Computational logic First Order Logic_part2

SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS, CHENNAI-600 089
COMPUTATIONAL LOGIC
Dr.J.Faritha Banu
SRM IST- Ramapuram

Topics Covered in this Presentation are
Scope and Bound in First Order Logic
Substitutions
Semantics of FL
Interpretation and Valuation of FL
Satisfiability and Validity
Metatheorems

Scope and Bound
 Scope: In a formula (∀𝑥 𝛼) or (∃𝑥 𝛼), the scope of a quantifier is the formula 𝛼.
 Example 1 : In the formula ∃x(Pxy ∧∀y Qy)
Scope of ∃ is ∃x(Pxy ∧∀y Qy)
Scope of ∀ is the formula ∀y Qy
 Bound: An occurrence of a variable x in 𝛼 is a bound occurrence iff this occurrence is within the
scope of some quantifier of the same variable.
 more than one occurrence of quantifiers using that variable, then this occurrence of the variable
is said to be bound by the rightmost among all these occurrences of quantifiers.
 Free Variable: An occurrence of a variable in a formula is called a free occurrence or free
variable if it is not bound.
 A formula having no free variables is called a closed formula or a sentence. A formula that is
not closed is called an open formula.

Scope and Bound
 Example 2: In the formula ∀x(∀yPxyz ∧ ∀xQx → ∃zPzzz) find the scope of quantifier
Scope of first ∀ is the whole formula
Scope of second ∀ is the formula ∀yPxyz
Scope of third ∀ is the formula ∀xQx
Scope of ∃ is the formula ∃zPzzz
 Example 3: ∃x(Px ∧ Qx → ¬Px ∨ Qy)
Scope of ∃ is the formula ∃x(Px ∧ Qx → ¬Px ∨ Qy)
Free Variable : y
Example 4: ∃x ( (Px ∧ Qx) → (¬Px ∨ Qy) )
Scope of ∃ is the formula ∃x ( (Px ∧ Qx) → (¬Px ∨ Qy) )
All occurrences of x are bound by this ∃x , Free variable : y
Here, x is a bound variable and y is a free variable
The formula is not a sentence; it is an open formula.

Scope and Binding
Example 5: In the formula ∀x (∃y P(x,y) ∧ ∀y (¬P (x,y) ∨ P(x,y) ) )
Scope of ∃ is the formula ∃y P(x,y)
Scope of second occurrence of ∀ is the formula ∀y (¬P (x,y) ∨ P(x,y) )
Scope of first occurrence of ∀ is the formula (∃y P(x,y) ∧ ∀y (¬P (x,y) ∨ P(x,y) ) )
First two occurrences of y are bound by ∃y;
Third, fourth, and fifth occurrences of y are bound by ∀y.
All occurrences of x are bound by ∀x.
All occurrences of all variables are bound. Therefore ,The formula is a sentence.

Substitutions
 In substitution, We write [x/t] for the substitution of the free variable x by the term t,
For a formula X,
X[x/t] denotes the resulting formula is obtained by replacing each free
occurrence of 𝑥 in formula X with 𝑡.
substitution must NOT affect bound occurrences of the variable.
Only free occurrences of a variable can be substituted by a term.
Such substitutions are sometimes called admissible substitutions.

Substitutions
Example for substitution:
1. (Px→Qx)[x/t] = Pt →Qt ( both are free variable)
2. (∀xPxy)[x/t] = ∀xPxy (not replaced x by t because no free var x it is bound)
3. ∀x(Pxy[x/t]) = ∀xPty ( here after replacement only its bound )
4. ∀x∃y((Px∧Qyx)→Rzy[y/t]) = ∀x∃y(Px∧Qyx)→Rzt (here after replacement only its bound)
5. ∀x∃y((Px∧Qyx)→Rzy)[y/t] = ∀x∃y(Px∧Qyx)→Rzy (all y is bound) So no replacement

Substitutions
 When writing natural deduction proofs in predicate logic, it is often useful to replace a variable
in a formula with a term.
 Suppose that the following sentences are true:
(∀𝑥 (Fish(𝑥) → Swim(𝑥))) (1)
Fish(Nemo) (2)
To conclude that Nemo can swim, we need to replace every occurrence of the variable x in the
implication (Fish(𝑥) → Swim(𝑥)) by the term Nemo.
This gives us (Fish(Nemo) → Swim(Nemo)) (3)
By modus ponens on (2) and (3),
we conclude that Swim(Nemo).
Formally, we use substitution to refer to the process of replacing 𝑥 by Nemo in the formula (∀𝑥
(Fish(𝑥) → Swim(𝑥))).

Semantics of FL
 Interpretation of first order formulas starts with fixing a nonempty set as the domain or universe
of that interpretation.
 All constants are interpreted as certain elements in the domain.
 All function symbols are interpreted as partial functions on the domain.
 All predicates are interpreted as relations over the domain.
 An assignment l, which associates variables to elements of the universe or domain of an
interpretation is called a valuation (or a variable assignment function).
 (Do not confuse this with the boolean valuation of propositions)
 Ex: Soumya is younger than her father.
 While assigning x to ‘Soumya’, you had taken l(x) = Soumya, l(f (x)) = Soumya’s father.
φ( f ) = ‘father of’, l( f (x)) = φ( f (l(x)).

