Philosophy 60                                                                  Name:   Sol  Ution
Homework 1

Instructions:  Complete the following problems.  Type your name.  Type your answers. Staple or paper clip multiple pages.  If you turn this in by e-mail, type the date and time you sent it.   10 pts. total.

Due:  2/14

1.   Make the minimum correction to the following formulas that will render them official formulas of sentential logic.  Write what is wrong with the formula and correction it as shown.  If something is already an official formula, write “ducky” in the correction column.  (2 pts. total, 1 pt for every three answers correct.)

	Original	Problem	Correction	
	(P v Q) & r	No capital letters, missing outside brackets	((pvq) & r)	
1	q ↔ (r & ¬¬s)	Missing outside brackets	(q ↔ (r & ¬¬s))	
2	((p & q))	Extra set of brackets	(p & q)	
3	p → (¬ q)	Brackets around negated formula, missing outside brackets.	(p → ¬ q)	
4	¬(¬p & p)		Ducky	
5	¬((¬p) → r)	Brackets around negated formula	¬(¬p → r)	
6	(p v (q v (p1 v r7)))		Ducky	

2.  For the following formula (which may be either official or unofficial), identify the scope of the connective indicated. (2pts. total, 1 pt for every three answers correct.)

	Formula	Connective	Scope	
	((p  v q) & r) → s	&	((pvq) & r)	
1	( r v ¬(r & r)	¬	¬(r & r)	
2	¬(p & q) v r	v	¬(p & q) v r	
3	p → (¬¬¬ q → t)	Second ¬	¬¬ q	
4	(p v q) ↔ (p → (k v t))	→	(p → (k v t))	
5	(r v (q & (s ↔ ¬s))	&	(q & (s ↔ ¬s))	
6	¬ (p & ¬q) v (p ↔ (q v t))	↔	(p ↔ (q v t)	

3.  Symbolize the following English passage as an argument with a conclusion using the sentence letters p, q & r in the obvious way.  (2 pts.)

Paulina or Queenie went to the store.  Paulina only goes to a store if she robs it.  Queenie doesn’t go to a store if Paulina robs it.  So, not both Paulina and Queenie went to the store. 

p v q
p → r
r → ¬q
 ¬(p & q)

4.  Use a truth table to determine whether the following formula is satisfiable, contradictory, or valid.  Highlight the column that provides the answer and provide your rationale.  (Table provided)  (2 pts.  1 pt. for proper truth table.  1 pt. for proper rationale.)

¬ (p v ¬p) → q

p	q	¬	(p	v	¬p)	→	q	
T	T	F	T	T	F	T	T	
T	F	F	T	T	F	T	F	
F	T	F	F	T	T	T	T	
F	F	F	F	T	T	T	F	

Rationale:  This formula is valid because every possible interpretation of the formula is true.  This is shown by all T’s in the column under the main conditional →.

5.  Use a truth table to determine whether the following formulas are logically equivalent.  Make your own table. Provide rationale  (2 pts.  1 pt. for proper truth table.  1 pt. for proper rationale.)

¬(p → q)
p & ¬q

p	q	¬	(p	→	q)	↔	(p	&	¬q)	
T	T	F	T	T	T	T	T	F	F	
T	F	T	T	F	F	T	T	T	T	
F	T	F	F	T	T	T	F	F	F	
F	F	F	F	T	F	T	F	F	T	

Rationale:  These formulas are logically equivalent because when they are connected with a ↔, the biconditional is a tautology, as shown by all T’s under the main connective ↔.