The Foundations: Logic and Proofs
Chapter 1, Part III:
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Section 1.6
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(p1 p2 pn ) q
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t : .
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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.
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: C(x) x ,B(x) x ,P(x) x .
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2. C1
3. C (2)
4.
5.4
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7.(2)
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9.8
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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
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