Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
First-Order Logic (FOL)
Ontological commitments of FOL:
• The world consists of objects i.e., things with individual
identities. Objects have properties that distinguish them
from other objects.
• Objects participate in relations with other objects. Some of
these relations are functions. Relations hold or do not hold.
These ontological commitments make FOL more powerful than PL.
FOL is here to stay!
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Syntax
The symbols of FOL (with equality) are the following:
• Parentheses: (, ).
• The logical connectives ¬, ∧, ∨, ⇒ and ⇔.
• A countably infinite set of variables. This set will be denoted
by V ars.
Examples: x, y, v, . . .
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Syntax (cont’d)
• The quantifier symbols: ∀, ∃
• A countably infinite set of constant symbols.
Examples: John, M ary, 5, 6, Ball, . . .
• The equality symbol: =
• Predicate symbols: For each positive integer n, some set
(possibly empty) of symbols, called n-place predicate symbols.
Examples: Happy(.), Brother(., .), Arrives(., ., .), . . .
• Function symbols: For each positive integer n, some set
(possibly empty) of symbols, called n-place function symbols.
Examples: F atherOf (.), Cosine(.), . . .
Logicians usually introduce only the connectives ¬ and ∨ and one
of the quantifiers.
&
Teqnht NohmosÔnh
%
M. Koumparˆkhc
$
'
FOL: Syntax (cont’d)
Terms are expressions of FOL that refer to objects. The set of all
terms will be denoted by T erms.
The following BNF grammar gives the syntax of terms:
T erm → ConstantSymbol | V ariable
| F unctionSymbol(T erm, . . . , T erm)
Examples:
John, x, F atherOf (John), W if eOf (F atherOf (x)), . . .
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Syntax (cont’d)
Atomic formulas are expressions of FOL that refer to simple
facts.
The following BNF grammar gives the syntax of atomic formulas:
AtomicF ormula → T erm = T erm
| P redicateSymbol(T erm, . . . , T erm)
Examples:
John = ElderSonOf (F atherOf (John)), Happy(John),
Lives(John, London), Arrives(John, Athens, M onday)
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Syntax (cont’d)
Well-formed formulas (wffs) are the most complex kind of
expressions in FOL. They can be used to refer to any complicated state
of affairs.
The following BNF grammar gives the syntax of wffs:
W f f → AtomicF ormula | ( W f f ) | ¬ W f f
| W f f BinaryConnective W f f
| ( Quantif ier V ariable ) W f f
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Syntax (cont’d)
Examples of wffs:
• ¬Loves(T ony, M ary)
• Loves(T ony, P aula) ∨ Loves(T ony, F iona)
• Loves(John, P aula) ∧ Loves(John, F iona)
• (∀x)(SportsCar(x) ∧ HasDriven(M ike, x) ⇒ Likes(M ike, x))
• (∃x)(SportsCar(x) ∧ Owns(John, x))
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Free Variables
The following recursive definition defines the notion of free variables
of a wff.
• If φ is an atomic formula, x occurs free in φ iff x is a symbol of φ.
• x occurs free in ¬φ iff x is a symbol of φ.
• x occurs free in φ ∧ ψ iff x is a symbol of φ or ψ. Similarly for the
remaining binary connectives.
• x occurs free in (∀v)φ iff x is a symbol of φ and x is different than
v. Similarly for ∃.
The opposite of free is bound.
Definition.
sentence.
&
If no variable occurs free in the wff φ, then φ is a
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Free Variables (cont’d)
Examples:
• x is free in Brother(x, John) but not in
(∀x)(Cat(x) ⇒ M ammal(x)).
• y is free in
(∀x)(F riend(x, y) ⇒ Loves(x, y))
but not in
(∀x)(∀y)(F riend(x, y) ⇒ Loves(x, y)).
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Semantics
The meaning of FOL formulas is provided by interpretations.
An interpretation is a mapping between symbols of FOL and
objects, functions or relations in the world. More precisely:
• An interpretation maps each constant symbol to an object in
the world.
