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.