Transcript The Foundations

Discrete Mathematics
Chapter 1.
The Foundations:
Logic and Proofs
Cheng-Chia Chen
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The Foundations
1. The foundations : logic and proofs
Logic and proofs:
 The basis of all mathematical reasoning
 has practical applications to the design of
computing machine
 give the precise meaning of math. statements.
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Foundations of Logic (§§1.1-1.4)
Mathematical Logic is a tool for working with
complicated compound statements.
It includes:
 A language(Vocabulary) for expressing them.
 A concise notation(Syntax) for writing them.
 A methodology for objectively reasoning about
their truth or falsity (Semantics and Axiomatics).
 It is the foundation for expressing formal proofs
in all branches of mathematics.
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Foundations of Logic: Overview
Propositional logic (§1.1-1.2):
 Basic definitions. (§1.1)
 Equivalence rules & derivations. (§1.2)
Predicate logic (§1.3-1.4)
 Predicates.
 Quantified predicate expressions.
 Equivalences & derivations.
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§1.1 Propositional Logic
Propositional Logic is the logic of compound
statements built from simpler statements
using so-called Boolean connectives.
Some applications in computer science:
 Design of digital electronic circuits.
 Expressing conditions in programs.
George Boole
 Queries to databases & search engines. (1815-1864)
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Proposition
Proposition: A statement that is either true
or false (,but not both)
 even if we do not know it is true or false now.
 Example:
1. Taipei is the capital of ROC. 2. Tokyo is the capital of Korea.
3. 1+1 = 2. 4. 2+2 = 3.
5. There are no integers x,y,z > 0 and n > 2 s.t.
xn + yn = zn ---- Fermat’s last theorem
6. “Every even number > 2 is the sum of two prime numbers.”
---Goldbach’s conjecture (1742)
==>
1. Statement (1)~(6) are propositions,
2. (1),(3) are true; (2),(4) are false.
3. Nobody knew (5) is true or false until 1994, but it must be
true or false, hence it is a proposition. (6) open problem.
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More Examples
Examples:
1. What time is it? (interrogative)
2. Read this carefully! (command)
3. x+1 = 2.
4. x+y = z.
5. 李登輝是中華民國總統
==>
1. (1) and (2) are not statements
2. (3),(4) are not propositions:
neither true nor false!
3. How about (5) ? // It depends on the time the sentence
refers to.
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Propositional variables and propositional constants
 In algebra, we use (numerical) variables to denote
items with an unspecified numberical value.
 e.g. 10 chickens and rabbits in a cage with 26 feet. ==>
#rabbits = ? and #chickens=?
 ==> let x = number of rabbits, y = number of chickens
 ==> x + y = 10 /\ 2x + 4y = 26.
 Likewise, In logic, we can use propositional variable
to denote an arbitrary proposition with unspecified
truth value.
 e.g., If it is raining, the ground is wet. since the ground is
wet, it is raining now.
 ==> Let p =def “It is raining”, q =def “the ground is wet”
 ==> if p, (then) q. Since q, p now. ==> ((pq) /\ q )  p.
 Propositional constants(literals): only two values:
 T(true), F(false).
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Topic #1.0 – Propositional Logic: Operators
Expressions, Operators and Operands
 表達式(expression), (運)算子(operator) 與 (運)算元
(Operand)
 Ex:
(2+x) * ( z – 5)
 Classifications of expressions
1. Primitve or compound expressions

Primitive expression: literals or variables

ex: 2, x, y, 50, T, F, p , q

Compound expressions: formed by one operator and
one or more operands.

ex: (2+x) * ( z – 5), (p q) || (! q)
2. Numeric or logical expressions
 Numerical expressions (數值表達式) are used for
expressing numerical values
 Logical(or boolean)expressions(formula; propositions)
(邏輯表達式) are used for expressing logical values.
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Operator / Connective
 More examples:
 2, 50, x, y, T, F, p, q
 - y, (2+x), (2+x) * y,
 p q, p && ! q, x < ( y – 4)
// 基本(數值;邏輯)表達式
// 複合(數值)表達式
// 複合(邏輯)表達式
 A compound expression is formed by using one
operator or connective (連接詞) to combine one or
more operands
 The number of operands an operator requires is called it
arity.
 Unary operators take 1 operand (e.g.,
− y );
 Binary operators take 2 operands (e.g., 3  4).
 Ternary: p? a:b In C and Java
 Propositional(Boolean; logical) operators operate
on logical (instead of numerical) operands.
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Some Popular Boolean Operators
Formal Name
Nickname
Arity
Symbol C/Java
Negation
NOT
Unary
¬ (~)
!
Conjunction
AND
Binary

&&
Disjunction
OR
Binary

||
Exclusive-OR
XOR
Binary

!=
Implication
IMPLIES
Binary

Biconditional
(equivalence)
IFF
Binary
↔
==
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Forming Compound propositions from simpler ones
using logical connectives
 Primitive (or atomic) propositions:


propositional variables : p, q ,r
propositional constants : T, F
 P, Q: two known propositions (operands)
 Then the followings are also propositions:
1. “P and Q”
( P /\ Q) : conjunction of P and Q
2. “It is not the case that p” (~P) : negation of P
3. “P or Q”
( P \/ Q) : disjunction of P and Q
4. “P implies Q” (P  Q) : implication
 P is called the antecedent(premise, hypotheses);
 Q is called the consequence(or conclusion)
 equivalent phrases:
– “P entails Q”, “Q if P”, “P only if Q”
5. “P iff Q”
(P ↔ Q) : equivalence of P and Q
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Truth conditions for compound propositions
 Problem: When will we say that a proposition, such
as (p /\ q -> r ), is true (or false) ?
1. P /\ Q is true iff both P and Q are true.
(i.e., if both P and Q are true, then, by definition, we say
(P/\Q) is true; on the other hand, if either P or Q is false,
then, by definition, we say (P/\Q) is false .)
2.
3.
4.
5.
P \/ Q is true iff either P or Q is true
~P is true iff P is not true (i.e., false)
P ↔ Q is true iff both are true or both are false.
PQ is true iff whenever P is true, Q is true
( i.e., it is not the case that P is true but Q is false )
 What about the truth condition for primitive
propositions?
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Truth conditions for primitive propositions
 Problem: when will you say that the sentence:
It_is_raining
is true ?
=> The proposition:” It_is_raining” is true iff the condition (or
fact) that the sentence is intended to state really
occurs(happens, exists) in the situation which the
proposition is intended to describe.
=>Example: Since it is not raining now(the current situation), the
statement It_is_raining is false (in the current situation). But if it
were raining now, then I would say that It_is_raining is true.
 Factors affecting the truth value of a proposition:
 the situation in which the proposition is used to describe.
 the meaning of the proposition.
 Note: Whether a primitive proposition is true or false is not the concern of
math/logic; we just assume it has been or could be assigned a truth value.
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Truth table
•The meaning(truth condition) of logical connectives
can also be represented by a truth table.
Note: We use 0, F, or false to represent false and
use 1, T, or true to represent ture.
A
B
~A
A/\B
A\/B
A->B
A<->B
0
0
1
0
0
1
1
0
1
1
0
1
1
0
1
0
0
0
1
0
0
1
1
0
1
1
1
1
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From natural language statements to formal logic expressions








Natural language statement
Today is Friday
I have a test today
It is not the case that today is
Friday
Today is not Friday
Today is Friday and
I have a test today
If today is Friday, then I have
a test today.
Today is Friday or I have a
test today.




symbolized logic expr.
p
q
~p
 ~p
 (P ∧ q)
 (pq)
 (p ∨ q)
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Other usual connectives
 Besides /\,\/, and <->, there are also some other
connectives which you will study in digital
system.
xor , nand, nor.
Rules: let I(x) denote the truth value of x.
 I(A  B)=def I(~(A<->B)) [ = 1 iff not I(A)=I(B) ]
 I(A nand B) =def I(~(A/\B)).
 I(A nor B) =def I(~( A \/B)).
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Nested Propositional Expressions
Use parentheses to group subexpressions:
“I just saw my old friend, and either he’s
grown or I’ve shrunk.” = f  (g  s)
 (f  g)  s would mean something different
 f  g  s would be ambiguous
By convention, “¬” takes precedence over
“” and “” takes precedence over “”.
 ¬s  f means (¬s)  f , not ¬ (s  f)
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The Implication Operator
The implication p  q states that p implies q.
I.e., If p is true, then q is true;
but if p is not true, then q could be either
true or false.
E.g., let p = “You study hard.”
q = “You will get a good grade.”
p  q = “If you study hard, then you will get
a good grade.” (else, it could go either way)
 in p  q , p is the antecedent and q is the
consequent of the implication.
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Implication Truth Table
p  q is false only when
p q pq
p is true but q is not true. F F
T
p  q does not say
F T T
that p causes q!
T F
F
p  q does not require
T T T The
that p or q are ever true!
only
False
E.g. “(1=0)  pigs can fly” is TRUE!
case!
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Examples of Implications
 “If this lecture ends now, then the sun will rise
tomorrow.” True or False?
 flase  true
==> true
 “If Tuesday is a day of the week, then I am a
penguin.” True or False?
 True  false ==> false
 “1+1=6 only if 2 = 7. ” True or False?
 False  false ==> true
 “3 + 4 = 7 implies 5 = 5” True or False?
 True  true ==> true
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English Phrases Meaning p  q
 “p implies q”
 “if p, then q”
 “if p, q”
 “when p, q”
 “whenever p, q”
 “q if p”
 “q when p”
 “q whenever p”
 “p only if q”
 “p is sufficient for q”
 “q is necessary for p”
 “q follows from p”
 “q is implied by p”
We will see some equivalent
logic expressions later.
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Converse, Inverse, Contrapositive
Some terminology, for an implication p  q:
Its converse is:
q  p.
Its inverse is:
¬p  ¬q.
Its contrapositive:
¬q  ¬ p.
Which of the above three has the same
meaning (logical equivalent; same truth
table) as p  q ?
 Ans: ¬q  ¬ p.
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How do we know for sure?
Proving the equivalence of p  q and its
contrapositive using truth tables:
p
F
F
T
T
q q
F T
T F
F T
T F
p
T
T
F
F
pq q p
T
T
T
T
F
F
T
T
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Boolean Operations Summary
We have seen 1 unary operator (out of
the 4 possible) and 5 binary operators
(out of the 16 possible). Their truth
tables are below.
p
F
F
T
T
q
F
T
F
T
p pq pq pq pq pq
T F
F
F
T
T
T F
T
T
T
F
F F
T
T
F
F
F T
T
F
T
T
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Some Alternative Notations
Name:
Propositional logic:
Boolean algebra:
C/C++/Java (wordwise):
C/C++/Java (bitwise):
not and or
  