Example: In one particular interpretation the symbol John
might refer to John Major, the British PM. In another
interpretation it might refer to the evil King John, king of
England from 1199 to 1216.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Semantics (cont’d)
• An interpretation maps each predicate symbol to a relation in
the world.
Example: In one particular interpretation the symbol
Brother(., .) might refer to the relation of brotherhood. In a
world with three objects, King John, John Major, and Richard
the Lionheart, the relation of brotherhood is defined by the
following set of tuples:
{ hKing John, Richard the Lionhearti,
hRichard the Lionheart, King Johni }
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Semantics (cont’d)
• An interpretation always maps the equality symbol to the
identity relation in the world. The identity relation is:
id = {ho, oi : o is an object in the world}
• An interpretation maps each function symbol to a functional
relation (or function) in the world.
Example: In one particular interpretation the symbol
F atherOf (.) might refer to the relation of fatherhood.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
FOL: Formal Semantics
An interpretation I is a function which makes the following
assignments to the symbols of FOL:
1. I assigns to the quantifier symbol ∀ a non-empty set |I| called
the universe or domain of I.
2. I assigns to each constant symbol c a member cI of the
universe |I|.
3. I assigns to each n-place predicate symbol P an n-ary relation
n
P I ⊆ |I| ; i.e., P I is a set of n-tuples of members of the
universe.
4. I assigns to each n-place function symbol F an n-ary function
n
F I on |I|; i.e., F I : |I| → |I|.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction
Definition. A variable assignment is a function s : V ars → |I|
for some set of variables V ars and interpretation I.
Let φ be a wff of FOL, I an interpretation and s : V ars → |I| a
variable assignment.
We will define what it means for I to satisfy φ with variable
assignment s. This will be denoted by
|=I φ[s] or I |= φ[s].
Intuitively |=I φ[s] if and only if the state of affairs denoted by φ is
true according to I (where any variable x which occurs in φ, stands
for s(x) wherever it occurs free).
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
The formal definition of satisfaction proceeds as follows:
Terms. We define the function
s : T erms → |I|
from the set of all terms T erms into the universe |I|. This function
is an extension of s, and maps each FOL term to the object in the
universe denoted by this term:
• For each variable x, s(x) = s(x).
• For each constant symbol c, s(c) = cI .
• If t1 , . . . , tn are terms and F is an n-place function symbol, then
s(F (t1 , . . . , tn )) = F I (s(t1 ), . . . , s(tn )).
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
Atomic formulas. The definition of satisfaction for atomic
formulas is as follows:
• For atomic formulas involving the equality symbol,
|=I t1 = t2 [s] iff s(t1 ) is identical to s(t2 ).
• For an n-place predicate symbol P ,
|=I P (t1 , . . . , tn )[s] iff hs(t1 ), . . . , s(tn )i ∈ P I .
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
Other wffs.
• |=I ¬φ[s] iff 6|=I φ[s] (i.e., iff |=I φ[s] is not the case).
• |=I (φ ∧ ψ)[s] iff |=I φ[s] and |=I ψ[s].
• |=I (φ ∨ ψ)[s] iff |=I φ[s] or |=I ψ[s].
• |=I (φ ⇒ ψ)[s] iff 6|=I φ[s] or |=I ψ[s].
• |=I (φ ⇔ ψ)[s] iff |=I φ[s] and |=I ψ[s], or 6|=I φ[s] and 6|=I ψ[s].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
• |=I (∀x)φ [s] iff for all d ∈ |I|, we have |=I φ[s(x|d)].
The function s(x|d) is defined as follows:

 s(y) if y =
6 x
s(x|d)(y) =
 d
if y = x
• |=I (∃x)φ [s] iff there exists d ∈ |I| such that |=I φ[s(x|d)].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example: the WW in FOL
4
Breeze
Stench
Breeze
3
Stench
PIT
Breeze
PIT
Gold
2
Breeze
Stench
Breeze
1
Breeze
PIT
START
1
2
3
4
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
If we want to formalize the WW in FOL, we can use the following
symbols:
• Constant symbols:
Agent, W umpus, Gold, Breeze, Stench, Rm11, Rm12, . . . , Rm44
• Function symbols:
– The unary function symbol N orthOf to denote the unique
room which is north of the room denoted by the argument
of the function. For example, the room north of room 11 is
room 21.
