CMSC 250 Fall 2001 Homework 5 solutions

1. (8 points. 2 points each all or none.)
   1. Therefore, .25 can be written in the form a/b for some a,b $\in$ Z,b$\neq$ 0.

   2. Therefore, 25 can be written in the form $4k+1$ for some $k\in{\bf {Z}}$

   3. Therefore, CMSC250 is hard.

   4. Therefore, $.5$ is not greater than 1.

2. (8 points. 2 points each, all or none.)
   1. Invalid - inverse error

   2. Invalid - converse error

   3. Valid - universal modus tollens

   4. Valid - universal modus ponens

3. (4 free points.) This can be done in many ways. The final conclusion must come at the end. They do not have to have the intermediate conclusions.
   1. No one has moved to Somerset Town during the past year.
      • Therefore, anyone who lives in Somerset Town today has lived there at least a year.

   2. Bob lives in Somerset Town today, and moved there on his 24th birthday.
      • Therefore Bob is over the age of 20.

      • Therefore Bob has lived in Somerset town at least a year.

   3. All people who lived in Somerset Town at least a year automatically become citizens of the town.
      • Therefore, Bob is a citizen of Somerset Town.

   4. All citizens of somerset town over the age of 20 can run for the town council.

   5. Therefore, Bob can run for the town council.

4. (12 points. 3 points each. 2 points for valid/invalid (no reason needed). 1 point for a correct picture.)
   1. Invalid - Not all animals can bite people.

      

   2. Valid - All dogs can bite people.

      

   3. Invalid - Some trucks may have wings.

      

   4. Valid - All natural numbers are complex numbers.

      

5. (8 points. 4 points each. 2 points each for showing the De Morgan's and negation of quantification steps. It is ok if they leave out reasons.)

   $\exists x \in D,\,\,(P(x) \vee Q(x) \vee R(x))$	Given
   $\sim (\sim \exists x \in D,\,\,(P(x) \vee Q(x) \vee R(x)))$	Double negation
   $\sim \forall x \in D,\,\, \sim (P(x) \vee Q(x) \vee R(x))$	Definition of negation
   $\sim \forall x \in D,\,\, (\sim P(x) \wedge \sim Q(x) \wedge \sim R(x))$	DeMorgan's law

   Hence,the given pair of statements are equivalent

   $\sim (\exists p \exists b\,\,((p\neq b) \wedge A(p,b))$	Given
   $\forall p \forall b \,\,\sim ((p\neq b) \wedge A(p,b))$	Definition of negation
   $\forall p \forall b \,\,\sim (p\neq b) \vee \sim A(p,b))$	DeMorgans's law
   $\forall p \forall b \,\,(p = b \vee \sim A(p,b))$	definition of negation

   Hence,the given pair of statements are Equivalent.

6. (10 points. -2 for each step that is incorrect, or not labeled with a reason and line numbers.)

   Show that the following argument is valid by deducing the conclusion from the premises. Give justification for each step.

   1. $\forall x\in D$, $P(x)\rightarrow Q(x)$.

   2. $\forall x\in D,\,\, R(x)\vee P(x).$

   3. $\forall x\in D,\,\, R(x)\rightarrow \sim S(x)$.

   4. $\exists y\in D,\,\, \sim Q(y)$.

   5. Therefore $\sim (\forall x\in D, \,\,S(x))$.

   1.	$\sim$Q(a)	$\exists$instantiation using (d).
   2.	$\sim$P(a)	Universal modus tollens using (a) and 1.
   3.	R(a) $\vee$ P(a)	Universal instantiation using 2.
   4.	R(a)	disjunctive syllogism using 2 and 3.
   5.	$\sim$ S(a)	Universal modus ponens using 4 and (c).
   6.	$\exists x \in D, \sim S(x)$	$\exists$ generalization using 5.
   7.	$\sim (\forall x \in D, S(x))$	Defination of negation using 6.

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