Mathematical Logic Final Exam 14th June PROPOSITIONAL LOGIC

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1 Mathematical Logic Final Exam 14th June PROPOSITIONAL LOGIC Exercise 1. [3 marks] Derive the following formulas via Natural Deduction: (A B) (A B) Solution. See slides of propositional reasoning part. A 1 A 2 E B c A B I B 3 (1) (A B) A B I (A B) E A c (2) B c (3) I A B Exercise 2. [4 marks] Provide (a) the definition of maximally consistent set of formulas and (b) show that if Γ is maximally consistent and Γ φ, then φ Γ. Solution. A set of formulae Γ is maximally consistent if it is consistent and any other consistent set Σ Γ is equal to Γ. A proof that if Γ is maximally consistent and Γ φ, then φ Γ is as follows: Since Γ φ, then Γ {φ} is consistent. In fact, assume that Γ φ and that Γ {φ} is not consistent, then Γ {φ}, and therefore Γ φ. But from the fact that Γ φ and Γ φ we can build a deduction Γ, which would mean that Γ is not consistent. Since Γ is maximally consistent from the hypothesis, then we have reached an absurdum and the assumption that Γ {φ} is not consistent cannot be true. Thus Γ {φ} is consistent. From the definition of maximally consistent we know that Γ {φ} Γ, and therefore we can conclude that φ Γ. Exercise 3. [4 marks] Provide the description of the steps and the output of the DPLL algorithm on the PL formula: (C A) ((A B) A) Solution. The formula needs to be rewritten in CNF first: ( C A) (C A) A (A B) By assigning a truth value to unit clauses: v(a) = false and by propagating them, the formula is simplified to: C B If we assume a DPLL with pure literal step, we can already observe that the formula is satisfiable. In fact, all variables appear pure. Otherwise, we need to select a branching literal. We can choose B as branching literal and call DPLL recursively on:

2 Mathematical Logic Final Exam 14th June DPLL( C B B) DPLL( C B B) Let us first solve the first. By assigning a truth value to unit clauses: v(c) = false, v(b) = true, and by propagating them, the formula is simplified to T, therefore DPLL returns true. There is no need to call (b) and the DPLL returns true. FIRST ORDER LOGIC Exercise 4. [7 marks] Formalize the following statements, by using only the following first order predicates: F (x) M(x) MW (x, y) P A(x, y, z) x is female x is a male x is married with y x plays against y in match z 1. everybody must be either a male or (exclusively) a female 2. Men are married (only) with women and vice-versa 3. One can be married with at most one person 4. Being married is a symmetric and irreflexive relations 5. Matches can be between two players (singles) or between two teams of two players (doubles) (as in the sport of tennis). That is, in a match one can play against one or two against two. 6. married people play always together in the same team (that is, they don t play against each other and they cannot play against someone without their partner) 7. Married couples always play doubles against other married couples Solution. Possible formalizations are: 1. everybody must be either a male or (exclusively) a female x(m(x) F (x)) 2. Men can be married to women and vice-versa xy(mw (x, y) ((M(x) F (y)) (F (x) M(y))))

3 Mathematical Logic Final Exam 14th June One can be married with at most a person xyz(m(x, y) M(x, z) y = z) 4. Being married is symmetric and irreflexive xy(mw (x, y) MW (y, x)) x( MW (x, x)) 5. Games can be between two players (singles) or between two teams of two players (doubles) (as in the sport of tennis). That is, in a game competition one can play against one or two against two. This statement can be formalized by imposing that: One cannot play against more than 2 people in a match (i.e., there are only singles and doubles) xywtz(p A(x, y, z) P A(x, w, z) P A(x, t, z) y = w y = t w = t If one plays against two different people then he/she has a team partner (ie. we are in a double) xywz(p A(x, y, z) P A(x, w, z) y w t(p A(t, y, z) P A(t, w, z) t x)) 6. married people play always in team xyz(mw (x, y) P A(x, y, z) t(p A(x, t, z) P A(y, t, z))) 7. Married couples always plays doubles against other married couples xyzt(mw (x, y) P A(x, t, z) wmw (t, w) P A(x, t, z)) Exercise 5. [4 marks] Prove soundness of the I rule of Natural Deduction: φ(x) x.φ(x) I Solution. Assume that the last rule used is I. Then the derivation tree is of the form

4 Mathematical Logic Final Exam 14th June Γ Π φ(x) x.φ(x) I with x not free in Γ. Let I, a be such that I = Γ[a]. From the inductive hypothesis we know that I = φ(x)[a]. Since x does not appear free in Γ, then I = Γ[a[x/d]] holds for all d. Therefore from the inductive hypothesis I = φ(x)[a[x/d]] holds for all d. Then for the definition of =, we have that I = x.φ(x)[a]. DESCRIPTION LOGIC Exercise 6. [3 marks] Given the TBox T = {C R.D, D E F, F E} and the ABox A = {C(a)}. Provide the expansion of A w.r.t. T, without normalizing T. Solution. Let us first compute the expansion of T, that is S = {F E, D E, C R.E}. We then expand A w.r.t. S, which requires adding the necessary assertions to A such that R.E(a), thus obtaining B = {C(a), R(a,b), D(b), E(b)}. Exercise 7. [4 marks] Consider the following ABox { } likes(alice, Bob), is-neighbour-of(bob, Claudia) clever(claudia) A = likes(alice, Claudia), is-neighbour-of(claudia, Darren), clever(darren) 1. Is A satisfiable? Provide a rationale for the answer. 2. Is Alice an instance of likes.(clever is-neighbour-of. clever) with respect to A? Provide a rationale for the answer. Solution. For the above: 1. The answer is YES. An ABox is satisfiable if consistent (w.r.t. the TBox). As there is no TBox, it is enough to observe that in A there are no contradictory assertions. 2. It corresponds to the following instance checking problem: A = likes.(clever is-neighbour-of. clever)(alice) This corresponds to verifying the following:

5 Mathematical Logic Final Exam 14th June I(Alice) {x y : (x, y) I(likes), y I(clever) B} where B = {y y : (y, z) I(is-neighbour-of), z I(clever)}. In other words A should contain the following assertions: Likes(Alice, y), clever(y), is-neighbour-of(y, z), clever(z). This is true for y = Claudia and z = Darren. Exercise 8. [4 marks] Translate the following natural language sentences in Description Logic at the best of its expressiveness: 1. If Anna goes to the party then Bob does not go 2. A good apple is neither dirty nor rotten 3. A parent is a person having at least one child 4. Companies that do not have female employees are discriminatory Solution. Possible formalizations are as follows: 1. ANNA-GOES BOB-GOES 2. GOODAP P LE DIRT Y ROT T EN 3. P ARENT P ERSON HasChild. 4. DISCRIMINAT ORY COMP ANY COMP ANY HasEmployee.F EMALE

SAT and DPLL. Espen H. Lian. May 4, Ifi, UiO. Espen H. Lian (Ifi, UiO) SAT and DPLL May 4, / 59

SAT and DPLL. Espen H. Lian. May 4, Ifi, UiO. Espen H. Lian (Ifi, UiO) SAT and DPLL May 4, / 59 SAT and DPLL Espen H. Lian Ifi, UiO May 4, 2010 Espen H. Lian (Ifi, UiO) SAT and DPLL May 4, 2010 1 / 59 Normal forms Normal forms DPLL Complexity DPLL Implementation Bibliography Espen H. Lian (Ifi, UiO)