Download HOMEWORK 12 1. Prove that for commutative operations, every left

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.
2.
3.
4.
It is an algebra, since ℘(A) is closed under symmetric difference.
Symmetric difference is associative
℘(A) contains an identity element, the empty set.
Each element in ℘(A) has an inverse element, namely, itself.