Download Math 470 Spring `05

Math 470 Spring ’05
HW #5 Solutions
Section 2.2
1. a), c), e), g)
3. b) is an atomic formula
c) the given formula and P(c)
d) the given formula and P(x) (but x should be in parentheses)
f) the given formula, (((y)P(z)) -> R(x,y)), ((y)P(z)), R(x,y), P(z)
5. a) no; y does not remain free
b) yes. (Note that only the last occurrence of x is free.)
c) yes , but nothing changes as x is not free
d) yes
6. Base case: A variable or constant symbol has 0 left and 0 right parentheses, and no
proper initial segments.
Induction: Suppose that f is an n-ary function symbol and t1, …, tn are terms with
equal numbers p1, …, pn of left and right parentheses, then f(t1, …, tn) has p1 + p2 +
…+pn + 1 left and right parentheses.
A proper initial segment of f(t1, …, tn) of length 2 or more has the left but not the right
parenthesis associated with f. If it ends inside one of the ti, it also includes a proper initial
segment of that ti. If this is of length at least 2, it has more left than right parentheses, so
the whole segment has one additional left and no additional right parentheses, so at least
2 more left than right.
Otherwise, it has exactly one more left than right parentheses.
8. Base case: Let  = an atomic formula R(t1, …, tn) . By exercise 6, each ti has an equal
number of left and right parentheses, say pi. Then  has p1+ …+ pn + 1 left parentheses
and the same number of right parentheses.
Induction: Assume the claim is true for subformulas. If  is (), it has one additional
left and one additional right parenthesis. The same is true for the binary connectives.
If  is ((v) ) or ((v) ), it has 2 more left and 2 more right parentheses than .
Section 2.3
1. a) c
b) f (x, d)
/ \
x
d
c)
g(f, x, d)
/ \
f(x, d) c
/ \
x
d
d)
h(y, g(z, f(f(c, d), g(x, z))))
/
\
y g(z, f(f(c, d), g(x, z)))
/
\
z
f(f(c, d), g(x, z))
/
\
f(c, d) g(x, z)
/ \
/ \
c d
x z
2. a) R(c, d)
/ \
c
d
b) R(f(x, y), d)
/
\
f(x, y)
d
/ \
x y
c) R(c, d) ^ R (f(x, y), d)
/
\
(a above)
(b above)
d) yz(R(x, f(c, d)) v P(h(y)))
\
z(R(x, f(c, d)) v P(h(y)))
\
R(x, f(c, d)) v P(h(y))
/
\
R(x, f(c, d)) P(h(y))
/
\
\
x
f(c, d)
P(h(y))
/ \
\
c d
h(y)
\
y
e) z(R(g(x, z, z)) -> P(y)) ^ P(z)
/
\
z(R(g(x, z, z)) -> P(y))
P(z)
/
\
(R(g(x, z, z)) -> P(y))
z
/
\
R(g(x, z, z))
P(y)
\
\
g(x, z, z)
y
/ | \
x z z
3. b) x and y
c) x and y’
d) x
e) x, y, final occurrence of z
4. The unique readability determines the tree.
A variable or constant symbol has a tree with a single node.
A term of the form f(t1, …, tn) has a tree with root f(t1, …, tn) and n children, one for
each of the ti, which are unique by induction.
7. Again this is done by induction, using the unique readability of formulas.
Section 2.4
1. First structure: Let A = N, c map to 0 , f to + and P to >= 0.
Second structure: same but map P to = 0.
2. Let A = N; p map to x >=0, q to x is even, f to the function taking x to 2x.
Then x(p(x) -> q(f(x)) is true; xp(x) is true and xq(x) is true,
3. x(p(x) ^ p(x))
4. Let A = N, map < to the actual < relation on N, map + to subtraction.
Then “x + 1” is mapped to x – 1, which is < x. “x + x” is mapped to x – x = 0.
“x+ x < x “ is false for x = 0.