Logic and Formal Language: Understanding Validity and Satisfiability, Assignments of Computer Science

The concepts of validity and satisfiability in formal logic through examples and problem-solving. It covers topics such as truth tables, implication, and quantifiers, and provides exercises to test understanding.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

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Question1:
A.
1) false
2) false
B.
Leave to you
C.
1) 6 2)4 3)4
Question2:
A.
(a) A valid sentence is one that is true in all models. The sentence True
is also valid in all models. So if αis valid then T rue |=αholds (because
both True and αhold in every model), and if the entailment holds then α
must be valid, because it must be true in all models in which True holds.
(b) False doesn’t hold in any model, so αtrivially holds in every model
that False holds in.
(c) Assume α|=β. Then consider any model m. If αis true in m, then
by our assumption, βis also true in m, so αβis true in m. Otherwise,αis
false in m, so αβis true in m. Thus αβis valid. Conversely, suppose
αβis valid. Then consider any model mwhereαis true. βmust also be
true in m, or else mwould not satisfy αβ. So α|=β.
(d) This reduces to c, because α ¬βis unsatisable just when αβis
valid.
B.
1) VALID. By truth table: both F alse F alse and Tr ue T rue are
true.
2) SATISFIABLE. By truth table: it is true when Smoke is true and
F ire is true, but false when S moke is false and F ir e is true.
3) VALID. By truth table: if Smoke is false, both expressions are false;
if Smoke is true then either the (Smoke F ire) is false (in which case the
whole implication is true) or F ire is true, in which case both sides are true.
4) SATISFIABLE. By truth table: true when both Big and Dumb are
true, but false when Big is false and and Dumb is true.
1
pf3
pf4

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Question1: A.

  1. false
  2. false B. Leave to you C.
  3. 6 2)4 3)

Question2: A. (a) A valid sentence is one that is true in all models. The sentence True is also valid in all models. So if α is valid then T rue |= α holds (because both True and α hold in every model), and if the entailment holds then α must be valid, because it must be true in all models in which True holds. (b) False doesn’t hold in any model, so α trivially holds in every model that False holds in. (c) Assume α |= β. Then consider any model m. If α is true in m, then by our assumption, β is also true in m, so α ⇒ β is true in m. Otherwise,α is false in m, so α ⇒ β is true in m. Thus α ⇒ β is valid. Conversely, suppose α ⇒ β is valid. Then consider any model m whereα is true. β must also be true in m, or else m would not satisfy α ⇒ β. So α |= β. (d) This reduces to c, because α ∧ ¬β is unsatisable just when α ⇒ β is valid.

B.

  1. VALID. By truth table: both F alse ⇒ F alse and T rue ⇒ T rue are true.
  2. SATISFIABLE. By truth table: it is true when Smoke is true and F ire is true, but false when Smoke is false and F ire is true.
  3. VALID. By truth table: if Smoke is false, both expressions are false; if Smoke is true then either the (Smoke ⇒ F ire) is false (in which case the whole implication is true) or F ire is true, in which case both sides are true.
  4. SATISFIABLE. By truth table: true when both Big and Dumb are true, but false when Big is false and and Dumb is true.

Question3: A.

  1. A ∧ ¬I 2) (S ∨ I) ⇒ D 3) (¬A ∧ ¬S) ⇒ ¬D B. Let the basic vocabulary be as follows: T akes(x; c; s): student x takes course c in semester s; P asses(x; c; s): student x passes course c in semester s; Score(x; c; s): the score obtained by student x in course c in semester s; x > y: x is greater than y; F and G: specic French and Greek courses; Born(x; c): person x is born in country c; P arent(x; y): x is a parent of y; Citizen(x; c; r): x is a citizen of country c for reason r; Resident(x; c): x is a resident of country c; F ools(x; y; t): person x fools person y at time t; Student(x), P erson(x), Man(x), Barber(x), Expensive(x), Agent(x), Insured(x), Smart(x), P o predicates satisfied by members of the corresponding categories.
  2. ∃x Student(x) ∧ T akes(x, F, Spring2001).
  3. ∀x, s Student(x) ∧ T akes(x, F, s) ⇒ P asses(x, F, s).
  4. ∃x Student(x)∧T akes(x, G, Spring2001)∧∀y y 6 = x ⇒ ¬T akes(y, G, Spring2001).
  5. ∀s∃x∀y Score(x, G, s) > Score(y, F, s).
  6. ∀xP erson(x)∧Born(x, UK)∧(∀yP arent(y, x) ⇒ ((∃rCitizen(y, UK, r))∨ Resident(y, UK))) ⇒ Citizen(x, UK, Birth).
  7. ∀xP erson(x)∧¬Born(x, UK)∧(∃yP arent(y, x)∧Citizen(y, UK, Birth)) ⇒ Citizen(x, UK, Descent).
  8. ∀xP olitician(x) ⇒ (∃y∀tP erson(y)∧F ools(x, y, t))∧(∃t∀yP erson(y) ⇒ F ools(x, yt)) ∧ ¬(∀t∀yP erson(y) ⇒ F ools(x, y, t)).

Question4: A. From 1, 2, and 3, we get S ∨ Z is true; From 4 and 5, we get R ∨ ¬Z is true; Then for all possible models, S ∨ R is true. So KB |= (S ∨ R). B. Leave to you.

Question5:

  1. Because x and y are in the same domain and y does not appear in φ. B. ∀x Male(x) ⇔ ¬F emale(x); ∀x, y Spouse(x, y) ⇔ (Male(x) ∧ f emale(y)) ∨ (Male(y) ∧ f emale(x)). C. Yes. Let x = A, y = G(C, C) and z = F (A, G(G(C, C), G(C, C))).

Question8: a) The expression in question means all people are movie stars and are envied. It should be ∀X (movie star(X) ⇒ envied(X)). b) The expression in question means all children are healthy or don’t like ice cream. It should be ∀X (child(x) ⇒ (healthy(x) ⇒ likes(X, ice cream))).

Question9:

  1. Valid.
  2. That is, algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every nonentailed sentence. First-order logic is semidecidable. Propositional logic is decidable.
  3. Propositional logic assumes that there are facts that either hold or do not hold in the world. Each fact can be in one of two states: true or false. First-order logic assumes more; namely, that the world consists of objects with certain relations among them that do or do not hold.
  4. It is declarative. Declarative approach: The agent’s initial program, before it starts to receive percepts, is built by adding one by one the sentences that represent the designer’s knowledge of the environment. Designing the representation language to make it easy to express this knowledge in the form of sentences simplifies the construction problem enormously. Procedural approach encodes desired behaviors directly as program code; minimizing the role of explicit representation and reasoning can result in a much more efficient system.
  5. Yes. That means the meaning of a sentence is a function of the meaning of its parts.
  6. Resolution can always be used to either confirm or refute a sentence.
  7. G¨odel’s completeness theorem.
  8. No. That means there are sentences that are entailed, but have no finite proof whithin the system. It is G¨odel’s imcompleteness theorem.