p pq +
! && ||
~ & |
xor implies



!=
^
iff

==
Logic gates:
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Bits and Bit Operations
A bit is a binary (base 2) digit: 0 or 1.
Bits may be used to represent truth values.
By convention:
0 represents “false”; 1 represents “true”.
Boolean algebra is like ordinary algebra
except that variables stand for bits, +
means “or”, and multiplication means
“and”.
 See chapter 11 for more details.
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Bit Strings
A Bit string of length n is an ordered series
or sequence of n0 bits.
By convention, bit strings are written left to
right: e.g. the first bit of “1001101010” is 1.
When a bit string represents a base-2
number, by convention the first bit is the
most significant bit. Ex. 11012=8+4+1=13.
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Bitwise Operations
Topic #2 – Bits
Boolean operations can be extended to
operate on bit strings as well as single bits.
E.g.:
01 1011 0110
11 0001 1101
11 1011 … Bit-wise OR
01 0001 … Bit-wise AND
10 1010 … Bit-wise XOR
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End of §1.1
You have learned about:
 Propositions: What
they are.
 Propositional logic
operators’




Symbolic notations.
English equivalents.
Logical meaning.
Truth tables.
 Atomic vs. compound
propositions.
 Alternative notations.
 Bits and bit-strings.
 Next section: §1.2
 Propositional
equivalences.
 How to prove them.
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§1.2 Propositional Equivalence
Two syntactically (i.e., textually) different
compound propositions may be semantically
identical (i.e., have the same true value in all
possible cases). We call them equivalent.
Learn:
 Various equivalence rules or laws.
 How to prove equivalences using symbolic
derivations.
 Analogy:
 x * (5 + y) and xy + 5x are always equal in algebra. But
how about
 p /\ (q \/ r) and (p/\q) \/ (p /\ r) ?
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Tautology, contradiction and contingency
Some definitions: Let P be any proposition,
then
1. P is a tautology(contradiction) iff P is true
(false) no matter what the truth values of the
propositional variables inside A are given.
2. If P is neither a tautology nor a
contradiction, then P is called a contingency.
Contingencies
Tautologies
(valid)
Contradictions
(unsatisfiable)
The geography of propositions
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Examples:
p \/ ~p, p  p : a tautology
 Both are true no matter p is true or false
p /\ ~p : a contradiction
 Always false no matter is true or false
p, p  ~p, ~p  p : a contingency
 p is true ==> p  ~p false
 p is false ==> p  ~p true
 Since p  ~p may be true and false
(depending on the truth value of p),
 it is a contingency.
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Logical Equivalence
Topic #1.1 – Propositional Logic: Equivalences
 proposition A is logically equivalent to
proposition B, written A  B, IFF the compound
proposition AB is a tautology.
Note:1.  is a logical symbol (connective) whereas
 is a meta logical symbol.
2. The textbook uses  instead of .
 Theroem : A  B iff
iff A and B contain the same truth values in all
rows of their truth tables.
Pf: A  B iff A  B is a tautology
iff every row of the true table for A  B are true
iff the values in column of A and in column of B in
each row of the truth table are equal
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Example
 Show that A: p /\ ~ q and B: ~(p  q) are logically
equivalent.
 Pf: By definition, it suffices to show that (A <->B )
(i.e., (p/\~q) <-> ~(p  q) ) is a tautology.
In the true table for A, B and A<->B:
p
q
A:p/\~q
B: ~(pq) A <->B :
(p/\~q) <-> ~(p  q)
0
0
0
0
1
0
1
0
0
1
1
0
1
1
1
1
1
0
0
1
We find the column for A <-> B are all 1s, since/hence
the columns for A and B should be equal.
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Properties of Logical Equivalence
 Theorem1: Logical equivalence relation is reflexive,
symmetric and transitive.
I.e., for all propositions A,B and C,
1. A  A
2. A  B implies B A and
3. A  B and B  C imply A  C.
 Theorem2[substitution theorem;等式代換定理]:
 If A  B and C[X] is a proposition containing X as a
subproposition, then C[A] and C[B] are logically
equivalent, where C[A] is the resulting proposition of
C with X in C replaced by A.
 ex: (p∨q)  (q∨p), C[X] =def ~(p ∧ X)
 => ~(p∧ (p∨q) )~(p∧ (q∨p))
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Determine tautologies, contradictions and contingencies
 Problems: Given a proposition A, how can you determine
whether it is a tautology, a contradiction or a contingency?
=>Exhausted evaluation! (by using truth table)
e.g., A = (p->q) <-> (~p \/q) a tautology ?
1. there are two primitive props: p, q.
2. So there are 4 possible assignments of truth values to p
and q.
p=0,q=0: p=0,q=1:
p=1,q=0:
p=1,q=1:
3. Under each assignment, we can evaluate the truth value of
A (in bottom up way) according to the evaluation rules of
boolean operators.
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Determine tautologies
isTautology(A) :// is proposition A a tautology?
1. Construct the evaluation dag for A.
2. Evaluate the trueth value of A for
each case given in the rows of the true table.
3. Return true iff A is evaluaetd to true in all
cases.
Note: A dag(directed acyclic graph) is basically
a tree except that children can be shared by
more than one parents.
(pq) ~p\/q :1
Pq:1
~p:1
p:0
~p \/q:1
q:0
The evaluation dag for (p->q)<-> (~p\/q)
p
q
~p p->q
~p\/q
p->q <-> ~p\/q
0
0
1
1
1
0
1
1
1
T
1
0
0
1
1
1
0
1
0
1
1
1
The truth table for pq <-> ~p\/q
1
0
1
A tautology
since all cells in
this column are
filled by 1’s.
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Determine tautologies
 Problem with evaluation dags:
 Need to replicate a dag for each possible case.
 Instead of using evaluation dag, we can place the
true value of each operand in its operator
column (or variable column if it is a prop
variable) in each row of the truth table.
p
q
(p -> q ) <-> ( ~ p
\/
q
0
0
0
1 0
1
0
0
1
0
1
1
1 0
0
1
) :
1
1
1
1
1
0
1
1
1
1
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Determine other logical problems
1. logical equivalence: is
AB?
 1. Using truth table: A  B if A and B have the same truth
values in all rows of their truth table.
 2. Since A  B iff A<->B is a tautology, we can call
isTautology(A<->B) to determinine if A B.
2. contradiction: A is a contradiction (恆假) iff ~A is a
tautology (恆真)
3. Satisfiability: A is satisfiable (可為真) iff ~A (可為假) is
not a tautology.
4. contingence: A is a contingence iff neither A nor ~A is
a tautology.
5. logical consequence: B is a logical consequence of A
[denoted A |= B] iff AB is a tautology.
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Reduce other logical problems to the taugology problem .
abstarct class LogicalSolver {
abstarct boolean isTautology(Prop A); // this need to be implemented by subclass
public boolean isContradiction(Prop A) {
return isTautology( new Not(A) ) ; }
public boolean isContingency(Prop A) {
return (! isTautology(A)) && ! isContradiction(A) ; }
public boolean isEquivalent(Prop A, Prop B) {
• Quiz: Instead of using isTautology()
return isTautology( new Iff(A,B) ) ; }
as a primitive function, can you use
public boolean isLogicalConseqence(B, A) {
isSatisfiable() (or
isLogicalConsequence()) as a
return isTautology( new Imply(A,B) ) ; }
primitive and implement other
Public booleanj isSatisfiable(Prop A) {
problems by function calls to it?
return ! isContradiction(A)! } }
Class MySolver extends LogicalSolver {
boolean isTautology(Prop A) { // my implemetation } }
// want to know if (p /\ ~q) and ~(p -> q) is equivalent ? =>
New MySolver().isEquivalent( new Prop(“p /\ ~ q”), new Prop(“~(p -> q)”))
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The Foundations
Topic #1.1 – Propositional Logic: Equivalences
Proving Equivalence via Truth Tables
Ex. Prove that pq  (p  q).
p
F
F
T
T
q
F
T
F
T
pq p q p  q (p  q)
F
T T
T
F
T
T F
F
T
T
F T
F
T
T
F F
F
T
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Example
 Show that ~(p \/q) and ~p /\~q are logically
equivalent.
pf: The truth table for both propositions are given
below. Since the columns for ~(p\/q) and ~p /\ ~q
are equal, by definition, they are equivalent.
p
q
p\/q
~(p\/q)
~p
~q
~p /\ ~q
T
T
T
F
F
F
F
T
F
T
F
F
T
F
F
T
T
F
T
F
F
F
F
F
T
T
T
T
Transparency No. 1-43
The Foundations
Exercise
 Show that pq  ~p \/ q
pf: The truth table for both propositions are given
below:
p
q
~p
pq
~p \/ q
Since the columns for pq and ~p \/ q are equal, by
definition, they are equivalent.
Transparency No. 1-44
The Foundations
Exercise
 Show that p\/(q/\r) and (p/\q) \/ (p/\r) are logically
equivalent.
p
q
r
q/\r
p\/(q/\r)
p/\q
p/\r
(p/\q)\/)p/\r)
• These columns can be skipped if you are familiar with the evalatiuon rules of
logical connectives.
Transparency No. 1-45
The Foundations
Equivalence Laws
Topic #1.1 – Propositional Logic: Equivalences
In algebra, we make heavy use of the
equality reasoning method of replacing
equal by equal:
 Ex: Given the known equalities:

