snhu Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit. PROBLEM 1 In the following question, the domain of discourse is a set of male patients in a clinical study. Define the following predicates: Ā· P(x) : x was given the placebo Ā· D(x) : x was given the medication Ā· M(x) : x had migraines Translate each of the following statements into a logical expression. Then negate the expression by adding a negation operation to the beginning of the expression. Apply De Morgan's law until each negation operation applies directly to a predicate and then translate the logical expression back into English. Sample question: Some patient was given the placebo and the medication. . Ex (P(x) A D(x)) . Negation: - 3x (P(x) \ D(x)) . Applying De Morgan's law: Vx (-P(x) V -D(x)) . English: Every patient was either not given the placebo or not given the medication (or both).
snhu (a) Every patient was given the medication or the placebo or both. Vx (D(x) V P(x)) Negation: Vx (D(x) VP(x)) =- Vx (D(x) V P(x)) De Morgan's Law: - Vx (D(x) V P(x)) = 3x (-D(x) A-P(x)) English: Some patient was not given the medication and was not given the placebo. (b) Every patient who took the placebo had migraines. (Hint: you will need to apply the conditional identity, p -> q = p V q.) Vx (P(x) -> M(x)) Conditional Identity: Vx (P(x)> M(x))= Vx (-P(x) V M(x)) Negation: Vx (-P(x) V M(x)) =- Vx (-P(x) V M(x)) De Morgan's Law: - Vx (-P(x) V M(x))= 3x ( -- P(x) A -M(x)) Double Negation: 3x( -- P(x)\-M(x))=3x (P(x) A -M(x)) English: Some patient took the placebo and did not have a migraine. (c) There is a patient who had migraines and was given the placebo. 3x(M(x) \ P(x)) Negation: 3x(M(x)\P(x)) == x(M(x) AP(x)) De Morgan's Law: - 3x(M(x) \P(x)) = Vx(-M(x) V-P(x)) English: Every patient did not have migraines or was not given the placebo.
snhu PROBLEM 2 Use De Morgan's law for quantified statements and the laws of propositional logic to show the following equivalences: (a) -Vx (P(x) A-Q(x)) = Ex (-P(x) V Q(x)) Start: - Vx (P(x) A-Q(x)) De Morgan's law for quantified statements: - Vx (P(x)A-Q(x)) = 3x (-P(x)A -Q(x) Now, Apply propositional logic: 3x (-P(x) \-Q(x) = = x (-P(x) V Q(x) Conclusion: - Vx(P(x)\Q(x))=Ax (-P(x) V Q(x) (b) -Vx(-P(x)-> Q(x)) = 3x (-P(x) A -Q(x)) Start: - x (-(Px) > Q(x) De Morgan's law for quantified statements: - Vx (-(Px) -> Q(x)) = Now, Apply conditional identity: 3x -(-P(x) -> Q(x)) = 3x -(P(x) V Q(x)) Now, Apply propositional logic: 3x-(P(x) VQ(x)) == x (-P(x) A-Q(x)) Conclusion: - Vx (-(Px) >Q(x)=x ((x)Q(x) (c) -3x(-P(x) V(Q(x) AR(x))) = Vx (P(x) A (-Q(x) V R(x)) ) Start: - 3x (-P(x) V (Q(x) AR(x))) De Morgan's law for quantified statements: - 3x (-P(x)V(Q(x)A-R(x))) =