– The unary function symbols SouthOf, W estOf, EastOf
with similar meanings.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
• Predicate symbols:
– The binary predicate Location will be used to denote the
location (i.e. room) of each object (agent, wumpus and
gold).
– The binary predicate P ercept will be used to denote the
percept (i.e., breeze or stench) in each room.
– The unary predicate Bottomless will be used to denote that
a room contains a pit.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
One could rightly object to having constant symbols R11 , R12 to
model the grid of rooms etc.
A better way to model the grid of rooms is to have a binary
function symbol Room(x, y) where x and y are the coordinates of
the room.
Try this as an exercise!
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
Let us now provide an interpretation I of the above symbols which
corresponds to the previous picture:
• The universe of I is the objects we see in the picture:
|I| = {agent, wumpus, gold, breeze, stench, rm11, . . . , rm44}.
• I makes the following assignments to constant symbols:
AgentI = agent, W umpusI = wumpus, GoldI = gold,
BreezeI = breeze, StenchI = stench,
Rm11I = rm11, . . . , Rm44I = rm44
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
$
'
Example (cont’d)
• I assigns to the unary function symbol N orthOf the function
N orthOf I : |I| → |I| which is defined as follows:
N orthOf I (rm11) = rm21,
N orthOf I (rm21) = rm22, . . . , N orthOf I (rm34) = rm44
• I assigns to the unary function symbols
SouthOf, W estOf, EastOf the function symbols
SouthOf I , W estOf I , EastOf I that are defined similarly with
N orthOf I .
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
• I assigns to the unary predicate symbol Bottomless the
following relation:
{hrm13i, hrm33i, hrm44i}
• I assigns to the binary predicate symbol Location the following
relation:
{hagent, rm11i, hwumpus, rm31i, hgold, rm32i}
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
• I assigns to the binary predicate symbol P ercept the following relation:
{hrm12, breezei, hrm14, breezei, hrm21, stenchi, hrm23, breezei,
hrm32, breezei, hrm32, stenchi, hrm34, breezei, hrm41, stenchi,
hrm43, breezei}
Note: To describe interpretation I we used words like agent, breeze, etc.
which start with a lowercase letter. These are not symbols of our FOL
language; they are just English words referring to what is in the picture.
Instead, we could have drawn little pictures to describe the elements of the
universe of the interpretation.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
We now give examples of satisfaction:
• |=I x = y[s] for any variable assignment s which maps x and y
to identical objects of the universe (e.g.,
s(x) = s(y) = wumpus). Why?
Because if s(x) = s(y) = wumpus then s(x) = s(x) = wumpus
is identical to s(y) = s(y) = wumpus.
• |=I Agent = Agent[s] for any variable assignments s.
This is trivial.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
• |=I Rm21 = N orthOf (Rm11)[s] for any variable assignment s.
Why?
Because s(Rm21) = Rm21I = rm21 and
s(N orthOf (Rm11)) = N orthOf I (s(Rm11)) =
= N orthOf I (Rm11I ) = N orthOf I (rm11) = rm21.
• |=I Rm21 = N orthOf (x)[s] for any variable assignment s such
that s(x) = rm11. Why?
Because s(Rm21) = Rm21I = rm21 and
s(N orthOf (x)) = N orthOf I (s(x)) =
N orthOf I (s(x)) = N orthOf I (rm11) = rm21.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Example (cont’d)
• |=I Bottomless(x)[s] for any variable assignment s such that
s(x) = rm13 or s(x) = rm33 or s(x) = rm44. Why?
Because if s(x) = rm13 then
hs(x)i = hs(x)i = hrm13i ∈ BottomlessI .
