Transcript Logic, part I - Cornell University

Discrete Structures
CS 2800
Prof. Bart Selman
[email protected]
Module
Logic (part 1)
Rosen, Chapter 1
1
Logic in general
Logics are formal languages for formalizing reasoning, in particular for
representing information such that conclusions can be drawn
A logic involves:
– A language with a syntax for specifying what is a legal expression in the
language; syntax defines well formed sentences in the language
– Semantics for associating elements of the language with elements of some
subject matter. Semantics defines the "meaning" of sentences (link to the
world); i.e., semantics defines the truth of a sentence with respect to each
possible world
– Inference rules for manipulating sentences in the language
Original motivation: Early Greeks, settle arguments based on
purely rigorous (symbolic/syntactic) reasoning starting from a given
set of premises. “Reasoning based on form, content.”
Example of a formal language: Arithmetic
E.g., the language of arithmetic
– x+2 ≥ y is a sentence; x2+y > {} is not a
sentence
– x+2 ≥ y is true iff the number x+2 is no less
than the number y
– x+2 ≥ y is true in a world where x = 7, y = 1
– x+2 ≥ y is false in a world where x = 0, y = 6
3
Simple Robot Domain
Consider a robot that is able to lift a block, if that block is liftable (i.e., not
too heavy), and if the robot’s battery power is adequate. If both of these
conditions are satisfied, then when the robot tries to lift a block it is
holding, its arm moves.
Feature 1:
Feature 2:
Feature 3:
BatIsOk (True or False)
BlockLiftable (True or False)
RobotMoves (True or False)
block
5
Simple Robot Domain
We need a language to express
the features/properties/assertions and constraints
among them; also inference mechanisms, i.e,
principled ways of performing reasoning.
block
Example logical statement about the robot:
(BatIsOk and BlockLiftable) implies RobotMoves
6
Binary valued featured descriptions
Consider the following description:
– The router can send packets to the edge system only if it supports the new
address space. For the router to support the new address space it is necessary
that the latest software release be installed. The router can send packets to the
edge system if the latest software release is installed. The router does not
support the new address space.
– Features:
• Router
– Feature 1 – router can send packets to the edge of system
– Feature 2 – router supports the new address space
• Latest software release
– Feature 3 – latest software release is installed
7
Binary valued featured descriptions
– Constraints:
• The router can send packets to the edge system only if it supports the new
address space. (constraint between feature 1 and feature 2);
• It is necessary that the latest software release be installed for the router to
support the new address space . (constraint between feature 2 and feature 3);
• The router can send packets to the edge system if the latest software release
is installed. (constraint between feature 1 and feature 3);
How can we write these specifications in a formal language and reason
about the system?
8
Propositional Logic
9
Syntax:
Elements of the language
Primitive propositions --- statements like:
Bob loves Alice
Alice loves Bob
P
Q
propositional symbols
(atomic propositions)
Compound propositions
Bob loves Alice and Alice loves Bob
PQ
( - stands for and)
10
Connectives
¬




- not
- and
- or
- implies
- equivalent (if and only if)
11
Syntax
Syntax of Well Formed Formulas (wffs) or sentences
– Atomic sentences are wffs:
Propositional symbol (atom)
Examples: P, Q, R, BlockIsRed, SeasonIsWinter
– Complex or compound wffs.
Given w1 and w2 wffs:
 w1
(w1  w2)
(w1  w2)
(w1  w2)
(negation)
(conjunction)
(disjunction)
(implication; w1 is the antecedent;
w2 is the consequent)
(w1  w2) (biconditional)
12
Propositional logic:
Examples
Examples of wffs
PQ
(P  Q)  R
PQP
(P  Q)  (Q  P)
 P
This is not a wff!
P
Note1: atoms or negated atoms are called literals; examples p and p are
literals. P  Q is a “compound statement or proposition”.
Note2: parentheses are important to ensure that the syntax is unambiguous.
Quite often parentheses are omitted. The order of precedence in propositional
logic is (from highest to lowest):  ,, , , 
13
Propositional Logic:
Syntax vs. Semantics
Semantics has to do with “meaning”:
it associates the elements of a logical language with
the elements of a “domain of discourse” (“the world”).
