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Lecture 11

1.5, 3.1 Methods of Proof

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Last time in 1.5 To prove theorems we use rules of inference such as: p, pq, therefore, q NOT q, pq, therefore NOT p. p AND q, therefore p FORALL x P(x), therefore for arbitrary c, P(c) EXISTS x P(x), therefore for some c, P(c) It is easy to make mistakes, make sure that: 1) All premises pi are true when you prove (p1 AND p2 AND...pn) q 2) Every rule of inference you use is correct. Some proof strategies: To proof pq 1) direct proof: assume p is true, use rules to prove that q is true. 2) indirect proof, assume q is NOT true, use rules to prove p is NOT true. To prove p is true: 3) By contradiction: assume p is NOT true, use rules to show that NOT pF i.e. it leads to a contradiction.

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Vacuous –Trivial Proofs Lets say we want to prove pq but the premise p can be shown to be false! Then pq is always true because (FT) = T and (FF) = T. This is a vacuous prove.

Old example: prove that for any set S: Proof: The following must be shown to be true: However: the empty set does not contain any elements and the premise is always false. Therefore the implication must always be true!

*S*∅⊆

( )*x x x S*∀ ∈∅→ ∈

Trivial Proof: We want to prove pq, and we can show that q is true. Then, because (TT) = T and (FT) = T we have proven the implication.

Example: P(n): a>=b a^n >= b^n for postive integers. Is P(0) true? P(0): a^0 >= b^0 is equivalent to 1>=1. Therefore, q is true and thus pq is true.

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Example Indirect Proof

Prove that: if n is an integer and n^2 is odd, then n is odd. Direct prove is hard in this case. Indirect proof: Assume NOT q : n is even. n = 2k n^2 = 4k^2 = 2(2k^2) is even, is not odd. Thus NOT q NOT p, pq

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Example of Proof by Contradiction
def: rational number is a number that can be written as a/b for integers a,b, where
b should not be 0. Reals that are not rational are irrational.
sqrt(2) is irrational.
(note: we are not proving an implication now, although we could
have written it as: if x=sqrt(2) x = irrational.)
Assume sqrt(2) is not irrational.
sqrt(2) = a/b where there is no common divisor (otherwise divide by this number).
2 = a^2 / b^2
2b^2 = a^2
a^2 = even
a = even
a = 2c
2b^2 = 4c^2
b^2 = 2c^2
b^2 = even
b=even
both a and b can be divided by 2! (contradiction) *p r r*¬ → ∧¬

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1.5 An indirect proof is in fact also a prove by contradiction: Assume p = T and (NOT q) = T indirect prove: (NOT q) (NOT p) Therefore: p AND (NOT p) = T, contradiction

Proof by cases: (p1 OR p2 OR p3 ... OR pn) q (if at least one of the premises hold, q follows) This is equivalent to (p1q) AND (p2q) AND ... AND (pnq) Example: Prove x^2 >= 0 p1: x<0 x^2 >=0 p2: x=0 x^2 >=0 p3: x>0 x^2 >=0

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Equivalence Proofs These are bidirectional statements of the form pq. Equivalent to proving 2 cases: pq qp since (pq) AND (qp) (pq) is a tautology.

Example: Prove integer n is odd if and only if (iff) n^2 is odd. Lecture 10: if n is odd n^2 is odd Lecture 11: if n^2 is odd n is odd.

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Lists of Equivalences

p1 p2p3.....pn This means they are all equivalent. There are C(n,2) pairs to prove! There is however a smarter way: Design one path through all pi’s that can bring you from any pi to any other then you are done:

p1 p2

p3

p4 p5

p6

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Theorems with Quantifiers
**Existence proofs:** Proofs of the form: There exists an element x such that.....
these proofs may be constructive (construct some x) or non-constructive.

( )*xP x*∃

**Uniqueness: **Proofs of the form: there exists a unique element x such that...

! ( ) ( ( ) ( ( )))*xP x x P x y y x P y*∃ = ∃ ∧∀ ≠ →¬

Example: Prove that ! ( 0)*x y x y*∀ ∃ + =
*proof*: for arbitrary x, y=-x makes the proposition true. Is it unique.
Assume it is not true and show contradiction:
Let say there is a r such that x+r=0 but r is not –x.
Then it follows that x+r=x+y r=y which contradicts our assumption.

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3.1

self reading

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The Halting Problem
Is it possible to design an algorithm that always predicts for a given program P
and a given input to that program I, if it will stop run forever?
Proposition: there isn’t
Proof.
Assume there is such a predictor H(P,I).
A program can be represented as a bit-string, and therefore as input to another
program. Imagine a program P that can take itself as input.
H(P,P) should predict if it stops: H = 1 if it stops, H=0 if it doesn’t stop.
Define a new program K(P) as follows:
that is K(P) loops forever if H(P,P) = 1
K(P) = stops if H(P,P) = 0
Can H(K,K) predict whether this program stops?
K(K) loops forever if H(K,K) = 1
K(K) stops if H(K,K) = 0
*So H(K,K) does not predict correctly which is in contradiction with the assumption!*

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Sequences & Summations

**definition:***A sequence is a function from the set of integers to a set S:
a1,a2,a3,.... (function from 1,2... ** a1,a2,...)
an is a term in the sequence which is sometimes denoted with {an} (not a set!)*

Example: {an} with an= 1/n over n=1,.... 1,1/2,1/3, ...

**Geometric progression: ***Sequence of the form: a,ar,ar^2,...,ar^n,...
a = initial term, r = common ratio are real numbers
*3,6,12,24,... (a=3,r=2): ratios are constant
**Arithmetic progression ***Sequence of the form a,a+d,a+2d,...,a+nd,...
a = initial term and d = common difference are real numbers
*3,5,7,9,11,...(a=3,r=2): differences are constant
**
A string: ***finite sequence a1,...,an.*

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Sums Notation

2 2 2

1 1 1

1 2 1

2

2 2

, , 1

,...,

( .... )

*n n n
*

*i a k m n
i m a m k m
n n n
*

*ij i m i m in
i m j m i m
*

*a a a a a
*

*a a a a
*

−

+ = = = −

+ = = =

= = =

= + +

∑ ∑ ∑

∑ ∑ ∑

notation:

single sum

double sum

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Summations

**Theorem
**

1

0

1 1

( 1) 1

*n
n
*

*j
*

*j
*

*ar a if rar r
n a if r
*

+

=

− ≠= −

+ = ∑

0

1

0

1

1

1

0 1

1

1 1

*n
j
*

*j
*

*n
j
*

*j
*

*n
k
*

*k
n
*

*n k
*

*k
n
*

*n
*

*S ar
*

*rS ar
*

*rS ar
*

*rS ar a ar
*

*rS ar a S
ar aS if r
*

*r
*

=

+

=

+

=

+

=

+

+

= ⇔

= ⇔

= ⇔

= − + ⇔

= − + ⇔

− = ≠

−

∑

∑

∑

∑

proof

trivial

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Summations
**Theorem:
**

0

( 1)( 1) 2

*n
*

*j
*

*n na jd n a d
*=

+ + = + +∑

0 1 2 ... (0 ) (1 1) (2 2) ...

( 1) 2

*n
n n n
*

*n n
*

+ + + + = + + + − + + − + +

=

sketch of proof: Split into 2 cases: even and odd n. In both cases argue as follows:

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