Similarly, for the other cases.
• |=I Location(Agent, Rm11)[s] for any variable assignments s. Why?
Because
hs(Agent), s(Rm11)i = hAgentI , Rm11I i = hagent, rm11i ∈ LocationI .
&
%
Teqnht NohmosÔnh
'
M. Koumparˆkhc$
Example (cont’d)
• |=I ¬Location(Gold, Rm44)[s] for any variable assignment s. Why?
Because
hs(Gold), s(Rm44)i = hGoldI , Rm44I i = hgold, rm44i 6∈ LocationI
therefore 6|=I Location(Gold, Rm44)[s] for any variable assignment s.
• |=I Location(Gold, Rm32) ∨ Location(Gold, Rm44)[s] for any
variable assignment s. Why?
Because
hs(Gold), s(Rm32)i = hGoldI , Rm32I i = hgold, rm32i ∈ LocationI
therefore |=I Location(Gold, Rm32)[s] for any variable assignment s.
&
%
Teqnht NohmosÔnh
'
M. Koumparˆkhc$
Example (cont’d)
• |=I (∃x)Location(x, Rm11)[s] for any variable assignment s. Why?
Because
hs(Agent), s(Rm11)i = hAgentI , Rm11I i = hagent, rm11i ∈ LocationI
thus |=I Location(x, Rm11)[s(x|agent)].
• 6|=I (∀x)Location(W umpus, x)[s] for any variable assignment s.
Why?
Because
hs(W umpus), s(Rm11)i = hW umpusI , Rm11I i =
hwumpus, rm11i 6∈ LocationI
thus 6|=I Location(W umpus, x)[s(x|rm11)].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
When we want to verify whether or not an interpretation satisfies a
wff φ with s, we do not really need all of the (infinite amount of)
information that s gives us. All that matters are the values of the
function s at the (finitely many) variables which occur free in s. In
particular, if φ is a sentence, then s does not matter at all. This is
made formal by the following theorem.
Theorem. Let s1 and s2 be variable assignments from V ars into
|I| which agree at all variables (if any) which occur free in the wff
φ. Then
|=I φ[s1 ] iff |=I φ[s2 ].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfaction (cont’d)
The previous theorem has the following corollary.
Corollary. Let φ be a sentence and I an interpretation. Then,
either
(a) I satisfies φ with every variable assignment s : V ars → |I|, or
(b) I does not satisfy φ with any variable assignment.
The above corollary allows us to ignore variable assignments
whenever we talk about satisfaction of sentences. Thus if φ is a
sentence and I an interpretation, we can just say that I satisfies
(or does not satisfy) φ.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Satisfiability
Definition. A formula φ is called satisfiable iff there exists an
interpretation I and variable assignment s such that |=I φ[s].
Otherwise, the formula is called unsatisfiable.
Examples:
The formulas
Location(W umpus, Rm31), Location(Agent, Rm11), (∃x)R(y, x)
are satisfiable. The following formulas are unsatisfiable:
P (x) ∧ ¬P (x), (∀x)P (x) ∧ ¬P (A)
Can you write an algorithm which discovers whether a given wff is
satisfiable?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Truth and Models
Definition. Let φ be a sentence and I an interpretation. If I
satisfies φ then we will say that φ is true in I or I is a model of φ.
Example: The interpretation I defined in the WW example is a
model of the following sentences:
Location(W umpus, Rm31), Location(Agent, Rm11),
(∃x)P ercept(Breeze, x)
Definition. An interpretation I is a model of a set of sentences
KB iff it is a model of every member of KB.
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Entailment
Definition. Let KB be a set of wffs, and φ a wff. Then KB entails
or logically implies φ, denoted by KB |= φ, iff for every
interpretation I and every variable assignment s : V ars → |I| such that
I satisfies every member of KB with s, I also satisfies φ with s.
