Difference between revisions of "Philosophy/PS2"
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(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. | (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). |
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Revision as of 22:45, 11 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. (a)
| p | q | r | p.-r | p.q | p.q ⊃ r | p.-r ∨ (p.q ⊃ r) |
|---|---|---|---|---|---|---|
| T | T | T | F | T | T | T |
| T | T | F | T | T | F | T |
| 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. (a)
| p | p | p$p |
|---|---|---|
| T | T | F |
| F | F | F |
(b)
| q | p | q$p |
|---|---|---|
| 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)$r |
|---|---|---|---|
| T | T | T | |
| T | T | F | |
| T | F | T | |
| T | F | F | |
| F | T | T | |
| F | T | F | |
| F | F | T | |
| F | F | F |
(e)
| p | q | r | (p⊃r) | (p⊃r)$q |
|---|---|---|---|---|
| T | T | T | ||
| T | T | F | ||
| T | F | T | ||
| T | F | F | ||
| F | T | T | ||
| F | T | F | ||
| F | F | T | ||
| F | F | F |
7. (a)
| p | p | r | (p⊃r) | (p⊃p) . (p⊃r) |
|---|---|---|---|---|
| T | T | T | ||
| T | T | F | ||
| F | F | T | ||
| F | F | F |
(b)
| p | r | q | (r⊃q) | (p⊃q) | (r⊃q) . (p⊃q) |
|---|---|---|---|---|---|
| T | T | T | |||
| T | T | F | |||
| T | F | T | |||
| T | 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) |
|---|---|---|---|---|---|---|
| T | T | T | ||||
| T | T | F | ||||
| T | F | T | ||||
| T | F | F | ||||
| 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) 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.
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.
