Faculty of Science and Engineering

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1 Faculty of Science and Engineering MATH1090 Solutions to Problem Set No 1 1. (3 marks) Consider the string ( ( (p q))), (a) Show that the above string is a well-formed formula (by the procedural definition). Consider the following (valid) formula-calculation:, p, q, (p q), ( (p q)), ( ( (p q))) Since the formula appears at some step of this formula calculation, it is therefore a wff. (b) Show the top-down parsing for this formula. (c) Show this formula in least parenthesized notation. (p q) 2. (3 marks) Prove by induction on formulae that the complexity of a wff equals the number of its left brackets. Proof. By Induction on formula A. Using notations introduced in the class, we want to prove comp(a)=left(a). Basis. If A is atomic, the complexity of A and the number of its left brackets are both zero. So we are okay. Page 1 of 6

2 Induction step. If A has complexity n, n > 0 I.H.: If formula X has complexity < n, comp(x)=left(x). (1) If A has the form ( B): Complexity of A is one more than complexity of B, in other words comp(a) = 1 + comp(b). Also, the number of left brackets of A is one more than the number of left brackets of B, or left(a) = 1 + left(b). We also know that the complexity of B is n 1 < n and I.H. applies to B. Therefore the complexity of B equals its number of left brackets, or comp(b) = lef t(b). We can conclude that comp(a) = left(a), in other words the complexity of A equals its number of left brackets. We are okay in this case. (2) If A has the form (BoC): Complexity of A is comp(a) = 1 + comp(b) + comp(c). Also, the number of left brackets of A is left(a) = 1 + left(b) + left(c). We also know that the complexity of B and C are both less than n and I.H. applies to B and C. Therefore by I.H., comp(b) = left(b) and comp(c) = left(c). We can conclude that comp(a) = left(a), in other words the complexity of A equals its number of left brackets. We are okay in this case. We are done! We proved the above statement by induction on formula A. 3. (3 marks) Prove that the last symbol of a Boolean formula is never the symbol. Hint. Use analysis of formula calculation, or prove by induction on formulae. First Proof. Proof by analysis of formula calculation. By definition, a wff is a string that is written in some step of formula calculation. We show that none of the rules of formula calculation can introduce the symbol as the last symbol of a formula. A formula can be written using one of these rules of formula calculation: (a) We may write a Boolean variable, or. None of these ends with. (b) We may write ( A) provided we have written A before. The last symbol in this case is ) and is not. (c) We may write (A B), {,,, } provided we have written A and B before. The last symbol in this case again is ) and is not. We showed that none of the steps in a formula calculation can introduce as the last symbol of a formula. Second Proof. Proof by induction on formula A. Basis. If A is atomic. Then A is p, or. None of these ends with. Page 2 of 6

3 Induction step. If has complexity n, n > 0. Then A is either of form ( B) or of form (B C), {,,, }. In both cases the formula ends with a ), and the last symbol is not a. We showed by induction on formula A that the last symbol of a formula is never the symbol. Note we did not need to use the I.H. for this proof. 4. (4 marks) Which of the following are tautologies? Show your work. Note that to prove that a schema is not a tautology, you must show an instance of it that is not a tautology. p q p q p q p q p q (p q) (p q) f f f f t f t f t t t f f t t t t t t t Since the above formula is true in all possible cases, it is therefore a tautology. p q As can be seen in he third column of above truth table, there are possible states for which p q is not true. It is therefore not a tautology. A B A B A B A B A B (A B) (A B) f f f f t f t t f f t f t f f t t t t t This schema is not a tautology, since there are possible instances of this schema that are not tautologies. for example if A is and B is, we have ( ) ( ) as an instance of above schema which is false and not a tautology. A (B C) A B A C Page 3 of 6

4 A B C A (B C) A B A C (2) (1) (6) (3) (5) (4) f f f t t t f t f f f t f f t f f t f t f f f t t f f f t t t t t t t t t f f t t t t t t t f t t f t t t t t t f t f t t t t t t t t t t t t t This schema is true in all possible states, it is therefore a tautology. 5. (4 marks) Use truth tables (with explanations) or truth table shortcuts to show if the following statements are valid. For any invalid schema, show an instance that is not a tautological implication. A, B = taut A B A B A B f f f t t f t t t Only the last row, shows a state that satisfies Γ = {A, B}. in this state, A B is also true. Therefore this is a valid tautological implication. p, q = taut q This is obviously a valid tautological implication. Any state that satisfies Γ = {p, q} satisfies q as well by definition. B A, A = taut B A B B A f f t f t f t f t t t t The third and fourth rows of the truth table satisfy Γ = {B A, A}. But in these states B is not necessarily true. To show an instance for which the tautological implication is not valid, consider A being and B being. Both and are true, but is false. We showed that this tautological implication is invalid. Page 4 of 6

5 A, A B C = taut C Assume A is, B is, and C is. Then A is true and (A (B C)) which is ( ( )) is true, but C is false. We showed an instance of the schema that is not a tautological implication. Therefore the schema is not a valid tautological implication. 6. (2 marks) Use the truth table shortcut method to show that C, A (B C) = taut A B Let s assume that the above is not a valid tautological implication. Therefore there exists a state v, for which the assumptions are true but the conclusion is false. In other words: v(c) = t (1) v(a (B C)) = t (2) { v(a) = t (3) v(a B) = f v(b) = f (4) v(b C) = t (By (2), (3)) (5) v(c) = f (By (4), (5)) (6) But (6) contradicts (1). Therefore our assumption must be incorrect, and the above statement is indeed a valid tautological implication. 7. (2 marks) Which of the following sets is satisfiable? {p q, q r, r p} Consider a state v for which v(p) = f, v(q) = t, v(r) = f In this state, all formulae in above set are true. Therefore this state satisfies above set and therefore the set is satisfiable. {p q, p p, } The formula p p is a contradiction and can never be true. Therefore no state can exist for which all formulae in above set are true. This set is not satisfiable. 8. (4 marks) Calculate the wff obtained after the following substitutions, if they are valid. Otherwise explain why the substitution is illegal. Show all steps for the first valid substitution only, and only the final answer for the rest. Page 5 of 6

6 p (q p)[p := ] = p (q[p := ] p[p := ]) = p (q ) (p q) p[p := t] This is an illegal substitution, since t is not a wff. ((p q) p)[(p q) := A] (where A is a formula) This is an illegal substitution, since we can only substitute for a Boolean variable, not a formula such as (p q) q[p := A B] (where A and B are formulae) = q (unchanged). Note this is a legal substitution. Page 6 of 6