Solutions to Exercises

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1 TDDC36 (LOGIC): EXAM Solutions to Exercises EXERCISE 1 1. Prove the following propositional formula: [ ( P Q) (P Q) R ] [ Q R ] (a) (2 points) using tableaux (b) (2 points) using Gentzen system (as provided in the book or during lectures - up to your choice). 2. Prove the following formula of predicate logic, where a is a constant: x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] (a) (3 points) using tableaux (b) (3 points) using resolution. SOLUTION TO 1(A) We first negate the formula: { [ ( P Q) (P Q) R ] [ Q R ] } ( P Q) (P Q) R [ Q R ] ( P Q) (P Q) R (Q R). Next, we construct a tableau for the negated formula: ( P Q) (P Q) R (Q R) ( P Q), (P Q) R (Q R) ( P Q), (P Q), R (Q R) ( P Q), (P Q), R, (Q R) (β-rule) ( P Q), (P Q), R, Q ( P Q), (P Q), R, R (β-rule) P, (P Q), R, Q P, P, R, Q (β-rule) P, Q, R, Q Q, (P Q), R, Q 1

2 2(9) SOLUTION TO 1(B) A possible proof: ( r) ( l) ( l) ( r) ( l) ( l) ( l) [ ( P Q) (P Q) R ] [ Q R ] ( P Q) (P Q) R Q R ( P Q), (P Q) R Q R ( P Q), (P Q), R Q R ( P Q), (P Q), R Q P, (P Q), R Q Q, (P Q), R Q P, Q, R Q axiom axiom P, P, R Q P, R P, Q axiom ( P Q), (P Q), R R axiom SOLUTION TO 2(A) We first negate the formula: { x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] } x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ]. Next, we construct a tableau for the negated formula: x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] x [ P (x, x, x) ], x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] x [ P (x, x, x) ], x y z [ P (x, y, z) P (z, x, a) ], z [ P (a, z, a) ] (γ-rule) P (a, a, a), x y z [ P (x, y, z) P (z, x, a) ], z [ P (a, z, a) ] (γ-rule) P (a, a, a), x y z [ P (x, y, z) P (z, x, a) ], P (a, a, a)

3 3(9) SOLUTION TO 2(B) We first negate the formula and obtain (see the solution for 2(a)): x [ P (x, x, x) ] x y z [ P (x, y, z) P (z, x, a) ] z [ P (a, z, a) ] so we have clauses: (1) P (x, x, x) (2) P (x, y, z) P (z, x, a) (3) P (a, z, a) A possible proof by resolution: (5) P (x, x, a) (res) applied to (1), (2) with unification x = y = z (6) FALSE (res) applied to (3), (5) with unification x = z = a.

4 4(9) EXERCISE 2 1. (4 points) Translate the following sentences into a set of propositional formulas: chose one of three roads: short, medium or long the short road is always crowded the medium road is not comfortable, but fast the long road is comfortable the chosen road should be comfortable. 2. (2 points) Assuming that fast roads are not crowded and crowded roads are not comfortable, hypothesize what roads can be chosen explain your reasoning informally. 3. (4 points) Prove your claim formally using a proof system of your choice (tableaux, Gentzen system or resolution. Please do not use truth table method, as this will give no points). SOLUTION TO 2.1 A possible translation: short medium long short crowded medium ( comfortable fast) long comf ortable comf ortable. (1) SOLUTION TO 2.2 The chosen road should be comfortable. Therefore it cannot be medium since medium roads are not comfortable. The chosen road cannot be short since short roads are crowded and crowded roads are not comfortable. Therefore the only road that can be chosen if the long one.

5 5(9) SOLUTION TO 2.3 To prove our hypothesis, we also need the translation of assumptions provided in point 2 of Exercise: fast crowded crowded comf ortable (2) We use resolution. We first transform (1) and (2) into a set of clauses: (1) short medium long (2) short crowded (3) medium comf ortable (4) medium fast (5) long comf ortable (6) comf ortable (7) fast crowded (8) crowded comf ortable. Now we negate formula and obtain: [(1) (2) (3) (4) (5) (6) (7) (8)] long [(1) (2) (3) (4) (5) (6) (7) (8)] long. Therefore, we consider clauses (1) (8) together with (9) long. A possible proof by resolution: (10) short medium (res) applied to (1), (9) (11) crowded medium (res) applied to (2), (10) (12) medium (res) applied to (3), (6) (13) crowded (res) applied to (11), (12) (14) comfortable (res) applied to (8), (13) (15) FALSE (res) applied to (6), (14).