An interpretation is a pair I = (D, φ)
Domain/ Universe of I Mapping
Function symbols partial functions
Predicates relations over the
domain
φ : P ∪ F → R
Collection of all relations and functions
0-ary n-ary
φ(P) - objects φ(P) ⊆ Dn
0-ary
φ(f) : Dn→ D
n-ary
Constant, Name
Dn is n-ary relations/
functions in D
Semantics of FL

Interpretation of FL
 An interpretation is a pair I = (D,φ), where D is a nonempty set, called the domain or universe of I, and φ
is a function associating function symbols with partial functions on D and predicates with relations on D.
 Further, φ preserves arity, that is,
(a) If P is a 0-ary predicate (a propositional variable), then φ(P) is a sentence in D, which is either true or
false.
(b) If P is ≈, then φ(P) is the equality relation on D, expressing “same as”, that is, φ(P) = {(d,d) : d ∈ D}.
Ex:Father (rajiv) ≈ Father (Sanjay)
(c) If P is an n-ary predicate for n≥1, other than ≈, then φ(P) is an n-ary relation on D, a subset of Dn.
(d) If f is a 0-ary function symbol (a constant, a name), then φ( f ) is an object in D; that is, φ( f ) ∈ D.
(e) If f is an n-ary function symbol, n≥1, then φ( f ) :Dn→D is a partial function of n arguments on D.

Valuation under the interpretation I
 A valuation under the interpretation I = (D, φ) is a partial function l that assigns terms to elements
of D, which is first defined for variables and then extended to terms satisfying:
1. If f is a 0-ary function symbol, then l(f) = φ(f).
2. If f is an n-ary function symbol and t1, . . . , tn are terms, then l(f(t1, . . . , tn)) = φ(f)(l(t1),
. . . , l(tn)).
3. A valuation l′ is called equivalent to a valuation l along the variable x iff l(y) = l′(y) for all
variables y = x.
4. For a valuation l, a variable x, and an object c ∈ D, we write l[x ↦ c] as
l[x ↦ c](x) = c; and for y ≠ x, l [x ↦ c](y) = l(y).

STATE
 A state I l is a triple (D, φ, l ), where I = (D, φ) is an interpretation
and l is a valuation under I.
We read Il ⊨ X as
I l satisfies X or as Il verifies X or as Il is a state-model of X
Two states Il and Il′ are said to be equivalent along the variable x iﬀ
the valuations l and l′ are equivalent along the variable x.

Equivalent
 Two formulas A, B are equivalent, written as A ≡ B iﬀ every state-model of A is
also a state-model of B and vice versa.
 That is, A ≡ B iﬀ for every interpretation I and every valuation l under I, we have
either
 Il ⊨ A and I l ⊨ B, or
 Il ⊭ A and I l ⊭ B

Semantics Comparison of FL and PL
First order Language Propositional Language
• Satisfiable
:
• Model:
• Valid:
State-model, might/might not have a model
For every valuation l under I, the state Il is a
state-model of A.
For every I and for every l under that I, the
state-model Il ⊨ A
• Model
• I(A) = 1
• Each I is its model

EXAMPLE Consider the formulas Pxy and Pc f (c) along with the following
interpretations I, J and valuations l, m :
I = (N,φ), Where J = (H,ψ), where
N is the set of all natural numbers, H is the set of all human beings
φ(P) = ‘is less than’, ψ(P) = ‘is a brother of’,
φ(c) = 0, ψ(c) = Rajiv,
φ( f ) = ‘successor function’, ψ( f ) = ‘mother of’,
l(x) = 0, l(y) = 1. m(x) = Rajiv, m(y) = Sanjay.
As we know, Rajiv and Sanjay are the grand sons of Pandit Jawaharlal
Nehru. Determine whether (a) I l ⊨ Pxy
(b) I l ⊨ Pc f (c)
(c) Jm ⊨ Pxy (d) Jm ⊨ Pc f (c).

Observe that f is a unary function symbol, and it is associated with the functions ‘successor of’ and
‘mother of’, which take only one argument. Similarly, P is a binary predicate, and it is associated with
binary relations ‘less than’ and ‘is a brother of’.
(a) To check whether the state I l = (N,φ, l) is a state-model of Pxy,
we see that I l ⊨ Pxy
iff (l (x), l (y)) ∈ φ(P)
iff (0,1) ∈ ‘less than’
0 < 1 is true in Natural numbers. Therefore, I l ⊨ Pxy .
Suppose l(x) = 2 and l(y) = 4 Suppose l(x) = 6 and l(y) = 3
iff (l (x), l (y)) ∈ φ(P) iff (l (x), l (y)) ∈ φ(P)
iff (2,4) ∈ ‘less than’ iff (6,3) ∈ ‘less than’
2 < 4. is true in N. Therefore, I l ⊨ Pxy . 6 < 3. is not true in N. Therefore, I l ⊭ Pxy .

