1. Methods of Proof: Overview
Deductive and inductive proofs
If and only if proofs
Review of sets
Examples of deductive proof using sets
Contrapositive of If H then C
Proof by contradiction
Induction on integers

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 implies 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 using either “if S(n) then S(n+1)” or “if S(n-1) then S(n)”

20. Setup equation designed to use the inductive hypothesis
I require a “structured” proof with the follow elements present and identified.
Proof of base case
Setup equation designed to use the 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 the equals sign.

21. Setup equations depends on approach
In approach “if S(n) then S(n+1)”, S(n+1) is on LHS of setup and inductive hypothesis is about S(n).
In approach “if S(n-1) then S(n)”, S(n) is on LHS of setup and inductive hypothesis is about S(n-1).
I will always specify which approach is to be used in a particular problem

22. Background on sums:
k=1 to n+1 k2 = k=1 to n k2 + ?
k=1 to n k2 = k=1 to n-1 k2 + ?

23. Background on sums:
k=1 to UL k = UL(UL+1)/2
If UL=n we get k=1 to n k = n(n+1)/2
If UL=n+1 we get ?
If UL=n-1 we get ?

24. Example: prove k=1 to n k = n(n+1)/2
using “if S(n-1) then S(n)”

25. IH: assume about S(n-1) what you want to prove about S(n)
Structured Proof of
k=1 to n k = n(n+1)/2
by if S(n-1) then S(n)
Base case n=1
k=1 to 1 k =?
1(1+1)/2 =?
IH: assume about S(n-1) what you want to prove about S(n)
Application of IH->
Setup equation
IH:

26. Assignment #1 Due 8-23-19 k=1 to n k2 = n(n+1)(2n+1)/6
Use the approach if S(n-1) then S(n) to give a structured proof that
k=1 to n k2 = n(n+1)(2n+1)/6

27. Induction in recursive definitions
Example: definition of a tree structure
Base case: a single node is a tree
Induction: the structure obtained by adding a node
to a tree is a tree
Prove: in a tree the #of nodes = #of edges+1
use approach if S(n-1) then S(n).

28. 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.
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.