Propositional Logic – we associate atoms with
propositions / assertions about the world (therefore
propositional logic).
14
Propositional Logic:
Semantics
Interpretation or Truth Assignment:
Assignment of truth values (True or False) to every proposition.
So if for n atomic propositions, there are 2n truth assignments or
interpretations.
This makes the representation powerful: the propositions
implicitly capture 2n possible states of the world.
15
Which ones are propositions?
–
–
–
–
–
–
Cornell University is in Ithaca, NY
Cornell University is in Palo Alto, CA
1+1=2
Note that a proposition can be False!
what time is it?
2 + 3 = 10
watch your step!
Aside:
We might associate the atom (just a symbol!)
BlockIsRed with the proposition: “The block is Red”, but we
could also associate it with the proposition “The table is
round” even though this would be quite confusing…
BlockIsRed has value True just in the case the block is red;
otherwise BlockIsRed is False. (Aside: computers
manipulate symbols. The string “BlockIsRed” does not
“mean” anything to the computer. Meaning has to come from
how to come from relations to other symbols and the
“external world”. Hmm.
How can a computer / robot obtain the meaning ``The
block is Red’’? The fact that computers only “push around
symbols” led to quite a bit of confusion in the early days or
Artificial Intelligence, Robotics, and natural language
understanding. Need: “Symbol Grounding”. How??
17
Propositional Logic:
Semantics
Truth table for connectives
Given the values of atoms under some interpretation, we can use a truth table to compute the
value for any wff under that same interpretation; the truth table establishes the semantics
(meaning) of the propositional connectives. (“Compositional semantics.”)
We can use the truth table to compute the value of any wff given the values of the constituent atom
in the wff. Note: In table, P and Q can be compound propositions themselves. Note: implication
not necessarily aligned with English usage.
18
Implication (p  q)
This is only False (violated) when q is False and p is True.
Related implications:
– Converse: q  p
– Contra-positive: q   p
– Inverse  p   q
Important: only the contra-positive of p  q is equivalent to p  q (i.e., has
the same truth values in all models); the converse and the inverse are
equivalent.
19
Implication (p  q)
Implication plays an important role in reasoning a variety of terminology
is used to refer to implication:
•
•
•
•
•
•
if p then q
if p, q
p is sufficient for q
q if p
q when p
a necessary condition for p is q (*)
•
•
•
•
•
•
p implies q
p only if q (*)
a sufficient condition for q is p
q whenever p
q is necessary for p (*)
q follows from p
Note: the mathematical concept of implication is independent of a cause and effect
relationship between the hypothesis (p) and the conclusion (q), that is normally
present when we use implication in English.
Note: Focus on the case, when is the statement False. I.e., p is True and q is False, should
be the only case that makes the statement false.
(*) Note: assuming the statement true, for p to be true, q has to be true
20
Notes: Bi-conditionals
(p  q)
Propositional Logic:
Semantics
Variety of terminology :
•
•
•
•
p is necessary and sufficient for q
if p then q, and conversely
p if and only if q
p iff q
p  q is equivalent to (pq)  (q p)
Note: the if and only if construction used in biconditionals is rarely used in
common language.
Example: “if you finish your meal, then you can play,” what is really meant is:
“If you finish your meal, then you can play” (i.e., (pq)) and
“You can play, only if you finish your meal” (i.e., (q p)).
21
Exclusive Or
Truth Table
P Q
PQ
_____________
T T
F
T F
T
F T
T
F F
F
P  Q is equivalent to (P ¬Q)  (¬PQ)
and also equivalent to ¬ (P  Q)
Use a truth table to check these equivalences.
22
Propositional Logic:
Satisfiability and Models
Satisfiability and Models
An interpretation or truth assignment satisfies a wff, if the wff is assigned the
value True, under that interpretation. An interpretation that satisfies a wff is
called a model of that wff.
Given an interpretation (i.e., the truth values for the n atoms) the one
can use the truth table to find the value of any wff.
23
The truth table method
(Propositional) logic has a “truth compositional semantics”:
Meaning is built up from the meaning of its primitive parts (just like English text).
