midterm-practice-sol

# midterm-practice-sol - Propositional Logic 1. (7.5 points)...

This preview shows pages 1–3. Sign up to view the full content.

Propositional Logic 1. (7.5 points) Write each of the sentences in propositional logic using two propositions p andq: a) For you to win the contest (p) it is necessary and su ffi cient that you have the only winning ticket (q) p ! q b) You can access the website (p) only if you pay a subscription fee (q). p " q c) Getting elected (p) follows from knowing the right people (q). q " p e) Whenever you get a speeding ticket (q) you are driving over 65 miles per hour (p). q " p f) You can see the wizard (p) only if the wizard is not in (q) and the wizard is not in only if you can see the wizard. p ! q 2. (12 points) Mark would like to determine the relative salaries of his friends based on two facts. First, he knows that if Francisca is not the highest paid of the three, then John is. Second he knows if John is not the lowest paid then Martha is paid the most. Is it possible to determine the relative salaries of Francisca, Martha, and John from what Mark knows? If so, who is paid the most and who is paid the least. Explain your reasoning. FHP – Francisca highest paid FLP – Francisca lowest paid JHP – John highest paid JLP – John lowest paid MHP – Martha highest paid MLP – Martha lowest paid ~FHP => JHP ~JLP => MHP Assume JHP : that would imply FHP (from 1 using modus tollens); So ~MHP and therefore JLP which contradicts assumption;

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Assume MHP: therefore ~JHP therefore FHP (from 1 using modus tollens); buththat contradicts the assumption. So assume FHP: therefore ~MHP and ~JHP; ~MHP=> JLP; so So the order is Francisca, Martha, John 3. (15 points) Use resolution refutation to prove: ( Q ! ¬ P ) ) ! (( Q ) ! P ) ! ¬ Q ) ROBERT – this is a homework question – so it should be discussed only after they hand it in!!!! {(Q ! ~p ) ! [( Q ! P) ! ~Q]} Negate it: ~{(Q ! ~p ) ! [( Q ! P) ! ~Q]} Convert to CNF ~{ ~(~Q or ~p) or [ ~ ( ~Q or P) or ~Q]} ~{ (Q and P) or [ (Q and ~P) or ~Q] } ~ (Q and P) and ~ (Q and ~P) and Q (~Q or ~ P ) and (~Q or P) and Q
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/22/2008 for the course INFO 3720 taught by Professor Gomes during the Spring '07 term at Cornell University (Engineering School).

### Page1 / 7

midterm-practice-sol - Propositional Logic 1. (7.5 points)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online