(b) (Small(a) ^ Cube(a)) –> (Large(b) ^ Tet(b)) – F

(d) ¬Between(c, a, b) ^ ¬FrontOf(c, a) ^ ¬FrontOf(c, b) – T

(e) Cube(b) –> Cube(a) – T

(f) SameShape(a, b) <–> (Cube(a) ^ Cube(b)) – T

(h) SameCol(b, c) ^ SameRow(c, d) ^ SameCol(d, a) – F

(i) SameRow(a, b) –> a=b – F

(j) a=b –> SameRow(a, b) – T

2.

(a) Since B is a tautological consequence of A there is no row of their truth table in which A is T but B is F; thus because of the truth table for material conditional, there is no row in which A –> B is false: in other words, it is a tautology.

(c) It is a tautology – the final column of its truth table has all Ts.

3.

(a) P = F       Q = F            R = T

(b) It's invalid.

4.

(b) Supposing that the conclusion is false, so that a is a small cube, then a is a cube, contrary to the premise. Thus, a is not a small cube, by indirect proof.

(c) You could give a world, but just giving some assignments of truth values would suffice:
Cube(a) = T      Cube(b) = F      Cube(c) = F

5.
(a) It's valid – every row in which (Tet(a) ^ Large(a)) v (Tet(a) ^ Small(a)) is T so is Large(a) v Small(a).

1. | (Tet(a) ^ Large(a)) v (Tet(a) ^ Small(a))
2. | –
3. | | (Tet(a) ^ Large(a))
    | | –
4. | | Large(a)
5. | | Large(a) v Small(a)
    |
6. | | Tet(a) ^ Small(a)
    | | –
7. | | Small(a)
8. | | Large(a) v Small(a)
    |
9. | Large(a) v Small(a)        v-Elim 1, 3-4, 6-8

Why don't you complete the proof by filling in the justifications for lines 4, 5, 7 and 8 yourselves?

Back to Quiz page

Back to 102 Home Page