Discrete math with computers_3

C.7 Predicate Logic

2.

F (1, 5) F (1, 6)

F (2, 5) F (2, 6)

F (3, 5) F (3, 6)

F (4, 5) F (4, 6)

3.

•There is an even number.

x. E(x)

•Every number is either even or odd.

x. (E(x) O(x))

•No number is both even and odd.

x. ¬(E(x) O(x))

•The sum of two odd numbers is even.

x. y. O(x) O(y) → E(x + y)

• The sum of an odd number and an even number is odd.

x. y. E(x) O(y) → O(x + y)

4.

•Chickens are birds.

x. C(x) → B(x)

•Some doves can fly.

x. D(x) F (x)

•Pigs are not birds.

x. P (x) → ¬B(x)

•Some birds can fly, and some can’t.

x. B(x) F (x) x. B(x) ¬F (x)

•An animal needs wings in order to fly.

x. (¬W (x) → ¬F (x))

•Chickens have more feathers than pigs do.

x. y. (C(x) P (y)) → M (x, y)

•An animal with more feathers than any chicken can fly.

x. A(x) ( y. (C(y) M (x, y))) → F (x)

5.

•x. ( y. wantsToDanceWith (x, y))

Everybody has someone they want to dance with.

•x. ( y. wantsToPhone (y, x))

There is someone whom everybody wants to call.

•x. (tired (x) y. helpsMoveHouse (x, y))

There is a person who is tired, and who helps everyone to move house.

6.

forall [1,2,3] (==2)

=1==2 /\ 2==2 /\ 3==2

=False /\ True /\ False

=False

forall [1,2,3] (< 4)

=(1 < 4) /\ (2 < 4) /\ (3 < 4)

=True /\ True /\ True

=True

7.

exists [0,1,2] (==2)

=0==2 \/ 1==2 \/ 2==2

=False \/ False \/ True

=True

exists [1,2,3] (> 5)

=(1 > 5) \/ (2 > 5) \/ (3 > 5)

=False \/ False \/ False

=False

8.

1.x.( y.p(x, y))

=x. x = 1 + 1 x = 2 + 1

=(1 = 1 + 1 1 = 2 + 1) (2 = 1 + 1 2 = 2 + 1)

=(False False) (True False)

=False True

=False

2.x.( y.p(x, y))

=x.(x = 1 + 1) (x = 2 + 1)

=(1 = 1 + 1 1 = 2 + 1) (2 = 1 + 1 2 = 2 + 1)

=(False False) (True False)

=False True

=True

3.x.( y.p(x, y))

=x.(x = 1 + 1 x = 2 + 1)

=(1 = 1 + 1 1 = 2 + 1) (2 = 1 + 1 2 = 2 + 1)

=(False False) (True False)

=False False

=False

4.x, y.p(x, y)

=x.(x = 1 + 1 x = 2 + 1)

(1 = 1 + 1 1 = 2 + 1) (2 = 1 + 1 2 = 2 + 1)

=(False False) (True False)

=False False

=False

9. The solution is based on showing that for any arbitrary value of x, say p, we can infer G(p); hence we can infer x.G(x).

Theorem 110. x.F (x), x.F (x) → G(x) x.G(x)

Proof.

x.F (x) x.F (x) → G(x)

{→E}

G(p)

{ I}

x.G(x)

10.

x. y.F (x, y) y. x.F (x, y)

F (p, q)

{ I}

y.F (p, y) x.F (x, q)

{ E}

x.F (x, q)

{ I}

{ E}

y. x.F (x, y)

11.

y. x.F (x, y) x. y.F (x, y)

WRONG!

Counterexample:

Let F (x, y) = x == y with U = {1, 2}

y. x.F (x, y)

=y.(1 = y 2 = y)

=(1 = 1 2 = 1) (1 = 2 2 = 2)

=(True False) (False True)

=(True True)

=True

but on the other side:

x. y.F (x, y)

=x.(x = 1 x = 2)

=(1 = 1 1 = 2) (2 = 1 2 = 2)

=(True False) (False True)

=False False

=False

The equations do not return the same result, therefore the sequent is not correct.

12.

Theorem 111. x.(F (x) G(x)) ( x.F (x)) ( x.G(x))

Proof.

{ I}

x.F (x) x.G(x)

13. There will be 1000 terms containing F . In general, if the quantifiers are nested k deep, and the universe contains n elements, then the innermost term will occur nk times.

16.

P → x.f (x) x.P → f (x)

P P → x.f (x)

{→E}

x.f (x)

{ E}

f (c)

{→I}

P → f (c){c arbitrary}

{ I}

x.P → f (x)

17.

•x.f (x) x.g(x) x.f (x) g(x)

Counterexample:

Let f (x) = x == 1 and g(x) = x == 2 with U = {1, 2}

x.f (x) x.g(x)

=(1 = 1 2 = 1) (1 = 2 2 = 2)

=(True False) (False True)

=False False

C.8 Set Theory

13.

