Methods of Proof
Definitions
• A theorem is a valid logical assertion which can
be proved using
– Axioms: statements which are given to be true
– Rules of inference: logical rules allowing the
deduction of conclusions from premises
• A lemma is a ‘pre-theorem’ or a result which is
needed to prove a theorem.
• A corollary is a ‘post-theorem’ or a result which
follows directly from a theorem.
Rules of Inference
H1
H2
Hn

C
H1, H2, … Hn are the hypotheses
We use conjunction: H1 ^ H2 ^ H3…
C is the conclusion.
“” means “therefore” or “it follows that”
Some Rules of Inference
p
pq
p  ( p  q)
Addition
pq
p
( p  q)  p
Simplification
(( p )  ( q ))  ( p  q )
Conjunction
( p  ( p  q ))  q
Modus Ponens
p
q
pq
p
pq
q
Mode that affirms
Example: Simplification
pq
p
( p  q)  p
Simplification
p:”it is below freezing”
q:”it is raining now”
• It is below freezing and raining now.
Therefore, it is below freezing.
Example: Modus Ponens
from Latin: mode that affirms
p
pq
q
( p  ( p  q ))  q
• In other words
If the hypothesis p is true
and the hypothesis (p->q) is true
Then I can conclude q
Example: Modus Ponens
• p : “n is greater than 3”
• q: “n2 is greater than 9”
• Assuming that p is true, and p q is
true, then
• if is n greater than 3, it follows that n2
is greater than 9.
More Rules of Inference
q
pq
 p
pq
(q  ( p  q ))  p
mode that denies
(( p  q )  ( q  r ))  ( p  r )
Hypothetical
Syllogism
(( p  q )  p )  q
Disjunctive
Syllogism
qr
pr
pq
p
q
Modus Tollens
Exercise
• If It rains today, then we will not have a
barbecue today. If we do not have a
barbecue today, then we will have a
barbecue tomorrow
• Therefore, if it rains today, then we will
have a barbecue tomorrow.
Validity of an Argument
• An argument is valid if
– whenever all hypotheses are true, the
conclusion is also true
• To prove that an argument is valid:
– Assume the hypotheses are true
– Use the rules of inference and logical
equivalences to determine that the
conclusion is true
Recap 1.2: Important
Equivalences
pTp
pFp
Identity
pTT
pFF
ppp
ppp
Domination
( p)  p
Double Negation
Idempotent
Recap 1.2: Important Equivalences
pqqp
pqqp
Commutative
(p  q)  r  p  (q  r)
(p  q)  r  p  (q  r)
Associative
p  (q  r)  (p  q)  (p  r)
p  (q  r)  (p  q)  (p  r)
(p  q)  p  q
(p  q)  p  q
Distributive
De Morgan’s
Recap 1.2: Important
Equivalences
p  (p  q)  p
p  (p  q)  p
Absorption
p  p  T
p  p  F
Negation
Example
• Consider the following logical argument:
– If horses fly or cows eat artichokes, then the
mosquito is the national bird. If the
mosquito is the national bird then peanut
butter tastes good on hot dogs. But peanut
butter tastes terrible on hot dogs. Therefore,
cows don’t eat artichokes.
• Assign propositional variables to each
component in the argument
Example
• Assignments:
p
q
r
s
Horses fly
Cows eat artichokes
The mosquito is the national bird
Peanut butter tastes good on hot dogs
• Represent the argument using the
variables
(p  q)  r
rs
s
 q
Hypotheses
Conclusion
Example
Assertion
Reasons
1. (p  q)  r
Hypothesis
2.
3.
4.
5.
6.
7.
8.
rs
(p  q)  s
s
(p  q)
p  q
q  p
q
Hypothesis
Hypothetical syll. on 1. and 2.
Hypothesis
Modus tollens on 3. and 4.
DeMorgan on 5.
Commutative on 6.
Simplification on 7.
We got our conclusion of “cows don’t eat artichokes”
Example
• Show that the following argument is valid:
– It is not sunny this afternoon and it is colder
than yesterday. We will go swimming only if it
is sunny. If we do not go swimming, then we
will take a canoe trip. If we take a canoe trip,
then we will be home by sunset. Therefore,
we will be home by sunset.
Rules of Inference for Quantified
Statements
xP( x)
 P(c) if c U
P(c) for an arbitrary c  U
 xP( x)
xP( x)
 P(c) for some element c  U
P(c) for some element c  U
 xP( x)
Universal instantiation
xP(x), then for any
C, therefore P(c) is true
Universal generalization
we select any
element c and P(c) is true
Therefore xP(x)
Existential instantiation
xP(x)
therefore for at
least one specific c,
P(c) is true
Existential generalization
we select a particular
element c and P(c) is true
Therefore, xP(x)
Example
• Everyone in the discrete math class has taken
a CS course. Marla is a student in the discrete
class. Therefore, Marla has taken a CS course.
