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Quiz Friday 8-29

Lectures: introduction and methods of proof

Text: pp 1-26

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Methods of Proof

Deductive proof: Used extensively in geometry &

trigonometry Not very important in this class

Inductive proof: Basis of most automata theory Proficiency in induction is a primary

objective of this class

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Deductive proof

Step-by-step argument from given information to a conclusion

“from the information that you gave me, I deduce that ….”

No internal assumptions are allowed Example: proving that sets are

equivalent

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Proving equivalences about sets

In automata theory frequently ask “Are sets constructed in different ways actually the same set?”

Elements of sets are usually strings; hence sets are called “languages”

Statement “set E = set F” means every element in E is in F and every element in F is in E

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Equivalent sets 2

set E = set F is example of “if and only if”

Element x is in E if and only if x is in F To prove E = F must prove 2 if…thenstatements

If x is in E then x is in F If x is in F then x is in E

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Background on sets If x A implies x B, then A is a

subset of B (written A B) If, in addition, A B then A is a proper

subset of B (written A B)  A B = intersection of sets A and B

defined by {x: x A and x B } A B = union of sets A and B defined

by {x: x A or x B } 

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Background on sets 2

A - B = difference of sets A and B defined by {x: x A and x B }

Let T be complement of S with respect to U then SU, TU, ST = U, and ST = null

Sets obey laws similar to numbers (commutative, associative, distributive, etc)

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Proof of distributive law of sets

R(ST) = (RS)(RT) Let E = R(ST) Let F = (RS)(RT) Must prove

if x is in E then x is in F if x is in F then x is in E

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Proof of distributive law of sets 2

if x R(ST) thenx R or x (ST) (def. of union)

x R or x S and x T (def. of intersection)

x RS (def. of union)

x RT (def. of union)

x (RS)(RT) (def. of intersection)

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Proof of distributive law of sets 3

if x (RS)(RT) thenx RS (def. of intersection)

x RT (def. of intersection)

x R or x S and x T (why?)

x R or x (ST) (def. of intersection)

x R(ST) (def. of union)

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Contrapositives Given hypothesis H and conclusion C,

one of following cases is true 1. H and C are both true 2. H is true and C is false 3. H is false and C is true 4. H and C are both false

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Only (2) makes “if H then C” false (3) does not apply to “if H then C” In both (1) and (4) “if H then C” is

true

Contrapositives 2 Contrapositive of “if H then C” is “if not

C then not H” Hypothesis is “not C” Conclusion is “not H” 4 cases of previous slide still apply

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Only (2) makes “if not C then not H” false

(3) does not apply to “if not C then not H”

In (1) and (4) “if not C then not H” is true

hence “if not C then not H” is equivalent to “if H then C” and may be easier to prove

Contrapositive 3

Contrapositive of “if x > 4 then 2x

> x2” is “if 2x < x2 then x < 4” Sometimes contrapositive is easier

to prove. Can this contrapositive be proved

by algebra?

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Proof by contradictionProving “if H and not C -> falsehood” is equivalent to proving “if H then C”

Example using complement of a set

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Proof by contradiction 2 T is the complement of S with respect to U

then SU, TU, ST = U, and ST = null If S is finite and U is infinite, then T is

infinite Hypothesis has 3 parts

T is complement of S S is finite U is infinite

Conclusion is “T is infinite” not C is “T is finite”

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Proof by contradiction 3

If T is finite, then ||T||=some integer m S is finite; therefore ||S||=some integer

n T is complement of S with respect to U;

therefore; ST = U and ST = null Hence, ||U||=m+n By definition, this means U is finite H and not C contradicts part of H This proves “if H then C” 17

Inductive proof Many types of induction

Induction on integers is most familiar

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Induction on integers S(n) is statement about integers we

want to prove Base case establishes truth of the

statement for n=is typically the smallest integer where statement is true

Inductive step proves statement for n>is typically by “if S(n) then S(n+1)”

“if S(n)” is called the “inductive hypothesis”

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How I grade inductive proofs

Proof of base case Statement of inductive hypothesis How inductive hypothesis is used

Basis for algebra that completes the proof

All algebraic operations are on the rhs of equals sign

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Assignment #1 Due 9-3-14

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Prove by induction on integers that

k=1 to n k = n(n+1)/2

Your proof must include the following:(1) truth of base case(2) statement of inductive hypothesis(3) correct application of inductive hypothesis

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Basis can consist of a range of i values, all of which must be used in the inductive step.

“if S(n-1) then S(n)” is equivalent to “if S(n) then S(n+1)”

Example: if S(n-1) then S(n) applied to assignment 1

Exceptions to typical approach

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Provej=1 to n j = n(n+1)/2

Base case n=1j=1 to 1 j = 1(1+1)/2

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ReviewInduction used 2 ways in the class

recursive definitionsproofs

Both have 3 partsbase caseassumption (IH)induction

Example: definition of tree and proof that n=e+1

Structural Induction Automata theory involves recursive

definitions that become the subject of inductive proofs.

Example: recursive definition of a tree Basis: single node is a tree IH: assume that T1, … Tk are trees Induction: connect then to the node that

is root of a recursively defined tree

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Structural induction 2

Prove: in every tree n = e +1, where n is the number of nodes and e is the number of edges

Basis: true for single-node tree IH: assume ni=ei+1 true for trees T1, …,

Tk Induction:

n=1+n1+n2+…+nk = 1+(e1+1)+…+(ek+1)

n=1+k+e1+…+ek = 1+e26

Switch automaton

Nodes are states; Arcs are transitions Arcs labeled by strings that induce

the indicated transitions Arrow indicates the start state

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off

push

on

Startpush

Data transfer automaton

Data induces transition to sending Loop keeps automaton in sending state

until transmission is done “Ready” and “Sending” are complex

states Many conditions must be met for

automaton to be in these states 28

Ready Sending

data in

done

timeout

Start

Mutual Induction An inductive hypothesis may have several

facets Example: all the conditions necessary to

put an automaton in a certain state If all facets depend on a common value of

an integer, then we have “mutual induction”

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Example of Mutual Induction

S1(n): automaton in the off state after n pushes iff n is even

S2(n): automaton is in the on state after n pushes iff n is odd

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off

push

on

Startpush

Prove the following about the “switch” automaton

Both S1(n) and S2(n) are iff

If n is even, then switch is off If switch is off, then n is even If n is odd, then switch is on If switch is on, then n is odd

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off

push

on

Startpush

Given that switch starts in “off” we must prove that after n pushes …

Prove : if n is even then switch is off

IH: if k is odd then switch is on If n is even then n-1 is odd By IH, switch is on after n-1 pushes One more push tunes switch off Similar proof for other 3 statements

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off

push

on

Startpush

Basis: n=0 true zero even by definition and switch starts in off

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Mutual Induction 3

“switch” automata is always in exactly one state. Not generally true

one-state character of switch makes this example of mutual induction artificial

Proof of S1(n) implies the truth of S2(n) Text treats as mutual induction

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off

push

on

Startpush

Loop invariant and induction

Loop invariant is statement about iterative algorithm

Prove the statement is true by induction

Show true before 1st iteration (base case or initialization)

If true on ith iteration (inductive hypothesis) remains true before i+1 iteration (maintenance)

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Loop invariant Termination

Aspect of loop invariant analysis that goes beyond general inductive proof

Having establish that the loop invariant is true, statement of the loop invariant when looping terminates proves the correctness of the iterative algorithm

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Example: Sort by insertion

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Validation is usual approach in code

development Test code for some simple cases

were hand calculation is practical If a “validated” code fails, usually

learn something important In CS 317 get some practice with

formal proofs

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