x + y = z and x + z = 0,

we have x + (x + y) = x + z = 0 .
Thanks to the substitution theorem, we can
do also equality reasoning for logical
expressions. Ex:
 since p \/ p  p, ~p /\ p  F
 ~p /\ ( p \/ p)  ~p \/ p  F
Transparency No. 1-46
The Foundations
Topic #1.1 – Propositional Logic: Equivalences
Equivalence Laws - Examples
Identity:
pT  p
pF  p
Domination:
pT  T
pF  F
Idempotent:
pp  p
pp  p
Double negation:
p  p
Commutative: pq  qp pq  qp
Associative:
(pq)r  p(qr)
(pq)r  p(qr)
Note. The textbook uses  instead of .
Transparency No. 1-47
The Foundations
More Equivalence Laws
 Distributive:
Topic #1.1 – Propositional Logic: Equivalences
p(qr)  (pq)(pr)
p(qr)  (pq)(pr)
 De Morgan’s:
(pq)  p  q
(pq)  p  q
 generalization:
 ~ (p /\ q /\ r …)  ~p \/ ~q \/ ~r \/ …
 Trivial tautology/contradiction:
p  p  T
p  p  F
Transparency No. 1-48
The Foundations
One Example:
Show that
~(p \/ (~p /\ q)) ~p /\ ~q
Proof:
1. By truth table (omitted)
2. ~(p \/ (~p /\ q)  deM
~p /\ ~(~p /\ q)  deM
~p /\ (~(~p) \/ ~q)  DbNeg [+substitution]
~p /\ (p \/ ~q)  Distr
(~p /\ p) \/ (~p /\ ~q)  TrivalContradition
F \/ (~p /\ ~q)  identity
(~p /\ ~q)
Transparency No. 1-49
The Foundations
Topic #1.1 – Propositional Logic: Equivalences
Defining Operators via Equivalences
Using equivalences, we can define operators
in terms of other operators.
Exclusive or: pq  (pq)~(pq)
pq  (pq)(q~p)
Implies:
pq  p  q
Biconditional: pq  (pq)  (qp)
pq  ~(pq)
Boolean Algebra uses   ~ as operations
Boolean Ring uses   T as operations
 ~p  p  T ; (p \/ q)  (pq  p  q)
 Note: pq above means (p /\ q)
Transparency No. 1-50
The Foundations
A longer Example
Topic #1.1 – Propositional Logic: Equivalences
 Check using a symbolic derivation whether
(p  q)  (p  r)  p  q  r.
(p  q)  (p  r) [definition of ]
(p  q)  (p  r) [Defn. of ]
(p  q)  ((p  r)  (p  r))
[DeMorgan’s Law + DoubleNeg]
 (p  q)  ((p  r)  (p  r)) [ commutes]
(q  p)  ((p  r)  (p  r))
Transparency No. 1-51
The Foundations
Example Continued...
Topic #1.1 – Propositional Logic: Equivalences
(q  p)  ((p  r)  (p  r)) [ associative]
 q  (p  ((p  r)  (p  r))) [distrib.  over ]
q  (((p  (p  r))  (p  (p  r))) [assoc.]
q  (((p  p)  r)  (p  (p  r))) [trival taut.]
q  ((T  r)  (p  (p  r))) [domination]
q  (T  (p  (p  r))) [identity]
 q  (p  (p  r))  cont.
Transparency No. 1-52
The Foundations
End of the Longer Example
Topic #1.1 – Propositional Logic: Equivalences
q  (p  (p  r))
[DeMorgan’s]
q  (p  (p  r))
[Assoc.]
 q  ((p  p)  r)
[Idempotent]
 q  (p  r)
[Assoc.]
 (q  p)  r
[Commut.]
 p  q  r
Q.E.D. (quod erat demonstrandum)
(Which was to be shown.)
Transparency No. 1-53
The Foundations
Topic #1 – Propositional Logic
Review: Propositional Logic (§§1.1-1.2)
Atomic propositions: p, q, r, …
Boolean operators: (~)     
Compound propositions: (p  q)  r
Equivalences: pq  (p  q)
Proving equivalences using:
 Truth tables.
 Symbolic derivations. p  q  r …
Transparency No. 1-54
The Foundations
Logical puzzles
Knight or knave:
Two kinds of inhabitants lived in an island
 Knights : always tell the truth
 Knaves : always lie.
Dialog of two people A and B lived in the
island:
 A : “You are a knight”,
 B: “ We are opposite types”
Question: What are the kinds of A and B ?
Transparency No. 1-55
The Foundations
Solution:
 Formalization :









Let p = “ A is a knight” ; ~p  “A is a knave”
Let q = “B is a knight” ; ~q  “B is a knave”
Observed facts:
1. since A says q, we have
p  q (1) /\ ~p  ~q (2)
2. since B says R where
R is (p /\ ~q) \/ ( ~p /\ q) ) (5) , we have
q  R (3) /\ ~q  ~R (4).
The problem then reduces to : what is the model of P= (1)/\(2)/\(3)/\(4) ?
 Analysis:





p true =(1)=> q true =(3)=> R true
=(5)=> R false (a contradiction!)
P false =(2)=>q false =(4)=> ~R => ~p /\ ~q. (consistent).
Conclusion: Both are knaves!!
(since P has only one model: [ p:false, q:false] ).
Transparency No. 1-56
The Foundations
Muddy Children puzzle ( reasoning about knowledge)
 Two children Son and Daughter getting dirty mud
on their foreheads.
 Father says to them: “At lest one of you has a muddy
forehead.” (1)
 F2S: Do you know whether you have a muddy
forehead?
 S : No,
 Then, F2D: Do you know whether you have a muddy
forehead?
 D: Yes.
 Problem: Why did S and D answer No and Yes,
respectively?
Transparency No. 1-57
The Foundations
Analysis
Let d = “D has a muddy forehead”
s = “S has a muddy forehead”
 Observed facts:
1. d \/ s. Hence know(S, d \/ s) and know(D, d \/ s)
2. Since both children can see each other’s forehead but
cannot see their own foreheads.
we have know(S, d), Hence S only knows
{d, d \/ s}, which cannot make him infer that s is true.
Hence S answered no when he was asked.
3. D’s inference when asked:
~d => Know(S, ~d) => Know(S, {~d, d \/ s }) => know(S, s)
=> S answered Yes.
Now since S did not answer Yes, ~d must be false (i.e., d
must be true), => D answered yes.
Transparency No. 1-58
The Foundations
1.3 Predicates (述詞) and quantifiers(量詞)
 Problems of Propositional logic:
 Cannot analyze the content of primitive
propositions.
 cannot group similar propositions
 cannot analyze quantifiers.
Ex: Consider the following situations:
1. Family = {f,m,c1,c2,c3 }
2. We are interested in who is whose brother.
3. Suppose in Family, ci is cj’s brother if i  j for i,j
=1..3.
Transparency No. 1-59
The Foundations
Inadequacy of propositional logic
Primitive propositions needed:




c1 is c2’s brother, c1 is f’s brother,
f is f’s brother, …
x is y’s brother, x,y  Family ={f, m, c1,c2,c3}
So we need 25 propositions.
New approach: Parameterized
proposition(Predicate).
 Use a (predicate) symbol B to symbolize the
statement:
 B(x,y)  x is y‘s brother.
 e.g., B(f,c1) : f is c1’s brother,

