Tutorial for First Order Predicate Logic I -Lecture6

8/13/2019 Tutorial for First Order Predicate Logic I -Lecture6

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Tutorial for Lecture #6 (First-order Predicate Logic I)

Let predicate fool(X,Y) means that X can fool Y, and suppose that the

universe of discourse is the set of all people in the world.

Express the following English sentences in First-order Predicate Logic

1) Everybody can fool Sompong.

2) Samak can fool everybody.

3) Everybody can fool somebody.

4) There is no one who can fool everybody.

5) Everyone can be fooled by somebody.

6) No one can fool both Taksin and Chuan.

7) Chaovalit can fool exactly two people.

8) There is exactly one person whom everybody can fool.

9) No one can fool himself or herself.

Solutions

1) ∀X fool(X, sompong).

2) ∀X fool(samak, X).

3) ∀X ∃Y fool(X,Y).

4) ¬∃X∀Y fool(X,Y).

5) ∀X ∃Y fool(Y, X).

6) ¬∃X ( fool(X, taksin) ∧ fool(X, chuan) )

7) ∃X,Y [ fool(chaovalit, X) ∧ fool(chaovalit, Y) ∧ X ≠ Y ∧

∀Z ( fool(chaovalit, Z) → ( Z = X ∨ Z = Y ) ) ]

8) ∃X (∀Y1 fool(Y1, X) ) ∧ ∀Z ∀Y2 ( fool(Y2, Z) → Z = X )

9) ¬∃X fool(X,X)



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FOL Logical Consequences

Examples

Suppose that

The set of constant symbols = { a, b, c },

The set of function symbols = { f, g }.

The set of variable symbols = { X, Y }.

∀ X p(X) |= p(t) for any term t. ( Universal Instantiation)

where p is a predicate.

Why?

Consider interpretation I = ( {0, 1, 2}, { P }, { F, G, 0, 1, 2}) where P, F, G

is assigned to p, f and g , respectively. Elements 0, 1 and 2 in the domain

are assigned to a, b and c, resp. Suppose that F(0) = 0, F(1) = 1 and

F(2) = 2, and similarly for G.

In the model theory,

∀ X p(X) means that for every element d ∈ {0, 1, 2}, P(d).

Thus, P(0), P(1), P(2), P(F(0)), P(F(1)), P(F(2)), P(G(0)), … hold in I .

So, ∀ X p(X) |= p(a)∧ p(b) ∧ p(c) ∧ p(f(a)) ∧ p(f(b)) ∧ p(f(c)) ∧ p(g(a)) ∧

…

Also, ∀ X p(X) |= p(X) ∧ p(Y)



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This is because any environment must map X and Y to some elements in

{0,1,2}.

• The following is not correct.

∀ X ∃ Y p(X,Y) |= ∃ Y p(Y,Y)

• Variable X is replaced by variable Y which already occurs under the

scope of a quantifier.

Definition

Given that A is a well-formed formula and t is a term. Term t is free for

variable X in A if no free occurrence of X falls within the scope of ∀ Y or

∃ Y where Y occurs in t .

Examplea) Y is not free for X in ∃ Y p(X,Y).

b) Z is free for X in ∃ Y p(X,Y).

• The following is the universal instantiation for general case.

∀ X P(X) |= P(t) for any term t that is free for X in P.

where P stands for any formula.

• The following is called the existential generalization.

p(c) |= ∃ X p(X) for any ground term c (Existential Generalization)

where p is a predicate.



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Example Valid formulae

a)∀ x ( study(x, english) ∧ study(x, programming) )

↔ ( ∀ x study(x, english) ) ∧ ( ∀ x study(x, programming) )

b) It is not true that

∀ x ( study(x, english) ∨ study(x, programming) )

→ ( ∀ x study(x, english) ) ∨ ( ∀ x study(x, programming) )

Consider the following: D = { john , marry }

study (john, english)

study (marry, programming)

Clearly, ∀ x ( study(x, english) ∨ study(x, programming) ) is true, but

( ∀ x study(x, english) ) ∨ ( ∀ x study(x, programming) ) is not true

c) It is not true that

( ∃ x study(x, english) ) ∧ ( ∃ x study(x, programming) )

→ ∃ x ( study(x, english) ∧ study(x, programming) )

Consider the following:

D = { john , marry }

study ( john, english)

study ( marry, programming)

Clearly, ( ∃ x study(x, english) ) ∧ ( ∃ x study(x, programming) ) is

true, but ∃ x ( study(x, english) ∧ study(x, programming) ) is not true

d) ∃ x ( study(x, english) ∨ study(x, programming) )

↔ ( ∃ x study(x, english) ) ∨ ( ∃ x study(x, programming) )



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e) ∃ x∀ y p(x,y) → ∀ y∃ x p(x,y)


∃ x∀ y study(x,y) → ∀ y∃ x study(x,y)

Subject = { english, programming, math }

Students = { john, marry, tom }

study (john, english)

study (john, programming)

study (john, math)

Cleary, ∃ x∀ y study(x,y) is true , so in∀ y∃ x study(x,y) is true

Example Satisfiability

a) ∃ x p(x)

b) ∀ x p(x)

c) ∀ x p(x) ∧ ¬∃ y p(y)

d) ∀ x,y ( p(x,y) → p(x,y) )

e) ¬∀ x∀ y ( ¬ p(x) ∨ p(y) )

Solution

a, b, d is satisfiable

c is not satisfiable

e is satisfiable because the formula is equivalent to

¬ ( ( ∀ x ¬ p(x) ) ∨ ( ∀ y p(y)) )


D = { a, b }

P(a) holds but P(b) do not where P is assigned to p

Logical Consequence ∀ x p(x) |= ∃ x p(x)

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Tutorial for First Order Predicate Logic I -Lecture6