Question 1.
(1)
Valid
(2)
Valid
Question 2.
From the first two statements, we see that if it is mythical, then it is immortal
(Mythical  Immortal); otherwise it is a mammal (Mythical  Immortal 
Mammal). So it must be either immortal or a mammal, and thus horned (Immortal 
Mammal  Horned). That means it is also magical (Horned  Magical). However,
we can’t deduce anything about whether it is mythical (Unicorn  Mythical).
Question 3.
(1)
x Politician(x)  (y t Person(y)  Fools(x, y, t)) 
(t y Person(y)  Fools(x, y, t)) 
(t y Person(y)  Fools(x, y, t))
(2)
x Person(x)  (y Policy(y)  Buys(x, y))  Smart(x)
Question 4.
We use a very simple ontology to make the examples easier:
a. Horse(x)  Mammal(x).
Cow(x)  Mammal(x).
Pig(x)  Mammal(x).
b. Offspring(x, y)  Horse(y)  Horse(x).
c. Horse(Bluebeard).
d. Parent(Bluebeard, Charlie).
e. Offspring(x, y)  Parent(y, x).
Parent(x, y)  Offspring(y, x).
(Note we couldn’t do Offspring(x, y)  Parent(y, x) because that is not in
the form expected by Generalized Modus Ponens.)
f. Mammal(x)  Parent(G(x), x) (here G is a Skolem function).
The proof tree is shown in the following figure. The branch with Offspring(Bluebeard,
y) and Parent(y, Bluebeard) repeats indefinitely, so the rest of the proof is never
reached.
Question 5.
a. x Horse(x)  Animal(x)
x, h Horse(x)  HeadOf(h,x)  y Animal(y)  HeadOf(h,y).
b. A. Horse(x)  Animal(x)
B. Horse(G)
C. HeadOf(H,G)
D. Animal(y)  HeadOf(H,y)
(Here A comes from the first sentence in a. while the others come from the
second. H and G are Skolem constants.)
c. Resolve D and C to yield Animal(G). Resolve this with A to give Horse(G).
Resolve this with B to obtain a contradiction.