B(c2,c2) : c2 is c2’s brother,

...
Transparency No. 1-60
The Foundations
Benefits of Predicates
Advantage of using predicates:
==>
1. need only 6 (1 predicate and 5 names)
symbols.
2. can group related propositions together.
( B(x,y) and others, for instance,
O(x)  x_is Old. )
3.The most important : allow representation of
quantified statement like: c1_has_a_brother (or,
there is an object x which is c1’s brother, $ x B(x, c1))
to be inferred from
c2_is_c1’s brother ( B(c2,c1) )
Transparency No. 1-61
The Foundations
Predicates
In the expression:
B(x,y)




B : predicate symbol (or proposition function)
x (y) : first (2nd) argument
B(x,y) : [atomic] proposition(or formula)
#arguments in B(X,Y): the arity of the predicate
symbol B.
Notes:
1. Each predicate symbol has a fixed arity.
2. B(x,y,c1) and B(x) are meaningless.
because used #arguments different from B’s arity.
Transparency No. 1-62
The Foundations
Predicates (cont’d)
3. The truth value of a proposition p(x1,...,xn)
depends on both p and real values of x1,...,xn.
Ex1: Let p(x)  “x >3 “. then
 P(4)  “4 > 3” has truth value T,
 P(2)  “2 > 3” has truth value F.
Ex2: Let Q(x,y)  “x = y +3” then
 Q(1,2) = “1 = 2 + 3” is false.
 Q(3,0) = “3 = 0 + 3” is true.
 Mathematically, we can define (the meaning of) P as a
propositional function [P]
 [P]: Un (n is P’s arity)  {T,F}with the intention that
 “P(x1,...,xn)” holds (is true) iff [P]([x1],...,[xn]) = T.
Transparency No. 1-63
The Foundations
Set representations of truth-functions
 Ex:





[B]: Family2 -->{T,F} s.t. [B](x,y) = T iff x =[ci], y=[cj], i  j.
P(x)  “x >3 “ , So [P]: N --> {T,F } s.t. [P](x) = T iff x > 3.
Q(x,y)  “x = y +3” , so
[Q] : N2 --> {T,F} with
[Q](x,y) = T iff x = y +3.
Note: [P] partition N into two sets, one with members
So, instead of represent
mapped to T and one mapped to F.
[P](x) = T. P as a truth function, we
equivalently represent P
{1,2,3}
{4,5,...}
as the set {x | [P](x) = T
} = {4,5,6,...}
[P](x) = F
Similarly, we can represent Q as {(x,y) | x = y + 3 }
N
Transparency No. 1-64
The Foundations
constant, function, variables and terms
 In the statement:
“y > min(x, 3)” --- (s1)
 x,y are called variables,
 3 is a constant (symbol),
 min is a function symbol with arity 2
 “min(3,2)” behaves more like x, 3 than “x > y”.
 So if let P(x,y)  “x > y”, then
 s1 can be represented as
P(y, min(x,3))
 we call any expression that can be put on the
argument position of an atomic proposition a term
 Obviously, constants and variables are terms;
moreover,if f is a function of arity n, and t1,...tn are n
terms, then so is f(t1,...,tn).
Transparency No. 1-65
The Foundations
Quantifiers and universe of discourse
Consider the sentence: p(x) =def “x > 0”.
 1. as stated before, p(x) is true or false depending on
the value of x.
 2. p(1), p(2),... : true;
p(0) , p(-1),...: false
Universe of Discourse (U,論域):
 The set of objects (domain) from which all variables
are allowed to have values.
 Ex: consider the sentence:
s2: ”all numbers are greater than 0”.
(for-all x “x > 0” ) or (" x p(x) )
 U = N+={1,2,3,...} => s2 is true. U = Z={...,-1,0,1,...}=>
s2 is false.
Transparency No. 1-66
The Foundations
Universal and existential quantifications
 A: any statement (or called formula)
 e.g., A = p(x,y) /\ q(x,f(3)),...
x: any variable,
Then
1. ("x A) is a new formula called the universal
quantification of A,
2. ("x A) means:
“A(x) is true for all values of x in the universe of
discourse.”
3. ($x A) is a new formula called the existential
quantification of A,
4. ($x A) means:
“A(x) is true for some value of x in the universe of
discourse.”
Transparency No. 1-67
The Foundations
Truth conditions for quantifications
quantification
When is (Qx A)
ture ?
("x A)
A is true for every
x
($x A)
There is an x for
which A is true.
When is (Qx A) false ?
(or ~Qx A true )
There is an x for which A
is false.
($x ~A)
A is false for every x.
("x ~A)
Logical equivalences: (direct from the above definition
and the defintion of not(~).
1. ~("x A) ($x ~A)
2.~($x A) ("x ~A)
Transparency No. 1-68
The Foundations
The Universal Quantifier "
Topic #3 – Predicate Logic
Example:
 Let the u.d.(universe of discourse) of x be
{1,2,3,4}
 Let P(x) be the predicate “x2 > 10.”
 Then the universal quantification of P(x), "x P(x),
is the proposition:
All numbers x (in u.d.) satisfy the predicate x2 >
10
 which is ture if
 P(1), P(2),P(3), P(4) all hold, i.e., (P(1) /\ p(2) /\(3) /\ p(4) ).
 But since P(2) = “22 > 10 “ is false, the universal
quantification is thus false.
Transparency No. 1-69
The Foundations
The Existential Quantifier $
Topic #3 – Predicate Logic
 Example:
 Let the u.d. of x be all natural numbers N = {0,1,2,3,4,…}
 Let P(x) be the predicate “x2 > 10.”
 Then the existential quantification of P(x), $x P(x), is the
proposition:
Some number x (in u.d.) satisfy the predicate x2 > 10
 which is ture iff
 one of P(0),P(1), P(2),P(3), P(4), … holds, i.e., (P(0) \/ p(1) \/ (2) \/ … ).
 since P(4) = “42 > 10 “ is true, the universal quantification is thus
true.
Transparency No. 1-70
The Foundations
Other quantifiers
 Quantifiers are about the number of objects in a
universe satisfying a property (unaray pricate).
Ex:
 "x even(x) . // all numbers are even
 $x odd(x). // there is at least one number which is
odd.
 Other quantifies possible ?







There is exactly one number which is >3 and < 5.
 $!x (x > 3 /\ x < 5)
There are exactly (or at least) 5 people in the shop.
 $5 x inTheShop(x); $5 x inTheShop(x)
Note: $2 x p(x) is equivalent to
$x$y (x != y /\ p(x) /\ p(y) /\ "z (p(z)  (z = x \/ z = y)))
Hence $!,$k ,$k ,$<k etc are essentially not necessary.
Transparency No. 1-71
The Foundations
Translation of NL statements into logical ones
 A basic skill you should learn in this course is to translate
statements in natural language into formal logical ones.
 Translate the following sentences into logical expressions:
1. Everybody likes Wu
1' Wang is a kind teacher.
2. Every student dislikes some teacher.
3. John gives every student a book.
4. Nobody likes everybody.
5. There is somebody Wu don’t like.
6. There is somebody no one likes.
7. There is exactly one person everybody likes.
8 Wu likes more than one person
9. Everyone likes himself.
10. Nobody likes those who don't like themselves (沒人喜歡不自
愛的人 ).
Transparency No. 1-72
The Foundations
Solutions
1. Everybody likes Wu
1': Wang is a kind teacher.
 "x like(x, Wu)
teacher(Wang) /\ kind(Wang)
2. Every student dislikes some teacher.
 "x ( student(x) -> $y (teacher (y) /\ ~like(x, y))
3. John gives every student a book.

"s ( student(s) -> $b (book(b) /\ give(John, s, b))
4. Nobody likes everybody.
 ~$x"y like(x, y)
or "x $y ~love(x, y)
rules:
1. verb  unary, binary predicate depending on its kind.:
ex: like -> binary, sleep, run ->unary, give ->ternary
2. proper noun  constant
ex: Wu, John,…
3. common noun, adjective  unary predicate
ex: teacher, student, kind -> unary predicate
4.不定代名詞 ( everybody/anybody/all/each/every … / : forall "
somebody/some … : exists $
none/ no … / nobody : ~$ )
Transparency No. 1-73
The Foundations
5.There is somebody Wu don’t like.
6. There is somebody no one likes.
 $y~ $ x like(x, y)
7.there is exactly one person everybody likes.
 $y (
 $y (
("x like(x, y) /\
("x like(x, y) /\
~$z (~z = y /\ "w like(w, z) ) )
"z (("w like(w, z))  z = y ) )
or
8 John likes more than one person.
$xy ( ~(x = y) /\ like(john,x) /\ like(john, y).
9. Everyone likes himself.
"y like(y, y)
10. Nobody likes those who don't like themselves (沒人喜歡不自愛的人 ).
~$x $y ( ~like(y,y) /\ like(x,y) )
"x "y ( ~like(y,y)  ~like(x,y) )
10' No boy likes any girl who don't like herself.
~$x (boy(x) /\ $y ( (girl(y) /\ ~like(y,y) ) /\ like(x,y) ))
"x (boy(x) -> "y ( (girl(y) /\ ~like(y,y))  ~like(x,y) )).
11. Everybody likes his father. 11' Everybody likes his parents.
"x (like(x, father(x)).
"x "y parent(y,x) > (like(x, y).
Notes: father is a function symbol; parent is a binary predciate.
Transparency No. 1-74
The Foundations
More examples
1. If a person is female and is a parent, then this person is
someone’s mother.
2. There is a woman who has taken a flight on every airline in the
world.
Sol: 1.let F(x) “x is female”; P(x) “x is a parent”  “$z parent(x,z)”
M(x,y)  “x is the mother of y”. Then
we have: "xPerson, (F(x) /\ P(x)  "x M(x,y) ) or
"x (Person(x)  (F(x) /\ P(x)  $y M(x,y) ))
2. Let take(w,f) =“w has taken f”; flight(f, a) =“f is a flight on a”. Then
we have: $y:Woman"a:airline"f:flight (take(w, f) /\ flight(f,a)), or
simply $y"a $f:flight (take(w, f) /\ flight(f,a)),
if the domain of y,a and f are restricted to all women, airlines and
flights in the world, respectively.
Transparency No. 1-75
The Foundations
Summary of the syntax of predicate logic
 Suppose we are given the following sets of
symbols (Vocabulary) :





