HOMEWORK 12 1. Prove that for commutative operations, every left identity element is also a right identity element. Let 1 be the left identity element for the operation ο. Then, the proof goes as follows: 1. ∀x∀y (x ο y = y ο x) By commutativity of ο 2. ∀x (1 ο x = x) By def. of left identity element 3. Show: ∀x (x ο 1 = x) UDerivation 4. Show: a ο 1 = a 5. 1 ο a = a 2, UI 6. a ο 1 = 1 ο a 1, UI 7. a ο 1 = a 5, 6, by identity logic (Leibniz’s Law?) 2. Show that no Boolean algebra can be a group. Any Boolean algebra (with one of its operations, say ∧) will satisfy the axioms G1 and G2 but will violate G3 and, therefore won’t be a group. Axiom G1 (the operation is associative) is satisfied, since the operations of a Boolean Algebra are, by definition of Boolean Algebra , associative. Axiom G2 (G contains an identity element) is also satisfied. The right identity element is 1: By the top law, a ∧ 1 = a, for any a. Since the operations of a Boolean Algebra are, by definition of Boolean Algebra, commutative, then 1 will also be the left identity element (see previous exercise). Since 1 is both a left identity element and a right identity element it is a two sided identity element, i.e., an identity element. Axiom G3, however, is violated. G3 says that every element will have an inverse element. However, 0 doesn’t have an inverse element, since by the top law, a ∧ 0 = 0, for any a. So, there is no element x such that x ∧ 0 = 1. Instructors’ note: This proof is correct for any Boolean algebra with more than one element. [The axioms class notes don’t explicitly require that 0 ≠ 1, although the axioms in the handout from the old Partee textbook do.] But for the ‘degenerate’ one-element Boolean algebra this proof wouldn’t work, and that 1-element algebra is a group; 1 = 0, so the sole member is also its own inverse. [We had forgotten that possibility in stating the problem as we did.] 3. Determine whether the set-theoretic operation ‘symmetric difference’ is commutative, associative and idempotent. Is there an identity element for this operation? What sets, if any, have inverses? A + B =def (A ∪ B) – (A ∩ B) a) Set-theoretic symmetric difference is commutative A + B =def (A ∪ B) – (A ∩ B) B + A =def (B ∪ A) – (B ∩ A) Since set theoretic union and set theoretic intersection are commutative, A + B = B + A and, therefore, the operation is commutative. b) Set-theoretic symmetric difference is associative Let A = {a,b,2}, B = {1,2} and C = {a,c} A + (B + C) = [A ∪ [(B ∪ C) – (B ∩ C)]] - [A ∩ [(B ∪ C) – (B ∩ C)]] = [{a,b,2} ∪ [{1,2,a,c} – ∅ ]] - [{a,b,2} ∩ [{1,2,a,c} – ∅ ]] {a,b,2,1,c} –{a,2} = {b,1,c} (A + B) + C = [[(A ∪ B) – (A ∩ B)] ∪ C] – [[(A ∪ B) – (A ∩ B)] ∩ C] {a,b,1,c} – {a} = {b,1,c} c) Set theoretic difference is not idempotent A + A =def (A ∪ A) – (A ∩ A) = ∅ So, A + A ≠ A, for any A Instructors’ small correction: So, A + A ≠ A, for any A ≠ ∅ d) The identity element is the empty set. A + ∅ =def (A ∪ ∅) – (A ∩ ∅) = A - ∅ = A Since the operation is commutative, A + ∅ = ∅ + A = A e) Each set is its own inverse, since A + A = ∅ (see c), above), which is the identity element. f) Given the set A = {a,b}, what sort of operational structure is formed by the power set of A with the operation of symmetric difference? A group 1. It is an algebra, since ℘(A) is closed under symmetric difference. 2. Symmetric difference is associative 3. ℘(A) contains an identity element, the empty set. 4. Each element in ℘(A) has an inverse element, namely, itself. .