51 Midterm Solutions

51 Midterm Solutions - Handout #51 May 18, 2010 CS103...

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

View Full Document Right Arrow Icon
Handout #51 CS103 May 18, 2010 Robert Plummer Midterm Solutions 1. Logic Proof (20 points) (a) In this part, you are proving one of the De Morgan theorems for quantifiers. You are not allowed to use (nor will you need) the other De Morgan theorems for quantifiers. 1. ¬ x P(x) 2. Let c be an arbitrary member of the domain 3. P(c) 4. x P(x) Existential generalization, 3 5. Contradiction, 1, 4 6. P(c) Negation elimination, 3-5 7. x ¬P(x) Universal generalization, 2-6 That is all there is to it! To prove that something is universally true [in this case ¬P(x)], we consider an arbitrary element of the domain and show that the thing we want to prove is true for that element. So the form is the subproof from line 2 to line 6. That is the only way we can do a universal generalization, and the universally quantified result is outside thesubproof. Here is a common erroneous proof: 1. ¬ x P(x) 2. ¬ x ¬P(x) 3. Let c be an arbitrary member of the domain 4. ¬P(c) 5. x ¬P(x) Universal generalization, 4 ERROR! 6. Contradiction, 2, 5 7. P(c) Negation introduction, 4-6 8. x P(x) Existential introduction, 7 ERROR! 9. Contradiction, 1, 8 10. x ¬P(x) Negation introduction, 2-9 You cannot do the universal generalization at step 5. The pattern is not the one shown in the first proof. Step 8 is also incorrect. The options here would have been the existential introduction inside the subproof, or universal introduction outside the subproof (because at this point you would have the right pattern). Neither helps.
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 Here is another incorrect proof: 1. ¬ x P(x) 2. ¬P(c) Existential instantiation, 1 ERROR! 3. x ¬P(x) Universal generalization, 2 It is not possible to instantiate the existential quantifier, since it has a negation in front. To do the instantiation, the first symbol has to be . You could get line 2 by first applying De Morgan's Law, but that is what you are tying to prove! (b) In this part, you are allowed to use De Morgan's laws freely, both the quantified and unquantified versions , but cite them as justification if you do. Note that 'b' is a constant in this proof. 1. x ( (A(x) ¬B(x)) C(b) ) 2. ¬ x C(x) 3. y B(y) 4. B(d) For some d, existential instantiation, 3 5. (A(d) ¬B(d)) C(b) Universal instantiation, 1 6. ¬A(d) ¬B(d) C(b) Table 7, alternate form of 7. ¬A(d) C(b) Disjunctive syllogism, 4, 6 8. x ¬C(x) De Morgan's Law, 2 9. ¬C(b) Universal instantiation, 8 10. ¬A(d) Disjunctive syllogism, 7, 9 11. x ¬A(x) Existential instantiation, 10 12.  x A(x) De Morgan's Law, 11 Most of you did well on this part.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 02/08/2011.

Page1 / 7

51 Midterm Solutions - Handout #51 May 18, 2010 CS103...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online