1. Object variables: x,y,z,…
2. Predicate symbols: p,q,r,…
3. function symbols : f, g, h,…
4. constant symbols: a,b,c,…
 Notes:
 1. (2,3,4) together is called a signature, which can be
designated by users according to their application while
(1) is usually predefined by the logic.
 2. Constant symbols may be viewed as function symbols
of zero-arity.
Transparency No. 1-76
The Foundations
The syntax of predicate logic (supplementary)
 A signature Sis a triple( F, P, ar) where F is a set of
function symbols and P is a set of predicate
symbols, and ar: P U F  N is the arity fuction.
(1) If f  F, then f is a function symbol of S of arity ar(f).
(2) If n = 0, then f is a constant.
(3) If p  P, => p is a predicate symbol of arity ar(p).
(4) If n = 0 => p is an atom or proposition.
(5) Fun(S) = F is the set of function symbols
(6) Pred(S) = P is the set of predicate symbols
Ex: Arith =({0, 1, suc + *} , { >, =}, ar )
String=({e, *},{ < }, ar) ...
Transparency No. 1-77
The Foundations
Terms (supplementary)
 Terms (object expressions) :
1. Every variable x or constant symbol b is a term
2. if f is a function symbol of arity k>0, and t1,…,tk are
known terms, then f(t1,…,tk) is a term.
 Ex:
 x: variable
 ---------------- 1
 x: term ; mother: functional symbol
 --------------------------------------- 2

mother(x) : term;
father : unary fun symbol
 -------------------------------------------------------------------------- --2

father(mother(x)) :term
Transparency No. 1-78
The Foundations
 Hierarchy of definitions:
 primitive symbols  terms  atomic formula 
(general) formula.
 Atomic Formuls:
 if p is a predicate symbol of arity k≥0 , and t1,…tk are
terms, then p(t1,…,tk) is an atomic formula.
 note: zero-arity predicate symbols are propositional
variables in propositional logic.
 Formula:
 1. Every atomic formula is a formula.
 2. If A and B are formulas, then

(A/\B), (A\/B), (A -> B), (A  B), (A<->B), and ~A are

formulas.
 3. if x is a variable and A is a formula, then

"X A and $X A are formula.
Transparency No. 1-79
The Foundations
Example
 Show that "x (boy(x) -> like(x, father(x)) )) is a formula.
pf:
….
…
….
------------------ ------------------x:term; boy:predicate x:term; father(x):term; like : predicate
--------------------------- AF ----------------------------------------------------AF
boy(x):atomic
like(x, father(x)) :atomic
---------- ---------- 1
------------------------------------------1
boy(x) :formula
like(x, father(x)) :formula
---------------------------------------------------------------------------2.3
(boy(x) -> like(x, father(x))) :formula
---------------------------------------------------------3.1
"x (boy(x) -> like(x, father(x)) ) :formula
Transparency No. 1-80
The Foundations
S-structures and variable assignments (supplementary)
 S : a signature.
 A (first order) S-structure M is a triple:
(D, {[f]}f ∈ Fun(S), {[p]}p ∈ Pred(S) ) where
 D is a nonempty set (of objects), called the domain of M
(,written DM).
 [ f ]: Dar(f) -> D is a function on D of arity n. [f] is called the
denotation(or meaning) of f in M (, or written fM).
 [p]  Dar(p) is a ar(p)-ary relations on D (a subset of Dar(p)). [p]
is called the denotation (or meaning) of p in M (,written pM).
 Variable M-assignment:
 Any mapping a: Var -> DM from variables to the domain of a
S-structure M is called a variable M-assignment (or simply
an assignment).
Transparency No. 1-81
The Foundations
Denotations of terms in structures (supplementary)
 syntactic objects
semantic objects (M, a)
----------------------------------------------------------------------function symbol(f)
function ([f] or fM)
constant(c)
domain object ([c] or cM)
predicate symbol(p)
relation ([p] or pM)
variable (X)
domain object (a(X))
Terms
?
Formula
?
Transparency No. 1-82
The Foundations
Denotation of terms in a S structure(supplementary)
 Given the pair (M,a), we can define a meaning
function (M,a) : T(S) -> DM inductively as
follows:
1. (M, a) (X) = a(X) for any variable X.
2. (M, a)(f(t1,..,tn)) = fM( (M, a)(t1),...,(M, a)(tn)),
where f is a function symbol of arity n, and t1,..,tn
are terms.
Note: (M,a)(c) = cM for constant symbol c (function
of arity zero)
Transparency No. 1-83
The Foundations
a: a variable assignment;
d: an object  D;
X: a variable,
=> a[X/d] is a new variable assignment
agreeing with a on all variables but X, for
which it is assigned d. Formally,
a[X/d] is defined as follows: for all var Y :
a[X/d] (Y) =def if Y = X then d else a(Y)
Transparency No. 1-84
The Foundations
Truth condition of formulas (supplementary)
The relation M, a |= A (A is true (satisfied)
under interpretation M and assignment a) is
defined inductively as follows:
 1. M, a |= p(t1,..,tn), iff pM((M, a)(t1),..,(M, a)(tn))
holds

i.e, ((M, a)(t1),..,(M, a)(tn))  pM.
 2. M, a |= ~A iff not M, a |= A
 3. M, a |= A \/ B iff M, a |= A or M, a |= B
 4. M, a |= A /\ B iff M, a |= A and M, a |= B
 5. M, a |= "X A iff M, a[x/d] |= A for all d  D.
 6. M, a |= $X A iff M, a[x/d] |= A for some d  DM.
Transparency No. 1-85
The Foundations
Logical notions (supplementary)
 A: a S-formula; T: a set of formulas
1. A is satisfiable iff there is S-structure M and an M-assignment
a s.t. M, a |= A.
2. A is valid iff for all S-structures M and for all M-assignment a,
M, a |= A.
3. A is valid in M iff for all M-assignments a, M, a |=A.
4. A is satisfiabe in M iff M, a |= A for some M-assignment a.
1' T is satisfiable iff there is a S-structure M and an Massignment a s.t. M, a |= A for all A in T.
2' T is valid in M iff for all M-assignments a, M, a |=A for all A in
T.
3' T is satisfiabe in M iff there is an assignment a s.t. M, a |= A
for all A in T.
Transparency No. 1-86
The Foundations
Semantical conclusion (supplementary)
 A [or T ] is valid iff for all S-structures M and for all Massignment a, M, a |= A [for all A ∈ T ].
 If A is valid in M,(denoted M |= A) we say M is a model of A.
 T: a (set of) formulas; A: a formula
T |= A (,A is a logical consequence of T) iff
Every model of T is a model of A.
 A and B are logically equivalent (denoted A B) iff A<-> B
is valid. (i.e., for any structure assignment pair (M, a), M, a
|= A iff M, a |= B.)
Transparency No. 1-87
The Foundations
syntactic objects and their meaning
syntactic object
constant
examples
a,b,c
semantical meaning
domain object
[a]
function symbol
f,g,h
domain function
[f] : Uarity  U
predicate symbol
p,q, r,…
domain relations
(or truth function)
[p] : Uarity  {t,f }
variable
x,y,z,…
? (determined by
assignment)
term
f(t1,…,tn)
domain object
[f] ([t1],…[tn])
atomic formula
(general) formula
f(t1,…,tn)
proposition (trut value) [p]
([t1],…[tn])
A,B,C
truth value [A] dertermined
by components.
Transparency No. 1-88
The Foundations
Some Terminology
Occurrences of variables
Free/Bound Variables
Transparency No. 1-89
The Foundations
Occurrences of variables in a formula
Let A = p(f(x),c) \/ q(x,y)
A contains 2 variables x, y, where x occurs
twice and y occurs once.
Transparency No. 1-90
The Foundations
Free and Bound Variables
Topic #3 – Predicate Logic
An expression like p(x) , where p is a
predicate symbol, is said to have a free
variable x (meaning, x is undefined).
A quantifier (either " or $) operates on an
expression having one or more free variables,
and binds one or more of those variables, to
produce an expression having one or more
bound variables.
 ex: "x p(x,y).
Transparency No. 1-91
The Foundations
Example of Bindings
Topic #3 – Predicate Logic
 p(x,y) has 2 free variables, x and y.
 "x p(x,y) has 1 free variable, and one bound
variable. [Which is which?]
 An formula with zero free variables is a sentence
(or proposition).
 An formula with one or more free variables is still
only a predicate: "x P(x,y)
 Note: we use expression and formula
interchangeably as logical expressions are
usually called formulas.
Transparency No. 1-92
The Foundations
Formalization of free variables
 A: a formula (or expression)
 fv(A) = the set of all free variables in A is defined as
follows:
 If A is atomic => fv(A) is the set of all variables in A
 fv(~A) = fv(A)
 fv(A/\B)=fv(A\/B)=fv(A->B)=fv(A<->B)= fv(A) U fv(B).
 fv("x A) = fv($xA) = fv(A) \ {x}.
 What is fv(A) where A is an atomic formula?
 can be defined by recursion also:
 1. var(t) = {t} if t is var
 2. var(t) = var(t1) U var(t2)…U var(tn) if t =