(b) I l ⊨ Pc f (c)
iff ( l(c), l(f (c)) ) ∈ φ(P)
( φ(c), φ( f (l(c)) )∈ φ(P)
iff (0, successor of 0) ∈ φ(P)
iff (0, 1) ∈ is less than
0 < 1 is true in N. Therefore I l ⊨ Pc f (c)
(c) Jm ⊨ Pxy
iff (m(x),m(y)) ∈ ψ(P)
iff (Rajiv, Sanjay) ∈ ‘is a brother of’
Rajiv is a brother of Sanjay. This latter sentence is true. Hence Jm ⊨ Pxy.

(d) Jm ⊨ Pc f(c)
iff (m(c),m( f (c))) ∈ψ(P)
iff (ψ(c),ψ( f )(ψ(c))) ∈ψ(P)
iff (Rajiv, mother of Rajiv) ∈ ‘is a brother of’
This results in the following
Rajiv is a brother of his own mother,
The above is not true. Thus, Jm ⊭ Pc f (c).

SATISFIABILITY AND VALIDITY
Let A be a formula. We say that A is satisfiable iff some state satisfies it; A is valid iff each state
satisfies it;
A is satisfiable iff some state satisfies I l ⊨ A.
Let Σ be a set of formulas, I an interpretation, and l a valuation under I. The state I l is a state-model
of (or satisfies, or verifies) the set Σ, written as I l ⊨ Σ, iff for each X ∈ Σ, I l ⊨ X.
The set Σ is called satisfiable iff Σ has a state-model, i.e., iff for some interpretation
I and for some valuation l under I, the state I l is a state-model of Σ.

Example A = ∀xPxf(x) satisﬁable?
I = (N, P′, f′), where P′ is the ‘less than’ relation and f′ is the function ‘plus 5’,
i.e., f′(n) = n+5. Let l be a valuation with l (x) = 3.
The state I l ⊨ ∀xPxf(x)
iﬀ for each n ∈ N, I l [x↦n] ⊨Pxf(x).
l [x↦n] = l [x↦3] maps x to 3
I l [x↦3] ⊨Pxf(x)
iff 3 < 3+5
Since 3 < 3+5 is true in N,
So, I l[x↦3] ⊨Pxf(x) Satisfiable
If l (x) = 10, also the formula satisfies, for that matter if we consider any other value ∈ N, the formula
satisfies. I interprets the formula as “Each natural number n is less than n + 5”

Example A = ∀xPxf(x) satisﬁable?
Since N is an infinite set, we have infinite models for the formula
In order to check if the formula is invalid we need to identify the a state that may falsify the
formula.
J = (N, P~, f′), where P~ is the ‘greater than’ relation. f′ is function ‘plus 5’
J l ⊨ Pxf(x) iﬀ for each n ∈ N, I l [x↦n] ⊨Pxf(x).
l [x↦3] ⊨Pxf(x) iff 3 > 3+5
As 3 ≯3+5
J l ⊭ Pxf(x) is Invalid
J interprets the formula as “Each natural number n is greater than n + 5

Metatheorems:
1. Relevance Lemma
Let A be a formula, VA be the set of all free variables of A, I = (D, φ) be an interpretation and l , l′
be valuations under I such that l(x) = l′(x) for every x ∈ VA. Then I l ⊨ A iff I l′ ⊨ A
2. Relevance Lemma for Sentences
Let A be a sentence, and let I be an interpretation.
(1) I ⊨ A iff I l ⊨ A for some valuation l under I. (2) Either I ⊨ A or I ⊨ ¬A.
An interpretation may neither satisfy an open formula nor satisfy its negation.
3. Monotonicity
Let Σ and Γ be sets of formulas and let X be a formula. Suppose that Σ ⊆ Γ.
If Σ ⊨ X, then Γ ⊨ X. If Σ is satisﬁable, then Γ is satisﬁable

Metatheorems:
4. Deduction Theorem
Let Σ be a set of formulas, and let X, Y be formulas. Then, Σ ⊨ X → Y iff Σ ∪ {X} ⊨ Y .
5. Reductio ad absurdum:
Let Σ be a set of formulas and let X be a formula.
Σ ⊨ X iff Σ ∪ {¬X} is unsatisfiable.
Σ ⊨ ¬X iff Σ ∪ {X} is unsatisfiable.

References
1. Arindama Singh," Logics for Computer Science", PHI Learning Private
Ltd,2nd Edition, 2018
2. Wasilewska & Anita, "Logics for computer science: classical and non-
classical", Springer, 2018
3. Huth M and Ryan M, Logic in Computer Science: Modeling and Reasoning
about systems‖, Cambridge University Press, 2005
4. Dana Richards & Henry Hamburger, "Logic And Language Models For
Computer Science", Third Edition, World Scientific Publishing Co. Pte.
Ltd,2018

Computational logic First Order Logic_part2

Computational logic First Order Logic_part2

In this document

More Related Content

What's hot

Similar to Computational logic First Order Logic_part2

Recently uploaded

Computational logic First Order Logic_part2