24
Inconsistency (Unsatisfiability) and Validity
Inconsistent or Unsatisfiable set of WFFs
It is possible that no interpretation satisfies a set of wffs
In that case we say that the set of wffs is inconsistent or unsatisfiable or a
contradiction
Examples of inconsistent sets:
1 {P  P}
2 { P  Q, P Q, P  Q, P Q}
Why??
Validity (Tautology) of a set of WFFS
If a wff is True under all the interpretations of its constituents atoms, we say that
the wff is valid or it is a tautology.
Examples:
1 PP
2 (P  P)
3 [P  (Q  P)]
4 [(P  Q) P) P]
25
Logical equivalence
Two sentences p an q are logically equivalent ( or ) iff p  q is a tautology
(and therefore p and q have the same truth value for all truth assignments)






Note: logical equivalence (or iff) allows us to make statements about PL, pretty
much like we use = in in ordinary mathematics.
26
Example Formalization
Consider the following description:
– The router can send packets to the edge system only if it supports the new
address space. For the router to support the new address space it is necessary
that the latest software release be installed. The router can send packets to the
edge system if the latest software release is installed. The router does not
support the new address space.
– Features:
• Router
– P - router can send packets to the edge of system
– Q - router supports the new address space
• Latest software release
– R – latest software release is installed
28
Formal:
• The router can send packets to the edge of system only if it supports
the new address space. (constraint between feature 1 and feature 2)
– If (router can send packets to the edge of system) then
(router supports the new address space )
PQ
• For the router to support the new address space it is necessary that the
latest software release be installed. (constraint between feature 2 and feature 3);
– If (router supports the new address space ) then
(latest software release is installed)
QR
• The router can send packets to the edge system if the latest software release
is installed. (constraint between feature 1 and feature 3);
If (latest software release is installed) then
(router can send packets to the edge of system)
RP
• The router does not support the new address space.
¬Q
Rules of Inference
and
Proofs
30
Propositional logic:
Rules of Inference or Methods of Proof
How to produce additional wffs (sentences) from other ones? What steps can we
perform to show that a conclusion follows logically from a set of hypotheses?
Example: Modus Ponens
P
PQ
______________
Q
The hypotheses (premises) are written in a column and the conclusions below the bar
The symbol  denotes “therefore”. Given the hypotheses, the conclusion follows.
The basis for this rule of inference is the tautology (P  (P  Q))  Q)
[aside: check tautology with truth table to make sure]
Verify!
In words: when P and P  Q are True, then Q must be True also.
(meaning of second implication)
Aside: P  Q  P  Q (implication elimination)
31
Propositional logic:
Rules of Inference or Methods of Proof
Example
Modus Ponens
If you study the CS 2800 material, then you will pass
You study the CS2800 material
______________
 You will pass
Nothing “deep”, but again remember the formal reason is that
((P ^ (P  Q))  Q is a tautology. (I.e., one cannot imagine
a world where this statement is does not hold!)
32
See Table 1, p. 66, Rosen.
Propositional logic:
Rules of Inference or Method of Proof
Rule of Inference
Tautology (Deduction Theorem)
Name
P
PQ
P  (P  Q)
Addition
PQ
P
(P  Q)  P
Simplification
P
Q
PQ
[(P)  (Q)]  (P  Q)
Conjunction
P
PQ
Q
[(P)  (P Q)]  P
Modus Ponens
Q
PQ
 P
[(Q)  (P Q)]  P
Modus Tollens
PQ
QR
 P R
[(PQ)  (Q  R)]  (PR)
Hypothetical Syllogism
(“chaining”)
PQ
P
Q
[(P  Q)  (P)]  Q
Disjunctive syllogism
PQ
P  R
QR
[(P  Q)  (P  R)]  (Q  R)
Resolution
Subsumes MP
Valid Arguments
or Proofs
An argument is a sequence of propositions. The final proposition is called the
conclusion of the argument while the other proposition are called the
premises or hypotheses of the argument.
An argument is valid whenever the truth of all its premises implies the truth of
its conclusion.
How to show that q logically follows from the hypotheses (p1  p2  …pn)?
Show that
(p1  p2  …pn)  q is a tautology
One can use the rules of inference to show the validity of an argument.
