Difference between revisions of "Philosophy/PS2"

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1. The schemata is not valid if we can find just one assignment of truth-values for which it is false.
+
1. The schema is not valid if we can find just one assignment of truth-values for which it is false.
−
(a) p = T, q = F, r = T makes the schemata false. Not valid.
+
(a) p = T, q = F, r = T makes the schema false. Not valid.
−
(b) The only way to make this false is for p ⊃ r to be false and p ⊃ q.r to be true. This fixes p to T and r to F. However fixing r to F makes p ⊃ q.r false as well (regardless of the value of q), so it is impossible to make the antecedent true and consequent false here. Thus schemata is valid.
+
(b) The only way to make this false is for p ⊃ r to be false and p ⊃ q.r to be true. This fixes p to T and r to F. However fixing r to F makes p ⊃ q.r false as well (regardless of the value of q), so it is impossible to make the antecedent true and consequent false here. Thus schema is valid.
−
(c) Making the schemata false requires each of the disjuncts being false. To make first one false, fix p to T and q to F (or the opposite, makes no difference). To make second one false, r must be made F since p is already fixed. Thus r and q have the same truth-value and the last disjunct evaluates to true. Again, impossible to make an assignment of truth-values giving us a false statement, so schemata is valid.
+
(c) Making the schema false requires all of the disjuncts to be false. To make first one false, fix p to T and q to F (or the opposite, makes no difference). To make second one false, r must be made F since p is already fixed. Thus r and q have the same truth-value and the last disjunct evaluates to true. Again, impossible to make an assignment of truth-values giving us a false statement, so schema is valid.
  
 
2. (i) An assignment of p = q = r = F gives a value of false to the entire schemata, so it is satisfiable but not valid. (ii) An assignment of p = T, q = r = F gives a value of false to the entire schemata so it is also not valid. Thus we cannot easily say if one implies the other based on validity.
 
2. (i) An assignment of p = q = r = F gives a value of false to the entire schemata, so it is satisfiable but not valid. (ii) An assignment of p = T, q = r = F gives a value of false to the entire schemata so it is also not valid. Thus we cannot easily say if one implies the other based on validity.
 
(i) does not imply (ii) as p = T, q = r = F is false for (ii) but true for (i). (ii) does not imply (i) as p = q = r = F is false for (i) but true for (ii).
 
(i) does not imply (ii) as p = T, q = r = F is false for (ii) but true for (i). (ii) does not imply (i) as p = q = r = F is false for (i) but true for (ii).
  
−
3. Neither the President is happy nor is Congress placated; The President is not happy or Congress is placated.
+
3. (a) No. p ⊃ q is false if p = T and q = F, but (p⊃r)⊃(q⊃r) is true with those assignments.
 +
(b) No. p ⊃ q is false if p = T and q = F, but (r⊃p)⊃(r⊃q) is true with those assignments (for example with r = F).
 +
(c) Yes. q = T, r = F, p = T makes the second schema false, and is the only assignment that does. It also makes the first schema false, so the first implies the second.
  
−
4.
+
4. For (2) to be false, q = T and u = F. We can try to find an assignment of the other variables that will make (1) true. For p ⊃ (q⊃r.(s∨t)) to be true, r = s = t = T. In this case, (-p.q.∨r.s∨r.t)⊃u becomes false, so the conjunct is false as well. Any other assignments of r, s, and t will make the first part of the conjunct false, thus making the entire conjunct false. Thus for all truth-values making (2) false, (1) is false as well, and (1) implies (2).
−
(a) George will cut down a tree . (George will marry Martha ∨ George will die a bachelor)
 
−
(b) (Abe will not become mayor . city will not prosper) ∨ (Abe will become mayor . Ava will become head of chamber of commerce)
 
−
(c) (Steve escapes the country ∨ Steve befriends Sally) ⊃ Steve will be safe
 
−
(d) (-Jerry stays in town . -Joan reappears) ⊃ Jan will triumph . Jan will convince Joe
 
−
(e) (-French object to pact . -Belgians object to pact) . Italian forces withdraw from Spain . attacks on British ships cease ⊃ Italo-British pact will take effect
 
−
(f) (-Mail-order campaign breaks Dripsweet monopoly . -mail-order campaign restores competition) ⊃ Jones will mortgage his home ∨ (Jones will sell his car . Jones will sell his boat)
 
