) = x

5 Theorem 7-1 x + x'y = x + y x + x'y = (x + x') (x + y) OR distributes over AND (P4) = 1 (x + y) Complement x + x' = 1 (P9a) = x + y Theorem 7-2 x(x' + y) = xy

6 Theorem 8-1 OR is associative x + (y + z) = (x + y) + z Let A = x + (y + z) and B = (x + y) + z To Show: A = B First, xa = x [x + (y + z)] Substitution of A = x Absorption x(x + y) = x (Thm 6-2) and, xb = x[(x + y) + z] Substitution of B = x(x + y) + xz AND distributes over OR (P8) = x + xz Absorption x(x + y) = x (Thm 6-2) = x Absorption x + xy = x (Thm 6-1) Therefore xa = xb = x Second, x'a = x'[x + (y + z)] Substitution of A = x'x + x'(y + z) AND distributes over OR (P8) = xx' + x'(y + z) AND is commutative (P7) = 0 + x'(y + z) Complement, x x' = 0 (P9b) = x'(y + z) 0 is the identity for OR (P2) and. x'b = x'[(x + y) + z] Substitution of B = x'(x + y) + x'z AND distributes over OR (P8) = (x'x + x'y) + x'z AND distributes over OR (P8) = (xx' + x'y) + x'z AND is commutative (P7) = (0 + x'y) + x'z Complement,, x x' = 0 (P9b) = x'y + x'z 0 is the identity for OR (P2) = x'(y + z) AND distributes over OR (P8) Therefore x'a = x'b = x'(y + z) Finally, A = A 1 = A(x + x') Complement, x + x' = 1 (P9a) = Ax + Ax' AND distributes over OR (P8) = xa + x'a AND is commutative (P7), twice = xb + x'a Substitution xa = xb = xb + x'b Substitution x'a = x'b = Bx + Bx' AND is commutative (P7), twice = B(x + x') AND distributes over OR (P8) = B 1 Complement, x + x' = 1 (P9a) = B Since A = x + (y + z) and B = (x + y) + z, we have shown that x + (y + z) = (x + y) + z

7 Theorem 8-2 AND is associative x(yz) = (xy)z Theorem 9-1 DeMorgan's Law 1 (x + y)' = x' y' By Theorem 1 (complements are unique) and Postulate P9 (complement), for every x in a Boolean algebra there is a unique x' such that x + x' = 1 and x x' = 0 So it is sufficient to show that x'y' is the complement of x + y. We'll do this by showing that (x + y) + (x'y') = 1 and (x + y) (x'y') = 0 (x + y) + (x'y') = [(x + y) + x'] [(x + y) + y'] OR distributes over AND (P4) = [(y + x) + x'] [(x + y) + y'] OR is commutative (P3) = [y + (x + x')] [x + (y + y')] OR is associative (Thm 8-1), twice = (y + 1)(x + 1) Complement, x + x' = 1 (P9a), twice = 1 1 x + 1 = 1 (Thm 2-1), twice = 1 Idempotent, x x = x (Thm 4-2) Also, (x + y)(x'y') = (x'y')(x + y) AND is commutative (P7) = [(x'y')x] + [(x'y')y] AND distributes over OR (P8) = [(y'x')x] + [(x'y')y] AND is commutative (P7) = [y'(x'x)] + [x'(y'y)] AND is associative (Thm 8-2), twice = [y'(xx')] + [x'(yy')] AND is commutative (P7), twice = [y' 0] + [x' 0] Complement, x x' = 0 (P9b), twice = x 0 = 0 (Thm 2-2), twice = 0 Idempotent, x + x = x (Thm 4-1) Theorem 9-2 DeMorgan's Law 2 (xy)' = x' + y'

8 Summary OR is closed for all x, y B, x + y B P1 0 is the identity for OR x + 0 = 0 + x = x P2 OR is commutative x + y = y + x P3 OR distributes over AND x + (y z) = (x + y) (x + z) P4 AND is closed for all x, y B, x y B P5 1 is the identity for AND x 1 = 1 x = x P6 AND is commutative x y = y x P7 AND distributes over OR x (y + z) = (x y) + (x z) P8 Complement (a) x + x' = 1 P9a Complement (b) x x' = 0 P9b Complements are unique Thm 1 x + 1 = 1 Thm 2-1 x 0 = 0 Thm 2-2 0' = 1 Thm 3-1 1' = 0 Thm 3-2 Idempotent x + x = x Thm 4-1 Idempotent x x = x Thm 4-2 Involution (x')' = x Thm 5 Absorption x + xy = x Thm 6-1 Absorption x (x + y) = x Thm 6-2 x + x'y = x + y Thm 7-1 x (x' + y) = xy Thm 7-2 OR is associative x + (y + z) = (x + y) + z Thm 8-1 AND is associative x(yz) = (xy)z Thm 8-2 DeMorgan's Law 1 (x + y)' = x' y' Thm 9-1 DeMorgan's Law 2 (xy)' = x' + y' Thm 9-2

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