Note that p1, p2, … q are generally compound propositions or wff.34
Proof Tree
Proofs can also be based on partial orders – we can represent them using a
tree structure:
– Each node in the proof tree is labeled by a wff, corresponding to a wff in
the original set of hypotheses or be inferable from its parents in the tree
using one of the rules of inference;
– The labeled tree is a proof of the label of the root node.
Example:
Given the set of wffs:
P, R, P  Q
Give a proof of Q  R
35
From premises: P, P Q, Q, R show: Q  R.
P
PQ
Tree Proof
R
MP
Q
Conj.
QR
What rules of inference
did we use?
36
Length of Proofs
Why bother with inference rules? We could always use a truth table
to check the validity of a conclusion from a set of premises.
But, resulting proof can be much shorter than truth table method.
Consider premises:
p_1, p_1  p_2, p_2  p_3, …, p_(n-1)  p_n
To prove conclusion: p_n
Inference rules: n-1 MP steps Truth table:
2n
Key open question: Is there always a short proof for any valid
conclusion? Probably not. The NP vs. co-NP question.
(The closely related: P vs. NP question carries a $1M prize.)
Beyond Propositional Logic:
Predicates and Quantifiers
38
Predicates
Propositional logic assumes the world contains facts that are true or false.
But let’s consider a statement containing a variable:
x > 3 since we don’t know the value of x we cannot say whether the
expression is true or false
x > 3 which corresponds to “x is greater than 3”
Predicate, i.e. a property of x
39
“x is greater than 3” can be represented as P(x), where P denotes
“greater than 3”
In general a statement involving n variables x1, x2, … xn can be
denoted by
P(x1, x2, … xn )
P is called a predicate or the propositional function P at the n-tuple
(x1, x2, … xn ).
40
When all the variables in a predicate are assigned values 
Proposition, with a certain truth value.
Predicate: On(x,y)
Propositions:
ON(A,B) is False (in figure)
ON(B,A) is True
Clear(B) is True
41
Variables and Quantification
How would we say that every block in the world has a property – say “clear”? We would have
to say:
Clear(A); Clear(B); … for all the blocks… (it may be long or worse we may have an
infinite number of blocks…)
What we need is:
Quantifiers
 Universal quantifier
x P(x)
- P(x) is true for all the values x in the universe of discourse
 Existential quantifier
x P(x)
- there exists an element x in the universe of discourse
such that P(x) is true
42
Universal quantification
Everyone at Cornell is smart:
x At(x,Cornell)  Smart(x)
Implicity equivalent to the conjunction of instantiations of P


…
At(Mary,Cornell)  Smart(Mary)
At(Richard,Cornell)  Smart(Richard)
At(John,Cornell) Smart(John)
43
A common mistake to avoid
Typically,  is the main connective with 
Common mistake: using  as the main connective with :
x At(x,Cornell)  Smart(x)
Means:
“Everyone is at Cornell and everyone is smart.”
44
Existential quantification
Someone at Cornell is smart:
x (At(x,Cornell)  Smart(x))
x P(x) “ There exists an element x in the universe of discourse such that
P(x) is true”
Equivalent to the disjunction of instantiations of P
(At(John,Cornell)  Smart(John))
 (At(Mary,Cornell)  Smart(Mary))
 (At(Richard,Cornell)  Smart(Richard))
 ...
45
Another common mistake to avoid
Typically,  is the main connective with 
Common mistake: using  as the main connective with :
x At(x,Cornell)  Smart(x)
when is this true?
is true if there is anyone who is not at Cornell!
46
Quantified formulas
If α is a wff and x is a variable symbol, then both x α and x α are
wffs.
x is the variable quantified over
α is said to be within the scope of the quantifier
if all the variables in α are quantified over in α, we say that we have a
closed wff or closed sentence.