  
−
5.
+
5. First schematize the statements.
−
(a)
+
p = prices are low
−
{|
+
q = sales are high
−
! p || q || r || p.-r || p.q || p.q ⊃ r || p.-r ∨ (p.q ⊃ r)
+
r = you sell quality merch
−
|-
+
s = your customers are satisfied
−
|T || T || T || F || T || T || T
+
 
−
|-
+
(i) becomes (p ⊃ q) . (r ⊃ s); (ii) becomes (p ∨ r) ⊃ (q ∨ s).
−
|T || T || F || T || T || F || T
+
(ii) is false when p∨r = T and q∨s = F. This means that q = s = F and p != r. q and s being false makes (i) false regardless of the values of p and r, so (i) implies (ii).
−
|-
 
−
|T || F || T || F || F || T || T
 
−
|-
 
−
|T || F || F || T || F || T || T
 
−
|-
 
−
|F || T || T || F || F || T || T
 
−
|-
 
−
|F || T || F || F || F || T || T
 
−
|-
 
−
|F || F || T || F || F || T || T
 
−
|-
 
−
|F || F || F || F || F || T || T
 
−
|}
 
−
(b)
 
−
{|
 
−
! p || q || r || s || -p.-q || q.-s || p.-r.s || -p.-q ∨ q.-s ∨ p.-r.s
 
−
|-
 
−
| T || T || T || T || F || F || F || F
 
−
|-
 
−
| T || T || T || F || F || T || F || T
 
−
|-
 
−
| T || T || F || T || F || F || T || T
 
−
|-
 
−
| T || T || F || F || F || T || F || T
 
−
|-
 
−
| T || F || T || T || F || F || F || F
 
−
|-
 
−
| T || F || T || F || F || F || F || F
 
−
|-
 
−
| T || F || F || T || F || F || T || T
 
−
|-
 
−
| T || F || F || F || F || F || F || F
 
−
|-
 
−
| F || T || T || T || F || F || F || F
 
−
|-
 
−
| F || T || T || F || F || T || F || T
 
−
|-
 
−
| F || T || F || T || F || F || F || F
 
−
|-
 
−
| F || T || F || F || F || T || F || T
 
−
|-
 
−
| F || F || T || T || T || F || F || T
 
−
|-
 
−
| F || F || T || F || T || F || F || T
 
−
|-
 
−
| F || F || F || T || T || F || F || T
 
−
|-
 
−
| F || F || F || F || T || F || F || T
 
−
|}
 
−
(c)
 
−
{|
 
−
! p || q || r || (p∨q)≡-r || p∨(-q ⊃ r) || (p∨q)≡-r . p∨(-q ⊃ r)
 
−
|-
 
−
|T || T || T || F || T || F
 
−
|-
 
−
|T || T || F || T || T || T
 
−
|-
 
−
|T || F || T || F || T || F
 
−
|-
 
−
|T || F || F || T || T || T
 
−
|-
 
−
|F || T || T || F || T || F
 
−
|-
 
−
|F || T || F || T || T || T
 
−
|-
 
−
|F || F || T || T || T || T
 
−
|-
 
−
|F || F || F || F || F || F
 
−
|}
 
  
 
6.
 
6.
−
(a)
+
(a) p = Smith was the murderer
−
{|
+
q = Jones was lying
−
! p || p || p$p
+
r = Jones met Smith last night
−
|-
+
s = murder took place after midnight
−
|T || T || F
+
P1: -p . -q ⊃ r
−
|-
+
P2: r ∨ -s ⊃ p
−
|F || F || F
+
P3: -p . q ⊃ -s
−
|}
+
C: p
 +
Thus, does (-p.-q ⊃ r) . (r∨-s ⊃ p) . (-p.q ⊃ -s) imply p?
 +
Say p = F. To make P3 true, q = T and s = F. This makes P2 false, so we cannot make the antecedent true. Thus the premises taken together do imply the conclusion.
 
(b)
 
(b)
 +
p = trains stop running
 +
q = airline prices will increase
 +
r = buses reduce their fares
 +
t = buses lose customers
 +
P1: p ⊃ q
 +
P2: -p ⊃ r
 +
P3: q ⊃ -t
 +
C: r ⊃ t
 +
To make conclusion false, we must set r = T and t = F. P2 and P3 will always be true, regardless of what values p and q take on. Thus it is possible for P1 to be true as well (p = q = T, for example), and the consequent can be false while the antecedent is true. Thus the premises taken together do not imply the conclusion.
 +
 +
7.
 