6 6(9) EXERCISE 3 Consider a relation R and properties: (a) x y z[(r(x, y) R(x, z)) R(y, z)] (b) x y [ R(x, y) u[r(u, x) R(u, y)] ] (c) x y[r(x, y) R(y, x)]. 1. (4 points) Check informally whether the conjunction of (a) and (b) implies (c). 2. (6 points) Verify your informal reasoning using resolution. SOLUTION TO 3(1) To check whether (c) holds assume that for arbitrarily chosen elements s and t we have that R(s, t) holds. To prove (c), we have to show that given assumptions (a) and (b), R(t, s) holds, too. 1. First, using (b), we observe that R(s, t) implies that there exists u such that R(u, s) and R(u, t) hold. 2. Second, using (a) we know that for all x, y and z, we have that R(x, y) and R(x, z) imply R(y, z). Now, for y = t, z = s and x = u we have that R(u, t) and R(u, s) imply R(t, s). 3. From the first point we have that R(u, s) and R(u, t) hold. From the second point we have that these imply R(t, s). Therefore we conclude that R(t, s). That way we have shown that indeed the conjunction of (a) and (b) implies (c).

7 7(9) SOLUTION TO 3(2) To apply resolution we negate the implication ((a) (b)) (c): { x y z[(r(x, y) R(x, z)) R(y, z)] x y [ R(x, y) u[r(u, x) R(u, y)] ] x y[r(x, y) R(y, x)]} x y z[(r(x, y) R(x, z)) R(y, z)] x y [ R(x, y) u[r(u, x) R(u, y)] ] x y[r(x, y) R(y, x)] x y z[(r(x, y) R(x, z)) R(y, z)] x y [ R(x, y) u[r(u, x) R(u, y)] ] x y[r(x, y) R(y, x)]. We deal with the following conjuncts: x y z[(r(x, y) R(x, z)) R(y, z)] (3) x y [ R(x, y) u[r(u, x) R(u, y)] ] (4) x y[r(x, y) R(y, x)]. (5) We first Skolemize (4) and (5): x y [ R(x, y) ( R(f(x, y), x) R(f(x, y), y) )] (6) R(s, t) R(t, s), (7) where s, t are constants. Formulas (3), (6) and (7) are transformed into the following clauses: (1) R(x, y) R(x, z) R(y, z) (2) R(x, y) R(f(x, y), x) (3) R(x, y) R(f(x, y), y) (4) R(s, t) (5) R(t, s). A possible proof by resolution: (6) R(f(s, t), s) (res) applied to (2), (4) with x = s, y = t (7) R(f(s, t), t) (res) applied to (3), (4) with x = s, y = t (8) R(f(s, t), z) R(t, z) (res) applied to (1), (7) with x = f(s, t), y = t (9) R(t, s) (res) applied to (6), (8) with z = s (10) FALSE (res) applied to (5), (9) which completes the proof.

8 8(9) EXERCISE 4 1. (2 points) Design a Datalog database for storing information about scientific papers, containing, among others, title, authors and information about direct references between papers (i.e., for each paper, information about papers listed in its bibliography section). By p q we denote that paper q is among references listed in paper p. We define that paper q is indirectly referenced by paper p (and denote this by p q) if there is k 1 and papers p 1, p 2,..., p k such that p p 1 p 2... p k 1 p k q. 2. (1 point) Provide an exemplary integrity constraint concerning. 3. (1 point) Provide an exemplary integrity constraint concerning. 4. Formulate in logic queries selecting: (a) (2 points) all pairs P 1, P 2 of papers both (co-)authored by John Smith such that P 1 P 2 (b) (4 points) all pairs P 1, P 2 of papers both (co-)authored by John Smith such that P 1 P 2. SOLUTION TO 4.1 The database contain relations: paper(id, title), where id is a (unique) identifier for a paper, title String is its title author(id, name), where id is a paper identifier and name String is a name of its (co-)author refers to(id1, id2), where id1, id2 are paper identifiers. Example: paper(11, Logic in Databases ) paper entitled Logic in Databases has id = 11 paper(22, F ixpoint Logics ) paper entitled F ixpoint Logics has id = 22 author(11, John Smith ) John Smith is a (co-)author of paper 11 author(11, Mary Jones ) Mary Jones is a (co-)author of paper 11 refers to(11, 22) paper 11 refers to paper 22.

9 9(9) SOLUTION TO 4.2 The relation of road intersection is irreflexive: x[ ref ers to(x, x)]. SOLUTION TO 4.3 Relation is transitive: x y z[ ( x y y z ) x z]. SOLUTION TO 4.4(A) Select all pairs P 1, P 2 of papers both (co-)authored by John Smith such that P 1 P 2 (recall that is an anonymous don t care variable): answer(id1, Id2) : paper(id1, ), author(id1, John Smith ), paper(id2, ), author(id2, John Smith ), ref ers to(id1, Id2). Now answer(x, Y ) provides all required pairs of papers. SOLUTION TO 4.4(B) Select all pairs P 1, P 2 of papers both (co-)authored by John Smith such that P 1 P 2. We need to define the relation P 1 P 2 (we give it the name indirectly ref): indirectly ref(id1, Id2) : refers to(id1, Id), refers to(id, Id 2). indirectly ref(id1, Id2) : ref ers to(id1, Id), indirectly ref(id, Id2). Now we can formulate the required query: answer(id1, Id2) : paper(id1, ), author(id1, John Smith ), paper(id2, ), author(id2, John Smith ), indirectly ref(id1, Id2).