1. Methods of Proof

2. Deductive proof: Inductive proof:
Used extensively in geometry & trigonometry
Not very important in this class
Inductive proof:
Basis of most automata theory
Proficiency in induction is a learning objective of this class

3. 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

4. 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

5. Equivalent sets 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 two if…then
statements
If x is in E then x is in F
If x is in F then x is in E

6. 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 }

7. More background on sets
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 SU, TU, ST = U, and ST = null
Sets obey laws similar to numbers (commutative, associative, distributive, etc)

8. Proof of distributive law of sets
R(ST) = (RS)(RT)
Let E = R(ST)
Let F = (RS)(RT)
Must prove
if x is in E then x is in F
if x is in F then x is in E

9. Proof of distributive law of sets: part 1
if x  R(ST) then
x  R or x  (ST) (def. of union)
x  R or x  S and x  T (def. of intersection)
x  RS (def. of union)
x  RT (def. of union)
x  (RS)(RT) (def. of intersection)
Therefore if x is in E then x is in F

10. Proof of distributive law of sets: part 2
if x  (RS)(RT) then x  RS (def. of intersection) x  RT (def. of intersection) x  R or x  S and x  T (logic?) x  R or x  (ST) (def. of intersection) x  R(ST) (def. of union) Therefore if x is in F then x is in E

11. Logic of “If H then C”
Given 2 statements, hypothesis H and conclusion C, exactly one of following is the case:
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
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

12. Contrapositives The contrapositive of “if H then C” is
“if not C then not H”
New hypothesis is “not C”
New conclusion is “not H”
4 cases of previous slide still apply

13. Logic of “If not C then not H”
Given statements not C and not H, exactly one of following is the case:
1. not C and not H are both true
2. not C is true and not H is false
3. not C is false and not H is true
4. not C and not H are both false
Only (2) makes “if not C then not H” false
(3) does not apply to “if not C then not H ”
In both (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

14. Contrapositive question
What is the contrapositive of “if x > 4 then 2x > x2”

15. Proof by contradiction
Proving that “if H then not C” is false is equivalent to proving “if H then C” Example using complement of a set

16. Proof by contradiction
Given: T is the complement of S with respect to U
Prove: 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”

17. Proof by contradiction
assume T is finite (i.e. not C)
then ||T||=some integer m
Given S is finite; therefore ||S||=some integer n
T is complement of S with respect to U; therefore; ST = U and ST = null
Hence, ||U||=m+n, which -> U is finite
Therefore “if H then not C” contradicts the part of H that U is infinite
This proves “if H then C”

18. Inductive proof Many types of induction
Induction on integers is probably the most familiar

19. 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”

20. Setup to use inductive hypothesis Statement of inductive hypothesis
I require a “structured” proof. The follow elements must be present and identified.
Proof of base case
Setup to use inductive hypothesis
Statement of inductive hypothesis
Application of inductive hypothesis
Algebra that completes the proof
All algebraic operations must be on the right-hand-side of equals sign

21. Assignment #1 Due 8-25-17 k=1 to n k = n(n+1)/2
Using the approach if S(n) then S(n+1), give a structured proof that
k=1 to n k = n(n+1)/2

22. “if S(n-1) then S(n)” is equivalent to
apply “if S(n-1) then S(n)” to assignment 1

23. Structured Proof of j=1 to n j = n(n+1)/2 by if S(n-1) then S(n)
Base case n=1
j=1 to 1 j =? 1(1+1)/2
Setup equation

24. Induction in recursive definitions
Example: definition of tree
Base case: a single node is a tree
Induction: the structure obtained by adding a node to
the root of a tree is a tree
Note: this definition is not general
What is the structure of tree it defines?
Prove: in a tree the #of nodes = #of edges+1
Hint: use if S(n-1) then S(n).

25. Structural Induction Prove: in every tree #of nodes = #of edges+1
Basis: true for single-node tree (no edges)
IH: assume #of nodes = #of edges+1 is true for a tree with n-1 nodes
Induction: As defined in the previous slide, a tree with n nodes can be obtained from a tree with n-1 nodes by attaching a node to the root.
Adding a node requires adding an edge; hence, the n-node tree has both one more node and one more edge than the n-1 node tree.
Therefore #of nodes = #of edges+1 applies to the n-node tree.

26. Mutual induction on integers
An inductive hypothesis may have several facets
If all facets depend on a common value of an integer, then we have “mutual induction on integers”

27. Example of mutual induction on integers
Given that the start state
of the switch automaton
is “off” prove the following:
off
push on
Start
S1(n): automaton is 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

28. Both S1(n) and S2(n) are iff
off
push on
Start
Given that switch starts
in “off” we must prove
that after n pushes …
If switch is off, then n is even
If n is even, then switch is off
If switch is on, then n is odd
If n is odd, then switch is on

29. Prove: if switch is off then n is even
push on
Start
Basis: in start state
switch is off and n=0,
which is even.
IH: if switch is on then k is odd
Let k=n-1
One more push tunes switch off
If n-1 is odd then n is even
Similar proofs for S2(n)

30. Prove: if n is even then switch is off
push on
Start
Basis: n=0 is true. Zero
is even by definition
and switch starts in 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

32. 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

33. 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+e

34. Switch automaton Nodes are states; Arcs are transitions
off
push on
Start
Nodes are states; Arcs are transitions
Arcs labeled by strings that induce the indicated transitions
Arrow indicates the start state

35. Data transfer automaton
Ready
Sending
data in
done
timeout
Start
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

36. Mutual Induction 3
off
push on
Start
“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

37. 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)

38. 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

39. Example: Sort by insertion

40. 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