{|
 
{|
−
! q || p || q$p
+
! p || q || r || p.q⊃r || p∨q⊃r || p⊃(q⊃r)|| q⊃(p⊃r) || (p⊃r).(q⊃r) || (p⊃r)∨(q⊃r)
−
|-
 
−
|T || T ||  
 
−
|-
 
−
|T || F ||  
 
−
|-
 
−
|F || T ||
 
−
|-
 
−
|F || F ||
 
−
|}
 
−
(c)
 
−
{|
 
−
! p || q || p$q || (p$q)$p
 
−
|-
 
−
|T || T ||
 
−
|-
 
−
|T || F ||
 
−
|-
 
−
|F || T ||
 
−
|-
 
−
|F || F ||
 
−
|}
 
−
(d)
 
−
{|
 
−
! p || q || r || (p$q)$r
 
 
|-
 
|-
−
| T || T || T ||  
+
|T || T || T || T      || T      || T      || T        || T            || T
 
|-
 
|-
−
| T || T || F ||  
+
|T || T || F || F      || F      || F      || F        || F            || F
 
|-
 
|-
−
| T || F || T ||  
+
|T || F || T || T      || T      || T      || T        || T            || T
 
|-
 
|-
−
| T || F || F ||  
+
|T || F || F || T      || F      || T      || T        || F            || T
 
|-
 
|-
−
| F || T || T ||  
+
|F || T || T || T      || T      || T      || T        || T            || T
 
|-
 
|-
−
| F || T || F ||  
+
|F || T || F || T      || F      || T      || T        || F            || T
 
|-
 
|-
−
| F || F || T ||  
+
|F || F || T || T      || T      || T      || T        || T            || T
 
|-
 
|-
−
| F || F || F ||  
+
|F || F || F || T      || T      || T      || T        || T            || T
 
|}
 
|}
−
(e)
+
Thus we can see that #1, 2, and 4 are equivalent to p.q⊃r while #3 is equivalent to p∨q⊃r
 +
 
 
{|
 
{|
−
! p || q || r || (p⊃r) || (p⊃r)$q
+
! p || q || r || r⊃p.q || r⊃p∨q || (r⊃p).(r⊃q) || (r⊃p)∨(r⊃q)
 
|-
 
|-
−
| T || T || T ||  
+
|T || T || T || T      || T      || T            || T
 
|-
 
|-
−
| T || T || F ||  
+
|T || T || F || T      || T      || T            || T
 
|-
 
|-
−
| T || F || T ||  
+
|T || F || T || F      || T      || F            || T
 
|-
 
|-
−
| T || F || F ||  
+
|T || F || F || T      || T      || T            || T
 
|-
 
|-
−
| F || T || T ||  
+
|F || T || T || F      || T      || F            || T
 
|-
 
|-
−
| F || T || F ||  
+
|F || T || F || T      || T      || T            || T
 
|-
 
|-
−
| F || F || T ||  
+
|F || F || T || F      || F      || F            || F
 
|-
 
|-
−
| F || F || F ||  
+
|F || F || F || T      || T      || T            || T
 
|}
 
|}
 +
Thus we can see that #1 is equivalent to r⊃p.q and #2 is equivalent to r⊃p∨q
  
−
7. (a)
+
8. (a)
 
{|
 
{|
−
! p || p || r || (p⊃r) || (p⊃p) . (p⊃r)
+
! p || φ(p,p,p)
−
|-
 
−
| T || T || T ||
 
 
|-
 
|-
−
| T || T || F ||
+
|T || T
 
|-
 
|-
−
| F || F || T ||
+
|F || F
−
|-
 
−
| F || F || F ||
 
 
|}
 
|}
−
(b)
+
 
 
{|
 
{|
−
! p || r || q || (r⊃q) || (p⊃q) || (r⊃q) . (p⊃q)
+
! p || q || φ(q,p,q)
 
|-
 
|-
−
| T || T || T ||
+
|T || T || T
 
|-
 
|-
−
| T || T || F ||  
+
|T || F || T
 
|-
 
|-
−
| T || F || T ||  
+
|F || T || T
 
|-
 
|-
−
| T || F || F ||
+
|F || F || F
−
|-
 
−
| F || T || T ||
 
−
|-
 
−
| F || T || F ||
 
−
|-
 
−
| F || F || T ||
 
−
|-
 
−
| F || F || F ||
 
−
|}
 
−
(c)
 
