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HW1S07
CS 336
1. Determine the truth value of the following statements
a. (
∀
x x
∈Ζ
: (
∃
y y
∈Ζ
: x
2
<y))=T
b. (
∃
y y
∈Ζ
: (
∀
x x
∈Ζ
: x
2
<y))= F
c. What you notice from parts a. and b.? The answers are different.
What are the implications? Order is important for nested quantifiers.
2. Rewrite these with equivalent statements with the negation only within
predicates.
a.
¬
(
∀
x x
∈Ζ
: (
∃
y y
∈Ζ
: x
2
<y)) = (
∃
x x
∈
Z: (
∀
y y
∈
Z: x
≥
y))
b.
¬
(
∃
y y
∈Ζ
: (
∀
x x
∈Ζ
: x
2
<y)) = (
∀
y y
∈
Z: (
∃
x x
∈
Z: x
≥
!y))
c. How might you check if your statements are equivalent? Check
the truth value of each. Are they? Yes.
3. Prove the following
a.
(b^(b
→
c))
→
c
Proof:
(b^(b
→
c))
→
c
↔
<
→
>
(b
∧
(
¬
b
∨
c) )
→
c
↔
<
→
>
¬
(b
∧
(
¬
b
∨
c) )
∨
c
↔
< Distributivity >
¬
((b
∧
¬
b)
∨
(b
∧
c) )
∨
c
↔
< Contradiction>
¬
( F
∨
(b
∧
c) )
∨
c
↔
< Commutativity;
∨
simplification >
¬
(b
∧
c)
∨
c
↔
< DeMorgan’s Law >
(
¬
b
∨
¬
c)
∨
c
↔
< Associativity >
¬
b
∨
(
¬
c
∨
c)
↔
< Commutativity; Excluded Middle >
¬
b
∨
T
↔
<
∨
simplification >
T QED
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View Full Documentb. p
∧
q
→
p
Proof:
p
∧
q
→
p
↔
<
→
>
¬
(p
∧
q)
∨
p
↔
< DeMorgan’s Law >
(
¬
p
∨
¬
q)
∨
p
↔
< Commutativity >
p
∨
(¬
p
∨
¬
q)
↔
< Associativity >
(p
∨
¬
p)
∨
¬
q
↔
< Excluded Middle >
T
∨
¬
q
↔
< Commutativity;
∨
simplification >
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 Spring '08
 Myers

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