Examples:
x [P(x)  R(x)]
x [P(x)(y [R(x,y)  S(x))]]
47
Properties of quantifiers
x y is the same as y x
x y is the same as y x
x y is not the same as y x
x y Loves(x,y)
– “Everyone in the world is loves at least one person”
y x Loves(x,y)
– “There is a person who is loved by everyone in the world”
Quantifier duality: each can be expressed using the other
x Likes(x,IceCream)
x Likes(x,IceCream)
x Likes(x,Broccoli)
x Likes(x,Broccoli)
48
Love Affairs
Loves(x,y) x loves y
Everybody loves Jerry
x Loves (x, Jerry)
Everybody loves somebody
x y Loves (x, y)
There is somebody whom somebody loves
y x Loves (x, y)
Nobody loves everybody
 x y Loves (x, y) ≡ x y Loves (x, y)
There is somebody whom Lydia doesn’t love
y Loves (Lydia, y)
Note: flipping quantifiers when ¬ moves in.
49
Love Affairs
continued…
There is somebody whom no one loves
y x Loves (x, y)
There is exactly one person whom everybody loves
(uniqueness)
y (x Loves(x,y)  z((w Loves (w ,z)  z=y))
There are exactly two people whom Lynn Loves
x y ((xy)  Loves(Lynn,x) Loves(Lynn,y) 
z( Loves (Lynn ,z) (z=x  z=y)))
Everybody loves himself or herself
x Loves(x,x)
There is someone who loves no one besides herself or himself
x y Loves(x,y) (x=y)
(note biconditional – why?)
50
Let Q(x,y) denote “xy =0”; consider the domain of discourse the real
numbers
What is the truth value of
a) y x Q(x,y)?
b) x y Q(x,y)?
False
True (additive inverse)
51
Statement
When True
When False
x y P(x,y)
y x P(x,y)
P(x,y) is true for
every pair
There is a pair for
which P(x.y) is
false
x y P(x,y)
For every x there is There is an x such
a y for which P(x,y) that P(x,y) is false
is true
for every y.
x y P(x,y)
There is an x such
that P(x,y) is true
for every y.
For every x there is
a y for which P(x,y)
is false
x  y P(x,y)
y  x P(x,y)
There is a pair x, y
for which P(x,y) is
true
P(x,y) is false for
every pair x,y.
Negation
Negation
Equivalent When is
Statement the
negation
True
When is
False
x P(x)
x P(x)
For every x,
P(x) is false
There is an x
for which P(x)
is true.
 x P(x)
x P(x)
There is an x For every x,
for which P(x) P(x) is true.
is false.
53
The kinship domain:
Brothers are siblings
x,y Brother(x,y)  Sibling(x,y)
One's mother is one's female parent
m,c Mother(c) = m  (Female(m)  Parent(m,c))
[uses function]
“Sibling” is symmetric
x,y Sibling(x,y)  Sibling(y,x)
54
Rules of Inference for Quantified
Statements
(x) P(x)
P(c)
Universal Instantiation
P(c) for an arbitrary c
(x) P(x)
Universal Generalization
(x) P(x)
Existential Instantiation
 P(c) for some element c
P(c) for some element c
 (x) P(x)
Existential Generalization
Example:
Let CS2800(x) denote: x is taking CS2800 class
Let CS(x) denote: x has taken a course in CS
Consider the premises x (CS2800(x)  CS(x))
CS2800(Ron)
We can conclude CS(Ron)
56
Arguments
Argument (formal):
Step
1 x (CS280(x)  CS(x))
2 CS280(Ron)  CS(Ron)
3 CS280(Ron)
4 CS(Ron)
Reason
premise
Universal Instantiation
Premise
Modus Ponens (2 and 3)
57
Example
Show that the premises:
1- A student in this class has not read the textbook;
2- Everyone in this class passed the first homework
Imply
Someone who has passed the first homework has not read the textbook
58
Example
Solution:
Let C(x) x is in this class;
T(x) x has read the textbook;
P(x) x passed the first homework
Premises:
x (Cx  T(x))
x (C(x)  P(x))
Conclusion: we want to show x (P(x)  T(x))
59
Step
1
2
3
4
5
6
7
8
9
x (Cx T(x))
C(a)  T(a)
C(a)
x (C(x)P(x))
C(a)  P(a)
P(a)
T(a)
P(a)  T(a)
x P(x) T(x)
Reason
Premise
Existential Instantiation from 1
Simplification 2
Premise
Universal Instantiation from 4
Modus ponens from 3 and 5
Simplification from 2
Conjunction from 6 and 7
Existential generalization from 8
Next: methods for proving theorems.