LLogics for DData and KKnowledgeRRepresentation

Exercises: First Order Logics (FOL)

Originally by Alessandro Agostini and Fausto GiunchigliaModified by Fausto Giunchiglia, Rui Zhang and Vincenzo Maltese

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Outline Introduction Syntax Semantics Reasoning Services

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Example of what we can express in FOL

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Kimba Simba

Cita

Hunts Eats

Monkey

LionNear

constantspredicatesrelations

INTRODUCTION :: SYNTAX :: SEMANTICS :: REASONING SERVICES

Write in FOL the following NL sentences “Fausto is a Professor”

Professor(fausto)

“There is a monkey”

∃x Monkey(x)

“There exists a dog which is black”

∃x (Dog(x) Black(x))

“All persons have a name”

∀x (Person(x) → ∃y Name(x, y))

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Write in FOL the following NL sentences “The sum of two odd numbers is even”

∀x ∀y ( Odd(x) Odd(y) → Even(Sum(x,y)) )

“A father is a male person having at least one child”

∀x ( Father(x) → Person(x) Male(x) ∃y hasChilden(x, y) )

“There is exactly one dog”

∃x Dog(x) ∀x ∀y ( Dog(x) Dog(y) → x = y )

“There are at least two dogs”

∃x ∃y ( Dog(x) Dog(y) (x = y) )

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The use of FOL in mathematics

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Express in FOL the fact that every natural number x multiplied by 1 returns x (identity):

∀x ( Natural(x) (Mult(x, 1) = x) )

Express in FOL the fact that the multiplication of two natural numbers is commutative:

∀x ∀y ( Natural(x) Natural(y) (Mult(x, y) = Mult(y, x)) )

Express the following problem in FOL

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“There are 3 blocks A, B, C in a stack. Each block can be either black or white. We know that the block A is on the bottom. A block is on the top if there are no blocks above it. It is possible to be on top if and only if the block is black”. A

BC

Constants = {a, b, c} Predicates = {Black, White, Top} Relations = {Above}

γ1 : ∃x Above(a,x)

γ2 : Top(y) Black(y) ∃z Above(z,y)

Constants = {a, b, c} Predicates = {Black, White, Top} Relations = {Above}

γ1 : ∃x Above(a,x)

γ2 : Top(y) Black(y) ∃z Above(z,y)

Is Γ = {γ1, γ2} satisfiable? Is there any structure M = <D,I>

and an assignment a such that M ⊨ γ1 [a] and M ⊨ γ2 [a]?

D = {A, B, C}

I(a) = A I(b) = B I(c) = C I(Black) = {A, C} I(White) = {B}

I(Top) = {C} I(Above) = {<B,A>, <C,B>}

Ia(y) = a(y) = C

NOTE: we can assign a different color to A and B. y is free.

Satisfiability of the problem of blocks

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ABC

Entailment Let be a set of FO- formulas, γ a FO- formula, we say

that

⊨ γ

(to be read entails γ)

iff for all the structures M and assignments a,

if M ⊨ [a] then M ⊨ γ [a].

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Entailment Given:

α : ∀x ∀y ∃z (p(x, y) → (p(x, z) p(z, y)))

β : ∀x ∀y (p(x, y)→ ∃z (p(x, z) p(z, y)))

Does α ⊨ β?

Yes, because in α we can put ∃z inside:

∀x ∀y (∃z p(x, y) → ∃z (p(x, z) p(z, y)))

z is not in the scope of ∃z p(x, y) and therefore we obtain β

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Entailment For all formulas of the form p(x, y):

1. Is ∃x∃y p(x, y) ⊨ ∃y∃x p(x, y)?2. Is ∃x∀y p(x, y) ⊨ ∀y∃x p(x, y)? 3. Is ∀x∃y p(x, y) ⊨ ∃y∀x p(x, y)? If no provide a counterexample.

Assume p(x, y) is x y. (1) Yes. The two formulas both says that there are at least two objects which are related via p.(2) Yes. The first formula says that there is an x such that for all y we have x y. We can take x = 0. The second formula says that for all y there exist an x such that x y. x = 0 is fine again.(3) In this case is No. The first formula says that for all x there exist a y such that x y. We can take y = Succ(x). The second formula says that there exists a y such that for all x we have x y. We should take y = +.

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Validity Is it possible to prove that:

⊨ ∀x (Person(x) Male(x)) [a] where D = {a, b, c}

We have only 3 possible assignments a(x) = a, a(x) = b, a(x) = cWe can translate it as follows:P: (Person(a) Male(a)) (Person(b) Male(b))

(Person(c) Male(c))

P: (Person(a) Male(a)) (Person(b) Male(b)) (Person(c) Male(c))

P: (Person-a Male-a) (Person-b Male-b) (Person-c Male-c)

We check that DPLL(P) returns false.

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