Logics for Data and Knowledge Representation Exercises: First Order Logics (FOL) Originally by Alessandro Agostini and Fausto Giunchiglia Modified by Fausto Giunchiglia, Rui Zhang and Vincenzo Maltese

Outline Introduction Syntax Semantics Reasoning Services Changed 2

Example of what we can express in FOL INTRODUCTION :: SYNTAX :: SEMANTICS :: REASONING SERVICES constants Cita Monkey predicates relations Eats Hunts Kimba Simba Lion Near 3

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))

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) ) 5

The use of FOL in mathematics 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)) ) 6

Express the following problem in FOL “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”. C B A Constants = {a, b, c} Predicates = {Black, White, Top} Relations = {Above} γ1 : ∃x Above(a,x) γ2 : Top(y)  Black(y)  ∃z Above(z,y) 7

Satisfiability of the problem of blocks 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. C B A 8

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]. 9

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 10

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 = +. 11

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) = c We 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. 12