D(x): x is in the discrete math class
C(x): x has taken a CS course
1. x(D(x)  C(x))
2. D(Marla)  C(Marla)
3. D(Marla)
4. C(Marla)
x(D(x)  C(x))
D(Marla)
 C(Marla)
Premises
Premise
Univ. Inst. using 1.
Premise
Modus ponens using 2. and 3.
Exercise
• A student in this class has not read the
book. Everyone in this class passed the
first exam.
• Therefore, Someone who passed the
first exam has not read the book
Fallacies
• Fallacies resemble rules of inference but
are based on contingencies rather than
tautologies. They are incorrect
inferences.
• Three common fallacies
– Affirming the Consequent
– Denying the Hypothesis
– Circular Reasoning (begging the question)
Fallacy of Affirming the Consequent
pq
(( p  q )  q )  p
q
p
• ((p  q)  q)  p is not a tautology and therefore not
a rule of inference.
• If you do every problem in this book, then you will
learn discrete mathematics. You learned discrete
mathematics. Therefore, you did every problem in this
book.
• Exercise. Compare with modus ponens and tollens
Fallacy of Denying the Hypothesis
pq
(( p  q )  p )  q
p
 q
• ((p  q)  p)  q is not a tautology and therefore
not a rule of inference.
• If you do every problem in this book, then you will
learn discrete mathematics. You did not do every
problem in this book. Therefore, you did not learn
discrete mathematics.
• Exercise. Compare with modus ponens and tollens
Circular Reasoning
• One or more steps of the proof are
based upon the truth of the statement
being proved.
• Also known as begging the question
Exercises
• 1, 3, 4, 7, 9, 11
Recap:
Some Rules of Inference
p
pq
p  ( p  q)
Addition
pq
p
( p  q)  p
Simplification
(( p )  ( q ))  ( p  q )
Conjunction
( p  ( p  q ))  q
Modus Ponens
p
q
pq
p
pq
q
Mode that affirms
Recap:
More Rules of Inference
q
pq
 p
pq
(q  ( p  q ))  p
mode that denies
(( p  q )  ( q  r ))  ( p  r )
Hypothetical
Syllogism
(( p  q )  p )  q
Disjunctive
Syllogism
qr
pr
pq
p
q
Modus Tollens
Rules of Inference for Quantified
Statements
xP( x)
 P(c) if c U
P(c) for an arbitrary c  U
 xP( x)
xP( x)
 P(c) for some element c  U
P(c) for some element c  U
 xP( x)
Universal instantiation
xP(x), then for any
C, therefore P(c) is true
Universal generalization
we select any
element c and P(c) is true
Therefore xP(x)
Existential instantiation
xP(x)
therefore for at
least one specific c,
P(c) is true
Existential generalization
we select a particular
element c and P(c) is true
Therefore, xP(x)
Methods of Proof
•
•
•
•
•
•
•
Direct proof
Indirect proof
Vacuous proof
Trivial proof
Proof by contradiction
Proof by cases
Existence proof
Proof Basics
• We want to establish the
truth of p  q
• p may be a conjunction of
other hypotheses
• p  q is a conjecture until
a proof is produced
p
q
p q
T
T
T
T
F
F
F
T
T
F
F
T
Direct Proof
• Assume the hypotheses are true
• Use rules of inference and any logical
equivalences to establish the truth of the
conclusion
• HOW TO PROVE:
– If p is true ,then q has to be true for p—>q to be
true
• Example: The proof we did earlier about cows
not eating artichokes was an example of a
direct proof
Example
• Give a direct proof of the theorem: “If n
is an odd integer, then n2 is an odd
integer”
(n is odd)  (n2 is odd)
• Using the following definition:
– If n is even, then exist an integer k such
that n=2k, and It is odd, if there exist and
integer k such that n=2k+1.
Example (Cont)
• Assume the hypothesis “n is odd” true:
– n is odd
• Since n is odd, then k n=2k+1
• Now, is the conclusion “n2 is odd” true?
•
n2 = (2k+1)2 = 4k2 +4k +1
= 2(2k2+2k)+1
= 2 (m) +1, where some integer m=2k2+2k
• Since n2 = 2(m)+1, then “n2 is odd” is true
• Proof complete
Indirect Proof
• A direct proof of the contrapositive
– Remember: pq is equivalent to ~q  ~p
– Proof ~q  ~p
• Assume that q is true i.e., q is false
• Use rules of inference and logical
equivalences to show that p is true i.e.,
p is false
Example
• Give an indirect proof to the theorem “if 3n+2
is odd, then n is odd”
(3n+2 is odd)  (n is odd)
p: 3n+2 is odd,
q: n is odd,
~p: 3n+2 is even
~q: n is even
The contrapositive is:
~(n is odd)  ~ (3n+2 is odd) , in other words
(n is even)  (3n +2 is even)
Example (Cont)
• Assuming the hypothesis (of the contra
positive) “n is even” true
• Then n=2k
• Now, is the conclusion (of the contrapositive)
“3n+2 is even” true?