(A B) ∩ C

=A ∩ B ∩ C

=A ∩ B ∩ C

=A ∩ (B C)

A − (B C)

=A ∩ (B C)

=A ∩ (B C)

(A ∩ B) (A ∩ B )

=A ∩ (B B )

=A ∩ U

=A

A (B − A)

=A (B ∩ A )

=(A B) ∩ (A A )

=(A B) ∩ U

=A B

A − B

=A ∩ B

=B ∩ A

=B ∩ A

=B − A

(A ∩ B) − (A ∩ C)

=(A ∩ B) ∩ (A ∩ C)

=(A ∩ B) ∩ (A C )

=((A ∩ B) ∩ A ) ((A ∩ B) ∩ C )

=(A ∩ B ∩ A ) (A ∩ B ∩ C )

=(A ∩ B ∩ C )

=A ∩ B ∩ C

=A ∩ (B − C)

A − (B C)

=A ∩ (B C)

=A ∩ (B ∩ C )

=A ∩ A ∩ B ∩ C

=(A ∩ B ) ∩ (A ∩ C )

=(A − B) ∩ (A − C)

A ∩ (A B)

=(A ∩ A ) (A ∩ B)

=(A ∩ B)

=A ∩ B

(A − B ) (A − C )

=(A ∩ B ) (A ∩ C )

=(A ∩ B) (A ∩ C)

=A ∩ (B C)

14.

smaller :: Ord a => a -> [a] -> Bool smaller x [] = True

smaller x (y:xs) = x < y

powerSet :: (Ord a, Eq a) => [a] -> [[a]] powerSet set = normalizeSet (foldr g [[]] set)

where g x acc =

[x:epset | epset <- acc,

not (elem x epset) && smaller x epset]

++acc

15.(A B) = A ∩ B

=(U − A) ∩ (U − B)

=((A A ) − A) ∩ ((B B ) − B)

=((A A ) ∩ A ) ∩ ((B B ) ∩ B )

16.

isSubset :: Eq a => [a] -> [a] -> Bool

isSubset set1 set2 = null [e | e <- set1, not (elem e set2)]

17.The arguments are in the wrong order.

18.The second definition of e shadows the first, so the result is the second set. One way to define it correctly is

intersection :: [a] -> [a] -> [a]

intersection set1 set2 = [e | e <- set1, e ‘elem‘ set2]

19.

union :: Eq a => [a] -> [a] -> [a] union set1 set2

=set1 ++ [e | e <- set2, not (elem e set1)]

20.Yes, when A B and C A.

21.Both unions contain C, so C must be empty, and A and B must be disjoint. For example, A = {1, 2, 3}, B = {4, 5, 6}, C = .

22.

A ∩ B

= { defn ∩ } {x|x A x B}

={ commutative } {x|x B x A}

={ defn ∩ }

B ∩ A

23.

A *** B = B *** A

24.

(A B) C

= { defn }

{x |(x A x B) x C}

= { associative }

{x|x A (x B x C)}

= { defn }

A (B C)

25.

A− B

={ double complement } {x|x A x B }

={ defn. − }

{x|x A ¬(x U x B )}

= { defn. complement }

{x|x A ¬(x U ¬(x B ))} = { defn. }

{x|x A ¬(x U ¬(x U x B))}

= { defn. complement }

{x|x A ¬(x U ¬(x U ¬(x B)))} = { defn. }

{x|x A ¬(x U (¬(x U ) (x B)))}

= { DM , double negation }

{x|x A (¬(x U ) ¬(¬(x U ) (x B)))} = { DM }

{x|x A (x U (¬¬(x U ) ¬(x B)))}

= { DM }

{x|x A (x U (x U x B))}

= { double negation, defn. } {x|x A (x U x B)}

={ null , defn. } {x|x A x U − B}

={ defn.− }

A∩ B

= { defn. complement, ∩ }

26.

A (B ∩ C)

= { defn ∩, }

{x|x A (x B x C)} = { over }

{x|(x A x B) (x A x C)} = { def . ∩, }

(A B) ∩ (A C)

27.

(A ∩ B)

={ defn. complement } {x|x U x (A ∩ B)}

={ defn. , ∩ }

{x|x U ¬(x A x B)} = { DM }

{x|x U (¬(x A) ¬(x B))} = { defn. }

{x|x U (x A x B)} = { over }

{x|(x U x A) (x U x B)}

= { defn. − }

{x|(x U − A) (x U − B)}

= { defn. complement, }

A B

C.9 Inductively Defined Sets

19.The base case appears at the end of the stream, so it cannot be used to start the inductive process of calculating successive elements. So, the stream does not have a printable value.

20.It will never terminate and return the accumulator in which the stream of naturals is constructed.

21.No. The problem is that the stream may present an infinite number of naturals before the one we are interested in can be reached. For example, in [1,3..] ++ [2,4..] all the odd naturals appear before any even number.

22.Given a number n, the set R of roots of n is defined as follows:

•n1 R