Examples:
{ Happy(John), (∀x)(Happy(x) ⇒ Laughs(x)) } |= Laughs(John)
{ W ellP aid(John), ¬W ellP aid(John)∨Happy(John) } |= Happy(John)
Can you give an algorithm that discovers whether a set of wffs entail a
wff?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Validity and Equivalence
Definition. A wff φ is valid iff for every interpretation I and
every variable assignment s : V ars → |I|, I satisfies φ with s.
Examples:
The formulas
P (A) ∨ ¬P (A), P (A) ⇒ P (A), (∀x)P (x) ⇒ (∃x)P (x)
are valid.
Can you write an algorithm which discovers whether a given wff is
valid?
Definition. Two formulas φ and ψ will be called logically
equivalent, denoted by φ ≡ ψ, iff φ |= ψ and ψ |= φ.
&
Teqnht NohmosÔnh
'
%
M. Koumparˆkhc $
Satisfiability, Equivalence and Validity
In the following theorems, φ and ψ are arbitrary wffs.
Theorem. φ |= ψ iff φ ⇒ ψ is valid.
Proof.
“Only if”
Let I be an arbitrary interpretation and s an arbitrary variable assignment
s : V ars → |I|. We will prove that |=I (φ ⇒ ψ)[s].
We consider two cases depending on whether |=I φ[s] or not.
Let us first assume that |=I φ[s]. Since φ |= ψ, we have |=I ψ[s]. Thus,
from the definition of satisfaction for implication, we have |=I (φ ⇒ ψ)[s].
Let us now assume that 6|=I φ[s]. From the definition of satisfaction for
implication, we can immediately conclude |=I (φ ⇒ ψ)[s].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Proof (cont’d)
“If”
Let I be an arbitrary interpretation and s an arbitrary variable
assignment s : V ars → |I|.
Let us also assume that |=I φ[s]. We will prove that |=I ψ[s].
Since φ ⇒ ψ is valid, we have |=I (φ ⇒ ψ)[s]. Since we also know
that |=I φ[s], we must have |=I ψ[s].
Thus, from the definition of entailment, we have φ |= ψ.
&
Teqnht NohmosÔnh
'
%
M. Koumparˆkhc
$
Satisfiability, Equivalence and Validity (cont’d)
Theorem. φ is unsatisfiable iff ¬φ is valid. Proof?
Theorem. φ ≡ ψ iff φ ⇔ ψ is valid. Proof?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Some Important Logical Equivalences
Let φ and ψ be wffs. Then:
1. ¬(φ ∧ ψ) ≡ ¬φ ∨ ¬ψ
2. ¬(φ ∨ ψ) ≡ ¬φ ∧ ¬ψ
3. φ ∧ ψ ≡ ¬(¬φ ∨ ¬ψ)
4. φ ∨ ψ ≡ ¬(¬φ ∧ ¬ψ)
5. φ ⇒ ψ ≡ ¬φ ∨ ψ
6. φ ⇔ ψ ≡ (φ ⇒ ψ) ∧ (ψ ⇒ φ)
Proofs?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Proof of φ ⇒ ψ ≡ ¬φ ∨ ψ
Let us first prove that
φ ⇒ ψ |= ¬φ ∨ ψ.
Let I be an arbitrary interpretation and s an arbitrary variable
assignment s : V ars → |I|.
Let us assume that |=I (φ ⇒ ψ)[s]. We will prove that |=I (¬φ ∨ ψ)[s].
From the definition of satisfaction for implication, we have 6|=I φ[s] or
|=I ψ[s].
If 6|=I φ[s] then, from the definition of satisfaction for negation, we have
|=I ¬φ[s]. Thus, from the definition of satisfaction for disjunction, we
have |=I (¬φ ∨ ψ)[s].
If |=I ψ[s] then, from the definition of satisfaction for disjunction, we
also have |=I (¬φ ∨ ψ)[s].
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Proof (cont’d)
Let us now prove that
¬φ ∨ ψ |= φ ⇒ ψ.
Let I be an arbitrary interpretation and s an arbitrary variable
assignment s : V ars → |I|.
