1)
(10 pts) Use the laws of logic to prove the following expression is a tautology: (Note: You must list
each rule you use. The only rules you can combine together in one step are the commutative and
associative rules to reorder and group terms more quickly. Hint: First try to “distribute” the not signs
before you simplify any other part of the expression.)
( (p
∨
r)
∨
( (q
∧
r)
∨
p) )
∨
¬
( (
¬
(p
∧
q) )
∧
p )
⇔
( (p
∨
r)
∨
( (q
∧
r)
∨
p) )
∨
¬¬
( p
∧
q)
∨
¬
p
⇔
(De Morgans)
( (p
∨
r)
∨
( (q
∧
r)
∨
p) )
∨
( p
∧
q)
∨
¬
p
⇔
(Double Negation)
(
¬
p
∨
p)
∨
( (p
∨
r)
∨
(q
∧
r)
∨
(p
∧
q))
⇔
(Commutative+Associative over or)
T
∨
( (p
∨
r)
∨
(q
∧
r)
∨
(p
∧
q))
⇔
(Inverse Law)
T
⇔
(Domination Law)
2)
(6 pts) True or False, circle the correct answer, no partial credit on this one.
a)
If A
⊂
B, then A
≤
B
TRUE
B contains all elements of A, thus it must have at least as many elements as A.
b)
If A={1,3,5,7} and B={2,4,6,8}, then
TRUE
A = A – B.
Since A and B are disjoint, the equality does hold.
c)
If A
∩
(
¬
B) = C, then C and B are
TRUE
disjoint sets.
We need to show that C
∩
B=
∅
. We have C
⊂¬
B, since C
⊂
(A
∩¬
B). Thus, for any element x,
if x
∈
C, then x
∈¬
B. But, C
∩
B is nonempty only if there exists an element x such that x
∈
C
and x
∈
B. But this element does not exist because all elements in C are also in
¬
B. Thus, we
must have C
∩
B=
∅
.
3)
(6 pts) Establish the following argument by the rules of implication. (Note: you must state the rule you
use (or premise) for each step in your proof.)
p
p
⇒
q
(q
∨
r)
⇒
s

s
1. p
(Premise)
2. p
⇒
q
(Premise)
3. q
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 Fall '09
 Set Theory, pts

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