−
{|
 
−
! p || r || q || (p⊃f(p,q,p)) || (q⊃f(p,q,p)) || (p⊃f(p,q,p)) . (q⊃f(p,q,p))
 
−
|-
 
−
| T || T || T ||
 
−
|-
 
−
| T || T || F ||
 
−
|-
 
−
| T || F || T ||
 
−
|-
 
−
| T || F || F ||
 
−
|-
 
−
| F || T || T ||
 
−
|-
 
−
| F || T || F ||
 
−
|-
 
−
| F || F || T ||
 
−
|-
 
−
| F || F || F ||
 
−
|}
 
−
(d)
 
−
{|
 
−
! p || r || q || ((p⊃q)⊃r) || ((q⊃p)⊃r) || ((p⊃q)⊃r) . ((q⊃p)⊃r)
 
−
|-
 
−
| T || T || T ||
 
−
|-
 
−
| T || T || F ||
 
−
|-
 
−
| T || F || T ||
 
−
|-
 
−
| T || F || F ||
 
−
|-
 
−
| F || T || T ||
 
−
|-
 
−
| F || T || F ||
 
−
|-
 
−
| F || F || T ||
 
−
|-
 
−
| F || F || F ||
 
 
|}
 
|}
  
−
8.
 
−
(a) Yes, as the truth-values for s#-t and -s#t are the same.
 
−
(b)
 
 
{|
 
{|
−
! s || t || u || (s#t) || (s#t)#u || t#u || s#(t#u)
+
! p || φ(q,p,q) || q || φ(p,φ(q,p,q),q)
 
|-
 
|-
−
| T || T || T ||  
+
|T || T         || T || T
 
|-
 
|-
−
| T || T || F ||  
+
|T || T         || F || F
 
|-
 
|-
−
| T || F || T ||  
+
|F || T        || T || F
 
|-
 
|-
−
| T || F || F ||
+
|F || F         || F || T
−
|-
 
−
| F || T || T ||
 
−
|-
 
−
| F || T || F ||
 
−
|-
 
−
| F || F || T ||
 
−
|-
 
−
| F || F || F ||
 
 
|}
 
|}
  
−
9. (a) Compare the values of the 2nd and 3rd rows; if they are equivalent, then # is commutative. Otherwise, it isn't.
+
(b) Yes; for φ(p,p,p) = F, p = F as well.
−
(b) Say P and Q are equivalent, i.e. both true. Then, for a connective to be anti-commutative, that means that P#Q is equivalent to -(Q#P). However, this is impossible as P#Q and Q#P are equivalent, so the negation of the latter cannot be equivalent to the former.
+
Yes; for φ(q,p,q) = F, p = F as well.
−
 
+
No; there are two cases in which φ(p,φ(q,p,q),q) = F, and p = F for only one of them. Thus p = T, q = F gives us φ(p,φ(q,p,q),q) = F while p = T, so implication does not hold.
−
10.
 
−
(a) No. Say p = "God exists", but there is actually no God. Many people believe God exists is true. Say p = "oceans are blue", which they are. Many people believe oceans are blue is true. Thus the truth value of the statement depends on more than just the truth-value of its single component.
 
−
(b) No. Say p = "the sun's gravitational field exerts an attractive force on all objects", q = "some frogs are green"; then the whole statement is false. However, say p remains the same and q = "the planets orbit around the sun"; then, the whole statement is true. Thus the whole statement can be both true and false if p & q are both true, and it is not truth-functional as the truth-values of p & q alone do not determine its truth value.
 
−
(c) Yes; this is simply the "not" connective, whose output depends solely on the truth value of the input.
 

Latest revision as of 19:29, 13 February 2009

1. The schema is not valid if we can find just one assignment of truth-values for which it is false. (a) p = T, q = F, r = T makes the schema false. Not valid. (b) The only way to make this false is for p ⊃ r to be false and p ⊃ q.r to be true. This fixes p to T and r to F. However fixing r to F makes p ⊃ q.r false as well (regardless of the value of q), so it is impossible to make the antecedent true and consequent false here. Thus schema is valid. (c) Making the schema false requires all of the disjuncts to be false. To make first one false, fix p to T and q to F (or the opposite, makes no difference). To make second one false, r must be made F since p is already fixed. Thus r and q have the same truth-value and the last disjunct evaluates to true. Again, impossible to make an assignment of truth-values giving us a false statement, so schema is valid.