f(t1,t2,…tn) or p(t1,…tn).
 Finally let fv(A) = var(A) if A is an atomic formula.
Transparency No. 1-93
The Foundations
Formalization of bound variables
 bd(A): the set of bound variables in A is defined as follows;
 1. If A is atomic => bd(A) = {}
 2. bd(~A) = bd(A)
 3. bd(A/\B) = d(A\/B) = bd(AB)
=bd(A<->B) = bd(A) U bd(B).
 4. bd( ∀x A) =
bd(A) if x ∉ fv(A).
bd(A) U {x} if x ∈ fv(A).
 Example A = ($ x p(x,y)) \/ q(x)
=> fv(A) = {x,y}; bd(A) = {x}
x occurs twice in A, one occurs free and the other occurs
bound.
 Def: A is a sentence iff fv(A) = {}
 Note: A is a proposition if A is a sentence
Transparency No. 1-94
The Foundations
Conventions
Topic #3 – Predicate Logic
 Sometimes the universe of discourse is restricted within
the quantification, e.g.,
 "x>0 P(x) is shorthand for
“For all x that are greater than zero, P(x).”
="x (x>0  P(x))
 $x>0 P(x) is shorthand for
“There is an x greater than zero such that P(x).”
=$x (x>0  P(x))
 "x:N $yZ (z < x) : For all natural numbers x, there is
an integer y such that y < x.
 = "x (N(x) $y (Z(y) /\ (z < x)) )
 "x"y"z …A can be abbreviated as "xyz…A
 $x$y$z…A can be abbreviated as $xyz…A.
Transparency No. 1-95
The Foundations
More to Know About Bindings
Topic #3 – Predicate Logic
"x $x P(x)
 x is not a free variable in
$x P(x), therefore the "x binding isn’t used.
("x P(x))  Q(x)
 The variable x in Q(x) is outside of the scope
of the "x quantifier, and is therefore free. Not
a proposition!
("x P(x))  ($x Q(x))
 This is legal, because there are 2 different x’s!
Transparency No. 1-96
The Foundations
Quantifier Equivalence Laws
Topic #3 – Predicate Logic
Definitions of quantifiers: If u.d.=a,b,c,…
"x P(x)  P(a)  P(b)  P(c)  …
$x P(x)  P(a)  P(b)  P(c)  …
From those, we can prove the laws:
"x P(x)  $x P(x)
$x P(x)  "x P(x)
Which propositional equivalence laws
can be used to prove this?
 DeMorgan’s Law
Transparency No. 1-97
The Foundations
More Equivalence Laws
Topic #3 – Predicate Logic
"x "y P(x,y)  "y "x P(x,y)
$x $y P(x,y)  $y $x P(x,y)
"x (P(x)  Q(x))  ("x P(x))  ("x Q(x))
$x (P(x)  Q(x))  ($x P(x))  ($x Q(x))
Transparency No. 1-98
The Foundations
Some Number Theory Examples
Topic #3 – Predicate Logic
Let u.d. = the natural numbers 0, 1, 2, …
“A number x is defined to be even, in symbols,
E(x), iff it is equal to 2 times some other number.”
 Define E(x)  $y x = 2 * y
“A number is prime, P(x), iff it’s greater than 1
and it isn’t the product of two non-unity numbers.”
 Define P(x)  (x>1  $yz (x=y * z  y1  z1)))
Transparency No. 1-99
The Foundations
Goldbach’s Conjecture (unproven)
“Every even number greater than 2
is the sum of two primes.”
"x [x>2  E(x)] →
$y $z ( P(y)  P(z)  y+z = x ).
Transparency No. 1-100
The Foundations
Calculus Example
Topic #3 – Predicate Logic
One way of precisely defining the
calculus concept of a limit, using
quantifiers:
(lim f ( x)  L) 
xa
 "  0 : $  0 : "x :



 (| x  a |  )  (| f ( x)  L |  )
Transparency No. 1-101
The Foundations
End of §1.3-1.4, Predicate Logic
Topic #3 – Predicate Logic
From these sections you should have
learned:




Predicate logic notation & conventions
Conversions: predicate logic  clear English
Meaning of quantifiers, equivalences
Simple reasoning with quantifiers
Upcoming topics:
 Introduction to proofs and proof-writing.
Transparency No. 1-102
The Foundations
1.5 Rules of inference
The following 3 sections are about proofs.
Argument (論述):
 Valid arguments
 Argument form (pattern)
Rules of inference (推論法則)
 For propositional reasoning
 For quantifier reasoning
Transparency No. 1-103
The Foundations
Arguments
 An argument is a sequence of statements made to
convince people of the truth of the final
conclusion.
 (用於使人確信結論為真的一系列邏輯敘述).
Definition 1.1: An argument in logic is a sequence of
propositions:
A1, A2,…,An (n > 0)
 The final one An is called the conclusion of the argument,
 Some of {A1,…,An} are called the premises of the
statements.
 An argument is basic if all propositions but the final one
are premises.
 Note: Our definition of argument is more general than that
given in the textbook, which corresponds to our basic
argument.
Transparency No. 1-104
The Foundations
Example of an argument
1. Have a passwd  can log onto the network
(premise)
2. Can log onto the network  can change grade (premise)
3. Have a passwd (premise)
Therefore :
4. can log onto the network (lemma)
5. can change the grade
(conclusion)
Notes: 1. 1..4 is a basic argument;
2. 1..5 is a non-basic argument.
Definition 1.2: An argument is valid if the truth of all its
premises implies the conclusion is true.
Note: Basic argument A1, A2,…,An is valid iff A1, A2,…,An-1
|= An or (A1/\ A2/\…An-1) An is a tautology.
Ex: the above argument (1..4 or 1..5) if valid since the
conclusion is true whenever all premises are true.
Transparency No. 1-105
The Foundations
Valid arguments and truth of conclusions
 Note: An argument is valid does not mean that its
conclusion must be valid since it is possible that some of
the premises are false.
 Ex: The following argument is valid
1. have access to the grading system can Change Grade
2. have access to the grading system
Therefore
3. can change the grade.
But (in the actual condition) the conclusion (3) is false since
premise (2) is false.
Note: A valid argument can guarantee the truth of its
conclusion only when all its premises are true.
Transparency No. 1-106
The Foundations
Argument forms
 In the basic argument :
1. have access to the grading system can Change Grade
2. have access to the grading system
Therefore
3. can change the grade.
Let p represent “have access to the grading system “ and
q represent “can Change Grade”
Then the basic argument has the form (or pattern):
1.
pq
2.
p
3.
q
 To save space, we may write the above form as :
 pq, p // q
Transparency No. 1-107
The Foundations
Argument form
 Definition 1.3 :
 An argument form in logic is a basic argument of which
some propositions contain propositional variables.
 If A is an argument form and A’ is the result of replacing
all occurrences of each variable by a same proposition,
then A’ is said to be an instance of the argument form A .
 Ex: The argument form : p q, p // q has instances:





