[SOLVED] The Foundations: Logic and Proofs

$25

File Name: The_Foundations:_Logic_and_Proofs.zip
File Size: 310.86 KB

5/5 - (1 vote)

The Foundations: Logic and Proofs

Chapter 1, Part III:
/

Copyright By Assignmentchef assignmentchef

Section 1.6

?

() () :

.

p1 ,p2, ,pnq
(p1 p2 pn ) q

:

p .
q .

,

,

(p (p q)) q

p .
q .

,

, .
(q(p q))p

p .
q .
r A.

,
,A

,,A .
((p q) (qr))(p r)

p .
q .

.
.

, .
(p(p q))q

p .
q .

.

,.

p .
q .

.

p .
q .

.
.

.

((p) (q)) (p q)

p .
q .
r .

.
.

.

((p r ) (p q)) (q r)

AliceAlice
JerryJerry

:

1.
2.1
3. 1
4.23

.
.
.
.
:
.

p : . q: .
r: . s: .
t : .

3.

1.
2.1
3.
4. 23
5.
6.45
7.
8.67

Note that a truth table would have 32 rows since we have 5 propositional variables.

pq,rq, rs

BAAAAB

Fido

.

, Fido

A.
a aA

A.
,A.

1: ,

: M(x) x L(x) x
,
:
1.
2.1
3.
4.23

.
.
.
: C(x) x ,B(x) x ,P(x) x .
.

1.
2. C1
3. C (2)
4.
5.4
6. )(3)(4)
7.(2)
8. ) (6)(7)
9.8

,a

P(x)Q(x)

Section 1.7

:

:
()

.

x > y, x y , x2 > y2
x y, x > y, x2 > y2 .

, c ,

, :

: p q

: q , p q

1=1

: p p q
2 + 2 = 5

[ , ]

: n
k n = 2k
k n = 2k + 1

: p q
: p, q
: n , n2
: n n = 2k + 1 k, :
n2 = (2k + 1)2 = 4k2 + 4k +1 = 2(2k2 + 2k) + 1= 2r + 1,
r = 2k2 + 2k , .
n , n2

: p q
: q pp q q p
: n 3n + 2 , n
: n , n = 2k k
3n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j j = 3k +1
3n + 2
q p ,p q n 3n + 2 , n

: p q
p, r r pp F , Tp
:224
: 223 7 21 p r r

: p q
pq, q p ()p p p
: 3n + 2 , n
: p 3n + 2 qn
pqn
kn=2k
3n + 2=3( 2k)+2=6k+2=2(3k+1)t=3k+1
3n+2=2t

p q, p q q p
: n , n n2
: p q q p. p q.

1 = 2

1. a=b
2. a2=a*b1a
3. a2 b2 = a*b b2(2)b2
4. (a-b)*(a+b) = (a-b)*b 3
5. a+b = b 4a-b
6. 2b = b(5)aba=b
7.2=1 (6)b

:

p qr, rq, p

Section 1.8

Example: a @ b = max{a, b} = a a b,
a @ b = max{a, b} = b.
a, b, c
(a @b) @ c = a @ (b @ c) (@ .)
Proof: a, b, and c .
6

Example: x y xy x+y , x y.
Proof: x y , , x x = 2m + 1 m
Case 1: y y = 2nn, x + y = (2m + 1) + 2n = 2(m + n) + 1 .
Case 2: y y = 2n + 1n, x y = (2m + 1) (2n + 1) = 2(2m n +m + n) + 1

yxy

.
c, P(c) .
(EG).
Example: :
Proof:1729 = 103+ 93= 123+ 13

(1877-1947)

Srinivasa Ramanujan
(1887-1920)

, c P(c)
Example: xy xy
Proof: 2 2 2 , x y xy, x = 2 y = 2. 2 2 ,x = 2 2y = 2 xy = (2 2)2= 2 (2 2)= 2 2= 2.

p q
.
(e.g.,).
.
p q, q p
q p p q

Example: 15123
Proof: n
Step n:Player11,23
Step n-1: Player2 4Player21,23
Step n-2: Player15,67 Player14
Step n-3: Player28
Step n-4: Player1 9,10, or 11
Step n-5: Player212
Step n-6: Player112

/docProps/thumbnail.jpeg

CS: assignmentchef QQ: 1823890830 Email: [email protected]

Reviews

There are no reviews yet.

Only logged in customers who have purchased this product may leave a review.

Shopping Cart
[SOLVED] The Foundations: Logic and Proofs
$25