2. (i) An assignment of p = q = r = F gives a value of false to the entire schemata, so it is satisfiable but not valid. (ii) An assignment of p = T, q = r = F gives a value of false to the entire schemata so it is also not valid. Thus we cannot easily say if one implies the other based on validity. (i) does not imply (ii) as p = T, q = r = F is false for (ii) but true for (i). (ii) does not imply (i) as p = q = r = F is false for (i) but true for (ii).

3. (a) No. p ⊃ q is false if p = T and q = F, but (p⊃r)⊃(q⊃r) is true with those assignments. (b) No. p ⊃ q is false if p = T and q = F, but (r⊃p)⊃(r⊃q) is true with those assignments (for example with r = F). (c) Yes. q = T, r = F, p = T makes the second schema false, and is the only assignment that does. It also makes the first schema false, so the first implies the second.

4. For (2) to be false, q = T and u = F. We can try to find an assignment of the other variables that will make (1) true. For p ⊃ (q⊃r.(s∨t)) to be true, r = s = t = T. In this case, (-p.q.∨r.s∨r.t)⊃u becomes false, so the conjunct is false as well. Any other assignments of r, s, and t will make the first part of the conjunct false, thus making the entire conjunct false. Thus for all truth-values making (2) false, (1) is false as well, and (1) implies (2).

5. First schematize the statements. p = prices are low q = sales are high r = you sell quality merch s = your customers are satisfied

(i) becomes (p ⊃ q) . (r ⊃ s); (ii) becomes (p ∨ r) ⊃ (q ∨ s). (ii) is false when p∨r = T and q∨s = F. This means that q = s = F and p != r. q and s being false makes (i) false regardless of the values of p and r, so (i) implies (ii).

6. (a) p = Smith was the murderer q = Jones was lying r = Jones met Smith last night s = murder took place after midnight P1: -p . -q ⊃ r P2: r ∨ -s ⊃ p P3: -p . q ⊃ -s C: p Thus, does (-p.-q ⊃ r) . (r∨-s ⊃ p) . (-p.q ⊃ -s) imply p? Say p = F. To make P3 true, q = T and s = F. This makes P2 false, so we cannot make the antecedent true. Thus the premises taken together do imply the conclusion. (b) p = trains stop running q = airline prices will increase r = buses reduce their fares t = buses lose customers P1: p ⊃ q P2: -p ⊃ r P3: q ⊃ -t C: r ⊃ t To make conclusion false, we must set r = T and t = F. P2 and P3 will always be true, regardless of what values p and q take on. Thus it is possible for P1 to be true as well (p = q = T, for example), and the consequent can be false while the antecedent is true. Thus the premises taken together do not imply the conclusion.

7.

p q r p.q⊃r p∨q⊃r p⊃(q⊃r) q⊃(p⊃r) (p⊃r).(q⊃r) (p⊃r)∨(q⊃r)
T T T T T T T T T
T T F F F F F F F
T F T T T T T T T
T F F T F T T F T
F T T T T T T T T
F T F T F T T F T
F F T T T T T T T
F F F T T T T T T

Thus we can see that #1, 2, and 4 are equivalent to p.q⊃r while #3 is equivalent to p∨q⊃r

p q r r⊃p.q r⊃p∨q (r⊃p).(r⊃q) (r⊃p)∨(r⊃q)
T T T T T T T
T T F T T T T
T F T F T F T
T F F T T T T
F T T F T F T
F T F T T T T
F F T F F F F
F F F T T T T

Thus we can see that #1 is equivalent to r⊃p.q and #2 is equivalent to r⊃p∨q

8. (a)

p φ(p,p,p)
T T
F F
p q φ(q,p,q)
T T T
T F T
F T T
F F F
p φ(q,p,q) q φ(p,φ(q,p,q),q)
T T T T
T T F F
F T T F
F F F T

(b) Yes; for φ(p,p,p) = F, p = F as well. Yes; for φ(q,p,q) = F, p = F as well. No; there are two cases in which φ(p,φ(q,p,q),q) = F, and p = F for only one of them. Thus p = T, q = F gives us φ(p,φ(q,p,q),q) = F while p = T, so implication does not hold.

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