raining  wet, raining // wet,
studyHard  GoodGrade, studyHard // goodGrade
pq , p // q
p \/ q  ~r, p \/ q // ~r,
but not
raining  wet, raining // cloudy (why?)
Transparency No. 1-108
The Foundations
Argument forms and arguments
 Def: An argument form A1,A2,…,An-1 // An in logic is valid
if (A1/\A2…/\An  A) is a tautology.
 Theorem : If an argument form A = (A1, .., //An) is valid
then all of its instances are valid.
Pf: This is a simple result of the substitution theorem of
propositional logic.
Let A’ = (A’1, .., A’n) be any instance of A. Then
A is valid iff B= (A1/\ A2/\…An-1) An is a tautology
=> B’ = (A’1/\ A’2/\…A’n-1) A’n is a tautology (substitution theorem)
=> argument A’ is valid.
Transparency No. 1-109
The Foundations
Why argument forms
 Analogy: When we say the equality x + x = 2 * x is valid, it
implies :
 #Students + #students = 2 * #Students ; #sons + #sons = 2 #sons
 ( #students / #classe) + ( #students / #classe) = 2 * + ( #students /
#classe)
 ( x – z = 5) + (x – z + 5) = 2 * (x – z + 5), …
 Similarly, since pq, p // q is valid then so are
 raining  wet, raining // wet,
 studyHard  GoodGrade, styudyHard // goodGrade
 p \/ q  ~r, p \/ q // ~r
 In other words, the validity of an argument form is independent
of how actual propositions are substituted for the variables in
the argument form. The argument form thus can be applied in
all applicable cases.
Transparency No. 1-110
The Foundations
Rules of inferences
 How to show the validity of an argument form:
 Solution 1: Truth table
 If (A1, .., An) is an argument form with propositional variables
p1,…,pm, then establish a truth table with 2m rows for the
target formula (A1/\A2…/\An-1)  An and then check if it is a
tautology.
 Issue: 2m is too large a number of rows to check validity even
for a small m (e.g., 20). However, practical argument forms
has fewer than 4 variables and thus is easy to check using
truth table.
 Solution 2: deduction:
 Establish the validity of a (complicated) argument form from
simpler valid known ones.
Transparency No. 1-111
The Foundations
Application of rules of inference
A rule of inference is an argument form that
is chosen to build the validity of other
arguments.
 A rule of inference is valid if it is a valid
argument form.
 An invalid rule of inference is called a fallacy.
Ex:
 Modus ponens (MP): pq, p // q
 Conjunction
: p, q
// p /\ q
 Abduction:
pq, q // p
(fallacy)
Transparency No. 1-112
The Foundations
Application of rules of inferences
Known assumptions (premises):
 snow  go skiing
 snow
Then
1. Snow
2. Snow  go skiing
3. Skiing
Is a valid argument.
 1, 2 are premises
 3 is the result of applying rule MP to 1 and 2.
Transparency No. 1-113
The Foundations
Modus Ponens & Tollens
 p
pq
q
 q
pq
p
Rule of modus ponens
(a.k.a. law of detachment)
“the mode of
affirming”
Rule of modus tollens
“the mode of denying”
Transparency No. 1-114
The Foundations
Some Inference Rules

p
 pq
Rule of Addition
 pq
p
Rule of Simplification

Rule of Conjunction
p
q
 pq
Transparency No. 1-115
The Foundations
Syllogism Inference Rules
 pq
qr
pr
hypothetical
syllogism
p  q
p
q
disjunctive
syllogism
p \/ q, ~p \/ r // q \/ r
Resolution
Transparency No. 1-116
The Foundations
Build arguments using rules of inference
Suppose we have the following
assumptions (premises):
“It is not sunny and it is cold.”
“We will swim only if it is sunny.”
“If we do not swim, then we will dance.”
“If we dance, then we will be home early.”
Given these premises, try to give a valid
argument with the conclusin “We will be
home early” using inference rules.
Transparency No. 1-117
The Foundations
Example cont.
Abbreviations:
 sunny = “It is sunny”; cold = “It is cold”;
swim = “We will swim”; dance = “We will dance”;
early = “We will be home early”.
Then, the premises can be written as:
(1) sunny  cold
(2) swim  sunny
(3) swim  dance
(4) dance  early
Transparency No. 1-118
The Foundations
Valid argument Example cont.
Step
1. sunny  cold
2. sunny
3. swimsunny
4. swim
5. swimdance
6. dance
7. danceearly
8. early
Proved by
Premise #1.
Simplification of 1.
Premise #2.
Modus tollens on 2,3.
Premise #3.
Modus ponens on 4,5.
Premise #4.
Modus ponens on 6,7.
Transparency No. 1-119
The Foundations
Fallacies
A fallacy is an inference rule which is not
valid.
(I.e., It is possible that all premises are true
but the conclusion is false).
Ex:
 fallacy of affirming the premise:
 pq, q // p

// q true, pq true but p false is possible
 Fallacy of denying the conclusion:
 pq, ~p // ~q ;
Transparency No. 1-120
The Foundations
Inference Rules for Quantifiers
 "x P(x) universal instantiation (UI)
P(t) (substitute any term t for x in P(-))
 Ex: from “全班考一百分” infer “小瑛考一百分”.

from "x (x2 > 0) infer 32 > 0 and (y+2)2 > 0 .
 P(t)
existential generalization (EG)
$x P(x) (substitute a new var x for term t in P(-) )
 Ex: from “小瑛考一百分” infer “有同學考一百分”.
 from (y + 2 )2 > (y+2)

infer
$x x2 > (y+2) or $x x2 > x.
 Note: Term t is either a constant c or a variable x or
function expression f(t1,…,tn)
Transparency No. 1-121
The Foundations
More Inference Rules for Quantifiers
 universal generalization(UG)
 P(g)
Ex: From: [令g 為班上任一學生，] g考一百分
"x P(x)
infer : 全班都考一百分。
Note: g is a constant referring to any general element of u.d.
I.e., g cannot be restricted to any property except being an
element of u.d.)
 Existential instantiation (EI)
 $x P(x)
P(c)
(substitute a new (Skolem) constant c )
Ex: From: “有同學考一百分”
infer [令c是考一百分的同學, 因此] c 考一百分。
Note:1. c must be a new constant
2. c denotes an object restricted to property
P
Transparency No. 1-122
The Foundations
Example
D(x) = “x has taken Discrete math course”
C(X) = “ x has taken a cs course”
Premises:
 D(Chen), " x (D(x) C(x))
Argument