Let us assume that |=I (¬φ ∨ ψ)[s]. We will prove that |=I (φ ⇒ ψ)[s].
From the definition of satisfaction for disjunction, we have |=I ¬φ[s] or
|=I ψ[s]. We consider these two cases below.
If |=I ¬φ[s] then, from the definition of satisfaction for negation, we
have 6|=I φ[s]. Thus, from the definition of satisfaction for implication,
we have |=I (φ ⇒ ψ)[s].
If |=I ψ[s] then we can immediately conclude |=I (φ ⇒ ψ)[s] from the
definition of satisfaction for implication.
&
Teqnht NohmosÔnh
'
%
M. Koumparˆkhc
$
Some Important Logical Equivalences (cont’d)
1. (∀x)φ ≡ ¬(∃x)¬φ
2. (∃x)φ ≡ ¬(∀x)¬φ
3. (∀x)¬φ ≡ ¬(∃x)φ
4. (∃x)¬φ ≡ ¬(∀x)φ
Proofs?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
Some Important Logical Equivalences (cont’d)
1. (∃x)(φ ∨ ψ) ≡ (∃x)φ ∨ (∃x)ψ
2. (∃x)(φ ∧ ψ) |= (∃x)φ ∧ (∃x)ψ
3. (∀x)φ ∨ (∀x)ψ |= (∀x)(φ ∨ ψ)
4. (∀x)(φ ∧ ψ) ≡ (∀x)φ ∧ (∀x)ψ
Proofs?
&
%
Teqnht NohmosÔnh
M. Koumparˆkhc
'
$
One of the Proofs
Prove that
(∃x)(φ(x) ∧ ψ(x)) |= (∃x)φ(x) ∧ (∃x)ψ(x).
Proof: Let I be an interpretation and s a variable assignment such
that
|=I (∃x)(φ(x) ∧ ψ(x))[s].
Then according to the definition of satisfaction for existential
statements, there exists a d ∈ |I| such that
|=I (φ(x) ∧ ψ(x))[s(x|d)].
Then according to the definition of satisfaction for conjunctive
statements, we have
|=I φ(x)[s(x|d)]
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and
|=I ψ(x)[s(x|d)].
Now from the definition of satisfaction for existential statements
again, we have
|=I (∃x)φ(x)[s]
and
|=I (∃x)ψ(x)[s].
Now from the definition of satisfaction for conjunctive statements
we have:
|=I (∃x)φ(x) ∧ (∃x)ψ(x)[s].
The proof is now finished.
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Representing Knowledge Using FOL
Definition. In knowledge representation, a domain is a section
of the world about which we wish to express some knowledge.
• The domain of family relationships.
• The domain of sets.
• The wumpus domain.
• The domain of web resources (HTML pages, images, programs
etc. on the WWW)
• ...
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Knowledge Engineering
The process of knowledge-base construction is called knowledge
engineering.
A knowledge engineer is someone who investigates a particular
domain, determines what concepts are important in that domain,
and creates a formal representation of the objects and relations in
that domain.
If you want to see how FOL can be applied to represent knowledge
in various sophisticated application domains, see Chapter 10 of the
AIMA book (we will not cover it in the course).
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Other Logics in Computer Science
FOL is certainly the most important logic in use today by
computer scientists. But there are others too:
• Second-order logic
• Modal logic (with operators such as “possible” and “certain”)
• Temporal logic (with operators such as “in the past”, etc.)
• Logics of knowledge and belief
• Logics for databases
• ....
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Readings
Chapter 8 of AIMA: First-Order Logic
Other formal presentations of FOL can be found in:
1. Any mathematical logic textbook. Most of the formal material
in these notes is from:
H.B. Enderton, “A Mathematical Introduction to Logic”,
Academic Press, 1972.
See the Web page of the course for more details.
2. M.R. Genesereth and N.J. Nilsson, “Logical Foundations of
Artificial Intelligence”, Morgan Kaufmann, 1987.
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