# Exercises for CS1512 Weeks 7 and 8 Propositional Logic 1 (questions + solutions)

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Exercises for CS1512 Weeks 7 and 8 Propositional Logic 1 (questions + solutions)

Exercise 1 1. Express each formula using only (at most) the connectives listed. In each case use a truth table to prove the equivalence. (Note: is exclusive `or`) a.Formula: p q. Connectives: {, }. b.Formula: p q. Connectives: {,, }. c.Formula: p q. Connectives: {, }. d.Formula: (p q) (( p) q). Conn: {, }.

Answer to Exercise 1. (Other answers possible) a. Formula p q. Connectives: {, }. Answer: p q b.Formula: p q. Connectives: {,, }. Answer: (p q) (q p) c. Formula: p q. Connectives: {, }. Answer: (p q) (q p) d. Formula: (p q) (( p) q). Conn: {, }. Answer: q (This was a trick question, since you dont need any connectives.)

Ex. 2. Which of these are tautologies? 1.p (q p) 2.p ( p p) 3.(q p) (p q) 4.(q p) (p q) 5.(p (q r)) (q (p r)) Please prove your claims, using truth tables. (Hint: Ask what assignment of truth values to p,q, and r would falsify each formula. In this way you can disregard parts of the truth table).

Answer to Ex.2 1.p (q p) Tautologous 2.p ( p p) Tautologous 3.(q p) (p q) Contingent 4.(q p) (p q) Tautologous 5.(p (q r)) (q (p r)) Tautologous 1,2,4,5 are known as `paradoxes of the material implication, because they contrast with implication in ordinary language.

Ex. 3. Reading formulas off truth tables Background: In class, a proof was sketched for the claim that every propositional logic formula can be expressed using the connectives {, }. The proof proceeded essentially by reading off the correct formula off the truth table of any given formula. Task: Use this meticulous method to construct a formula equivalent to p q.

Answer to ex. 3. Steps: 1.Construct the truth table of p q. 2. Mark those two rows in the table that make p q TRUE. 3.Corresponding with these two rows, construct a disjunction of two formulas, one of which is (p q), and the other (q p). 4.Use the De Morgan Laws to convert this disjunction (p q) (q p) into the quivalent formula ( (p q) (q p)) [5. Use truth tables again to check that these two formulas are indeed equivalent.]

Question 4a In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {, } is also functionally complete?

Question 4a In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {, } is also functionally complete? Answer: It is sufficient to show that both p and (p q) are expressible using the two connectives in {, }. The first of these two (i.e., negation) is obvious, and the second (i.e., conjunction) can be expressed as follows (p q) ( p q), using one of De Morgans laws.

Question 4b In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {NAND} is also functionally complete? [Explanation: (p NAND q) is TRUE iff (p q) is FALSE. This connective is also called the Sheffer stroke and written (p|q).)

Question 4b In class, it was proven that {, } is a functionally complete set of connectives. Making use of this result, can you prove that {NAND} is also functionally complete? Answer: the reasoning is similar to that in 4a, but it may be a bit trickier to find the right formulas: p can be expressed as p p|p. (It helps to do negation before conjunction!) The formula p q is equivalent to (p|q)|(p|q) (in other words, the negation of p|q) It might be useful to prove these equivalences using truth tables.

Question 4c Given this result, why do we bother defining and using more than one connective? Answer: Many things can be expressed more succinctly and transparently with a larger vocabulary of connectives (as your answer to 4b will show). Also, it becomes easier to let your formulas resemble the things we say in e.g. English.

Question 5 a.(r c) d b.r ( c d) Comparison of truth tables shows (a) and (b) to be equivalent. Both formulas are false if and only if (r true,c true,d false).

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