3n+2 = 3(2k)+2=6k+2
=2(3k+1)
=2(m), where m =3k+1
• Then “3n+2 is even” is true
• Proof complete
Vacuous Proof
• If we know one of the hypotheses in p is false
then pq is vacuously true.
• F  T and F  F are both true.
• Example:
– If I am both rich and poor, then hurricane Katrina
was a mild breeze.
– The hypotheses (pp) form a contradiction,
therefore q follows from the hypotheses vacuously.
• Sometimes used to proof theorems using
UNIVERSAL QUANTIFICATION
Example
Show that P(0) is true where P(n): If n > 1, then n2
> n.
P(n): (n>1)  (n2 > n)
P(0): (0>1)  ( 02 > 0)
Since the hypothesis (0>1) is false, P(0) is
automatically true.
Note that we do not even pay attention to the
conclusion “02 > 0”
Trivial Proof
• If we know q is true, then pq is true
• F  T and T  T are both true.
• Example:
– If it’s raining today then the empty set is a
subset of every set.
– The assertion is trivially true independent of
the truth value of p.
Example
• Show that P(0) is true where P(n): If a  b > 0,
then an  bn.
P(n): (a  b > 0)  (an  bn)
P(0): (a b > 0)  (a0  a0), in other words
P(0): (a b > 0)  (1  1),
Since the conclusion (1  1) is true, hence P(0) is
true.
Note that we do not even pay attention to the
hypothesis “(a b > 0) “
Proof by Contradiction
(reduction to the absurd )
• We want to proof that pq, but… what if we can
proof that ~p implies a contradiction (i.e., q is FALSE
no matters what, or an absurd)??
• Mathematical definition of the proof
– Find a contradiction q such that
pq  pF  T
• Consequently, if we show the contradiction, then the
assumption ~p is wrong (FALSE), so p must be true
• To prove that p is true, you have to show that p
leads to a contradiction i.e., you have to prove that
pF is true.
Example
• Prove that √2 is irrational
– P: “√2 is irrational”
• What if ~p is true, “√2 is rational”
– Does this lead to a contradiction???
• If √2 is rational, then a,b integers
such that √2 =a /b (assuming r:”a and
b have no common factors”)
• √2 = a/ b , then 2 = a2/b2, then 2b2=a2
Example cont
• Since a2=2 (b2), then a2 is even, therefore a
is even, then
• 2b2 = (2c)2, for some integer c
• 2b2=4c2, so b2=2c2
• Thus, b is even too.
• If a and b are even, then they have at least
one common factor (2), so the assumption r
is contradicted: p-> (r^~r)
• Therefore, ~p is false, p is true
» “√2 is irrational is true
Proof by Contradiction
(Cont..)
• An indirect proof of an implication pq can
be rewritten as a proof by contradiction.
• Assume that both p and q are true.
• Then use a direct proof to show that
q  p
• This leads to the contradiction pp.
• HOMEWORK. Example 22 (pg 67) If 3n+2 is odd,
then n is odd.
Proof by Cases
• Break the premise of pq into an equivalent
disjunction of the form p1p2pn
• Then use the equivalence
[(p1p2pn)  q]  [(p1q)(p2 q)   (pnq)]
• Each of the implications pi q is a case.
• You must
– Convince the reader that the cases are inclusive (i.e.,
they exhaust all possibilities)
– Establish all implications
Proof by Cases (Cont..)
• If n is an integer, then n2+1 is positive.
p: n is an integer
q: n2+1 is positive
• p  (p1p2p3)
where, p1: n = 0
p2: n > 0
p3: n < 0
• Now prove that ((p1q)(p2 q)(p3 q)) is
true i.e., all the cases (p1q), (p2 q), and
(p3 q) are true.
Proving an equivalence
• To prove that pq, you need to show
that pq is a tautology.
• You can do that by showing that pq
and qp are both true since,
pq  [(pq)(qp)]
• Example: The integer n is odd if and
only if n2 is odd.
Existence Proof
• The proof of xP(x) is called an existence proof.
– Constructive
– Non-constructive
• Constructive existence proof
– Find an element c in the universe of discourse such
that P(c) is true
• Non-constructive existence proof
– Do not find c, rather, somehow prove xP(x) is true
– Generally, by contradiction
• Assume no c exists that makes P(c) true
• Derive a contradiction