D(Chen)
" x (D(x) C(x))
D(Chen)  C(Chen)
C(Chen)
premise
premise
UI
MP
Transparency No. 1-123
The Foundations
Example
Premises:
 1. A student in this class has not read the book,
 2. Everyone in this class passed the exam
Expected conclusion:
 Someone passing the exam has not read the book.
Abbreviations:
 C(x) : x is in this class
 R(x) : x has read the book
 P(x) : x has passed the exam.
Transparency No. 1-124
The Foundations
 The arguments
1.
2.
3.
4.
5.
6.
7.
8.
9.
$x (C(x) /\ ~R(x) )
C(a) /\ ~R(a)
C(a)
"x (C(x) P(x) )
C(a) P(a)
P(a)
~R(a)
P(a) /\ ~R(a)
$x (P(X) /\ ~R(x)).
premise 1.
EI from 1
Simplification from 2
premise 2
UI 4
MP from 3,5
Simplification from 2
Conjunction 6,7
EG from 8
Transparency No. 1-125
The Foundations
Topic #3 – Predicate Logic
Example (skipped)
Definitions:
c :≡ Confucius ;
H(x) :≡ “x is human”;
M(x) :≡ “x is mortal”.
Premises:
H(c)
"x H(x)M(x)
Conclusion:
M(c)
Confucius is human.
All humans are mortal.
Confucius is mortal.
Transparency No. 1-126
The Foundations
The argument (skipped)
Topic #3 – Predicate Logic
Some valid conclusions you can draw:
H(c)
H(c)M(c)
premise
[Instantiate universal.] If Confucius is human
then he is mortal.
H(c)  M(c)
Socrates is inhuman or mortal.
H(c)  (H(c)  M(c))
Confucius is human, and also either inhuman or mortal.
(H(c)  H(c))  (H(c)  M(c))
[Apply distributive law.]
F  (H(c)  M(c))
[Trivial contradiction.]
H(c)  M(c)
[Use identity law.]
M(c)
Confucius is mortal.
Transparency No. 1-127
The Foundations
Example (universal genealization (UG )
 The squares of all odd numbers are odd.
 "x (Odd(x)  Odd(x2))
pf: Let g be any number. // note: cannot assume additional
property for g except for its being an integer; g is a universal
constants.
[ Assume g is odd, then,
g = 2k + 1 for some k,
g2 = 4kk + 4k + 1 = 2(2kk + 2k) + 1,
g2 is odd. ] // [ … ] is a subproof for 
=> Odd(g)  Odd(g2).
=> "x (Odd(x)  Odd(x2)).
Transparency No. 1-128
The Foundations
Example2
 "x$y x < y. // no largest number
pf: Let a be any number.
=>a < a + 1 ( number property; true of all numbers)
(but not a2 < a + 10 ; not a universal number prop )
=> $y a < y (EG)
=> "x$y x < y (UG)
 $x"y x  y //least number
pf: Let a be any natural number.
=> 0  a ( property of natural numbers)
=> "y 0  y (UG)
=> $x"y x  y (EG)
Transparency No. 1-129
The Foundations
A fallacy
Find an argument for $x P(x) |- "x P(x)
What’s wrong with the proof: ?
 1. $x P(x)
 2. P(c)
 3. "x P(x)
--- premise
--- EI, where c is a restricted object.
--- UG (x) why?
 Ans: 23 is wrong! c is not a general element! It
requires to satisfy property P and thus is not a
general element.
Transparency No. 1-130
The Foundations
Proofs and proof methods
Proof terminology
Formal proofs
Formal proof examples
proof methods
Transparency No. 1-131
The Foundations
Proof Terminology
Theorem [定理]
 A statement that can be proven to be true.
Axioms[公設], postulates, hypotheses,
premises
 Assumptions (often unproven) defining the
structures about which we are reasoning.
Rules of inference [推論法則]
 Patterns of logically valid deductions from
hypotheses to conclusions.
Transparency No. 1-132
The Foundations
More Proof Terminology
Lemma[引理] - A minor theorem used as a
stepping-stone to proving a major theorem.
Corollary - A minor theorem proved as an
easy consequence of a major theorem.
Conjecture - A statement whose truth value
has not been proven. (A conjecture may
be widely believed to be true, regardless.)
Theory – The set of all theorems that can
be proven from a given set of axioms.
Transparency No. 1-133
The Foundations
Graphical Visualization
inference rule
A Particular Theory
A proof
The Axioms
of the Theory
…
Various Theorems
Transparency No. 1-134
The Foundations
Formal Proofs
 A formal proof of a conclusion C, given premises
p1, p2,…,pn is a sequence of statements :
 A1,…,An=C,
each Ai is either a premise or the result of applying
some inference rule to previously-proven
statements.
I.e., for each k = 1 .. n, either
1. Ak is some premise Pj, or
2. there is an instance (D1,…,Dm, E) of a rule with
{D1,…,Dm}  {A1,…,Ak-1} and Ak = E.
Transparency No. 1-135
The Foundations
More about proofs
Notes:
 1. We use B |-Th A (or B1,…,Bm |-Th A)
 to mean the existence of a proof of A with
B1,…,Bm as premises. (Th can be omitted if there
is no worry of confusion)
 2. If {} |-Th A (or simply |-Th A) , we say A is a
theorem and each Ai (i <n) is called a lemma.
 3. If B can be inferred from A directly, it is called a
corollary of theorem A; both lemmas and
corollaries are theorems.
Transparency No. 1-136
The Foundations
Soundness of a proof system
Soundness theorem: If
1. there is a proof of C from a set of premises P= {P1,… Pn }
(I.e., P1,…Pn |- C ), and
2. all inference rules used are sound,
then the proof is valid, i.e.,
conclusion C must be a consequence of {P1,… Pn}.
(i.e., P1,P2,…Pn |= C )
 This theorem holds not only for propositional logic and
predicate logic but also for all logical systems.
Transparency No. 1-137
The Foundations
 Corollary : If |-Th C, then |= C. I.e., Every theorem
is valid).
pf: |-Th C means {} |-Th C.
Hence {} |= C.
But all interpretations are model of {},
C thus is valid.
Transparency No. 1-138
The Foundations
Example
Topic #3 – Predicate Logic
Definitions:
s :≡ Socrates (ancient Greek philosopher);
H(x) :≡ “x is human”;
M(x) :≡ “x is mortal”.
Premises:
H(s)
Socrates is human.
"x H(x)M(x)
All humans are mortal.
 Conclusion:
M(s)
Socrates is mortal.
Transparency No. 1-139
The Foundations
The proof
Topic #3 – Predicate Logic
Some valid conclusions you can draw:
H(s)M(s)
[Instantiate universal.] If Socrates is human
then he is mortal.
H(s)  M(s)
Socrates is inhuman or mortal.
H(s)  (H(s)  M(s))
Socrates is human, and also either inhuman or mortal.
(H(s)  H(s))  (H(s)  M(s))
[Apply distributive law.]
F  (H(s)  M(s))
[Trivial contradiction.]
H(s)  M(s)
[Use identity law.]
M(s)
Socrates is mortal.
Transparency No. 1-140
The Foundations
Techniques for proving implication theorems : pq
 Different ways of proving a theorem: p implies q.
 Vacuous proof: Prove that ~p. used rule : [ ~p //p->q ]
 Ex: All negative natural numbers are 0!
 Trivial proof: Prove that q.
[q // p->q ]
 Direct proof: Prove that if p then q. [p |- q implies |- p->q]
suppose p,
 ...,
conclude q
--------------------- (I)
hence pq
 Indirect proof: (proof by contraposition)
 Prove that "~q implies ~P" [~q->~p // p->q]
 Other forms: [ ~q  p // ~p  q ];
[ q  ~p // p  ~q ]
Transparency No. 1-141
The Foundations
Proof by contradiction and proof by cases
 Proof by contradiction:
 To prove P, it suffices to show that

~P -> F (false)
 rules used: [~p ->F // p]
 Proof by cases:
 To prove that "p \/ q implies r " it suffices to show that

p->r and q -> r.
 rules used: [p->r, q->r // (p\/ q) ->r.]
 Ex: Prove that p\/q  q \/ p.
Transparency No. 1-142
The Foundations
Direct Proof Example
 Definition: An integer n is called odd iff n=2k+1 for
some integer k; n is even iff n=2k for some k.
 Axiom: Every integer is either odd or even.
 Theorem: (For all numbers n) If n is an odd integer,
then n2 is an odd integer.
Proof: Let n be any integer.
[Assume n is odd, then
n = 2k+1 for some integer k. Thus,
n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1.
Therefore n2 is of the form 2j + 1 (with j the integer
2k2 + 2k), thus
n2 is odd. ]
Transparency No. 1-143
The Foundations
Another example of direct proof
Show that |- (p(q r)) ((pq) (pr)).
Pf:
1. [Assume p (q r)
2. [assume pq
3.
[assume p
4.
q
MP+2,3
5.
q r
MP+1,3
6.
r]
MP+4,5
7.
pr ]
3~6 is a proof of pr
8. (pq )  (pr) ] 2~7 is its proof
9. (p(q r)) ((pq) (pr)). 1~8 is its proof.
Transparency No. 1-144
The Foundations
Indirect Proof Example
 Theorem: (For all integers n)
If 3n+2 is odd, then n is odd.
 Proof: Suppose that the conclusion is false, i.e.,
that n is even.
Then n=2k for some integer k.
Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1).
Thus 3n+2 is even, because it equals 2j for
integer j = 3k+1.
So 3n+2 is not odd.
We have shown that ¬(n is odd)→¬(3n+2 is odd),
thus its contra-positive (3n+2 is odd) → (n is odd)
is also true. □
Transparency No. 1-145
The Foundations
Vacuous Proof Example
 Theorem: (For all n) If n is both odd and even,
then n2 = n + n.
 Proof: The statement “n is both odd and even” is
necessarily false, since no number can be both
odd and even.
So, the theorem is vacuously true. □
lemma : n could not be both even and odd.
pf: Suppose n is both odd and even, then n = 2k = 2m
+ 1 for some integer m and k. Then 2(k-m) = 1.
Then k != m and hence 2 | 1. But this is impossible.
Transparency No. 1-146
The Foundations
Trivial Proof Example
 Theorem: (For integers n) If n is the sum of two
prime numbers, then either n is odd or n is even.
 Proof: Any integer n is either odd or even.
So the conclusion of the implication is true
regardless of the truth of the antecedent.
Thus the implication is true trivially. □
Transparency No. 1-147
The Foundations
Proving existence theorem
Methods for proving $x p(x):
Constructive proof: find an object (or term)
a, s.t. P(a).
 rules used : p(a) // $x p(x)
;EG
Nonconstructive proof: a proof of $x P(x)
w/o knowing what object satisfies p.
 Ex: proof by contradiction:

Show that ~$x p(x)  F. or

("x ~p(X))  F.
Transparency No. 1-148
The Foundations
Example of existence proofs
Ex 20: [constructive proof] Show that there are n consecutive
composite integers for every integer n >0. (I.e. for all n $x
(x+1,x+2,...x+n) are all composites.
Sol: Let x = (n+1)! +1.
=> x+i = (n+1)! + (i+1) = (i+1)( (n+1)!/(i+1) +1) is composite for
i = 1,..,n. QED.
Ex 21: [nonconstructive proof] For all n >0 $ prime number > n.
Sol: by contradiction. Assume $n s.t. all prime number < n.
Let m = n! +1. ==> gcd(k, m) = 1 for all k ≤ n.
=> all primes cannot divide m
=> m is a prime > n
=> a contradiction. QED.
Note: We cannot know a prime > n from the proof.
Transparency No. 1-149
The Foundations
Example of proof by cases...
 [No largest prime: ] Given n>0, prove there is a
prime p>n.
pf: Consider x = n!+1. Since x>1, we know
(x is prime)(x is composite).
Case 1: x is prime. Obviously x>n, so let p=x and
we’re done.
Case 2: x has a prime factor p. But if pn, then
x = (n!/p) x p + 1 => x  1 (mod p).
Then p does not divide x, a contradiction.
So p>n, and we’re done.
Transparency No. 1-150
The Foundations
Limits on Proofs
 Some very simple statements of number theory
haven’t been proved or disproved!
 E.g. Goldbach’s conjecture: Every integer n≥2 is
exactly the average of some two primes.
 "n≥2 $ primes p,q: n=(p+q)/2.
 For every formal proof system, there are true (or
false )statements of number theory which can
never be proved (or disproved) (Gödel
incompleteness theorem).
Transparency No. 1-151