“All judges are lawyers” translates “x(J(x)->L(x))”
“Some lawyers are judges” translates “x(L(x)  J(x))”
Why the differences?
What is the translation of “x(J(x)  L(x))”?
What is the translation of “x(L(x) -> J(x))”?
The Island Of Knights and Knaves (again!)
Remember: knights always tell the truth and knaves always
lie. Remember: an inhabitant is a knight if and only the
inhabitant’s statements true and an inhabitant’s is a knave
if and only if the inhabitant is a knave.
K(x) <-> S(x)
~K(x) <-> ~S(x)
A, B, and C are knights or knaves
Suppose A says, “I am a knave but B is not a knave.” What
are A and B?
Suppose A says “Either I am a knave or B is a knight.” What
are A and B?
A says “All of us are knaves.”
B says “Exactly one of us is a knight.”
What are A, B, and C?
Venn diagrams provide an easy, visual way of representing
set relations. Venn diagrams, with difficulty, represent
the relations among many sets. That is why we have to learn
the algebra of sets.
What is a subset of a set?
A is a subset of B if and only if every element of A is an
element of B
AB <-> x(xA->xB)
This definition of subset does not exclude the possibility
that A = B. Two sets of are equal iff they have the same
members.
We can show that two subsets, A and B, are equal if we can
show that AB and BA. That is A=B iff AB and BA. In
plain English, the set A equal the set B iff A is a subset
of B and B is a subset of A. This makes sense. A is a
subset of B means that every element of A is an element of
B and B is a subset of A means that every element of B is
an element of A.
Properties of Inclusion
AA
AB and BC implies that AC
AB and BA implies that A=B
Algebra of Sets
A(BC) = (AB)C
AB=BA
A(BC)=(AB)(AC)
A=A
AAc=(the universe set)
A(BC)=(AB)C
AB=BC
A(BC)=(AB)(AC)
A=A
AAc=
If, for all A, A union B is A then B is the empty set
If A union B is the universe set and A intersection B is
the empty set then B is the complement of A
The complement of the complement of A is A
The complement of the universe set is the empty set
A union A is A
A union the universe set is the universe set
The union of A with the intersection of A and B is A
The complement of the union of A and B is the intersection
of the complement of A and the complement of B
The complement of the universe set is the empty set
A intersection A is A
A intersection the empty set is the empty set
The intersection of A with the union of A and B is A
The complement of the intersection of A and B is the union
of the complement of A and the complement of B.
The following statements are equivalent
AB
AB=A
AB=B
Is the empty set a subset of every set?
Is the intersection of two sets A, B a subset of the union
of A and B?
What is the difference between AB and AB?
Give two examples where AB and where AB
Is 2  {1,2,3}?
Is {1,2} a subset of {{1,2,3},{1,3},1,2}?
Is {1,2} {{1,2,3},{1,3},1,2}?
Can you think if a set that is a member of itself?
Give an
B, B is
A = {1}
B = {1,
C = {1,
example of sets A,B,C such that A is an element of
an element of C and A is not an element of C?
{1,2}, {1,2,3}}
{1}, {1,2}, {1,2,3}}
For more examples
See Stoll, pages 9, 12, 15 and 16
See Rosen, pages 119 and 120, 130 to 133
Inclusion/Exclusion
Consider a universe set of all the people in the world. Two
subsets of this set are the set of bike riders and the set
of swimmers.
Let U = set of all people in the world
Let b = set of bike riders in the world
Let s = set of swimmers in the world
Does U = b + s?
No! Because b + s does not include people who neither ride
bikes nor swim.
Does U = (b + s) + (b + s)c
Yes! Why? The union of a set and its complement is always
the universe. U = A + Ac
U = b + s + bcsc by DeMorgan’s Laws
(A + B)c = AcBc
If something is not in at least one of two sets that it
must be in neither of the sets. If it is not in either of
the two sets then it must be in the complement of the two
sets. If it is not in A and it is not in B then it is in
the complement of A and the complement of B. Which is
another way of saying that it is in intersection of the
complements of the two sets.
Are these sets disjoint? b and bcsc are disjoint and s and
bcsc are disjoint but we cannot guarantee that b and s are
disjoint as there may be bike riders who are also swimmers.
How do we represent bike riders who are not swimmers and
swimmers who are not bike riders? We use set difference.
b – s and s – b
but b – s = bsc and s – b = bcs. Why?
b – s is, by definition the set of bike riders who are not
swimmers. Therefore, elements of the set b - s are elements
of the set b and elements of the set sc, again by
definition. Therefore, b – s = bsc.
From another perspective,
U = s + sc
b = bU = b(s + sc) = bs + bsc
Since b – s is the set of elements of b that are not in s,
b – s = bsc
Similarly,
s = bs + bcs
Therefore,
U = bs + bsc + bcs + bcsc
This is a partition of U because the subsets are pair-wise
disjoint.
Everybody is a bike rider and a swimmer, a bike rider and a
non-swimmer, a non-bike rider and swimmer, or a non-bike
rider and a non-swimmer.
|b + s| = |b| + |s| - |bs|