MATHEMATICS 114L
SPRING 2014
Solutions to Exercises
2.3(b) and 2.3(d)
Parts (b) and (d) of Exercise 2.3 have are logical implication questions
with yes answers. Below are some examples of how one might explain these
answers. More informal explanations tha
MATHEMATICS 114L
SPRING 2014
Solutions for Final Exam
1. Let be the set of premises, and let be the possible conclusion. Dene a
model A as follows. Let A = cfw_0, 1, and let QA = cfw_(0, 1), (1, 0), (0, 0), (1, 1)
(= A2 ). is true in A. is false in A, sin
MATHEMATICS 114L
SPRING 2014
Extra Problem 1
Let be a formula, let x be a variable, and let t be a term. Assume that,
for any model A and any variable assignment s, if s agrees with s except
that if s (x) = dens (t) then
A
tvs () = tvs (x; t).
A
A
Prove t
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises
2.1, 2.7, 3.1, 3.2, 3.3, and 3.4
Exercise 2.1: Let P be the property of being a formula such that
tvs1 () = tvs2 () for all variable assignments s1 and s2 such that s1 () =
A
A
s2 () for every variable x
MATHEMATICS 114L
SPRING 2014
Solutions for 1st Midterm
1. Note that
x is valid
for all A for all s, tvs (x) = T
A
for all A for all s and s , if s (y) = s(y) for all y = x, then tvs () = T
A
for all A for all s , tvs () = T
A
is valid.
2. (a) |= . Let A
MATHEMATICS 114L
SPRING 2014
Solutions for 2nd Midterm
1. Assume (a). Let be a set of sentences and let be a sentence. Assume
that |= . Since is true in every model in which is true, cfw_
is not satisable. By (a), let be a nite subset of cfw_ that is no
MATHEMATICS 114L
SPRING 2014
Solutions for More Practice Problems for 2nd Midterm
1. By the Lemma 5.11, multiplication is represented in Q by v1 v2 = v3 . It
follows that the function b bb is represented in Q by v1 v1 = v2 . Since a
is a square b(b < S(a)
MATHEMATICS 114L
SPRING 2014
More Practice Problems for 2nd Midterm
1. Use Lemma 5.4 to show that the set of squares (i.e., of numbers of the
form a2 ) is representable in Q.
2. Give a formula that represents the set of squares in Q.
3. Let f : N3 N, g1 :
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises 6.6, 6.7, and 6.8
Exercise 6.6. The new system would not be sound. The formula
v1 = v2 v2 v1 = v2
would be deducible from no premises, and this formula is not valid.
Invalid sentences would also be deduc
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises 5.1 and 5.2
Exercise 5.1. Since the axioms of Q are true in N,
Q |= is true in N,
for every sentence of LA . Thus in both this exercise and Exercise 5.2. we
have to prove only only the part of the bicond
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises 4.3 and 4.4
Exercise 4.3: First assume that our system of deduction for L is complete. Let be a consistent set of sentences. Let be any sentence. The
consistency of implies that ( ). By Completeness, |=
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises 6.1 and 6.2
Exercise 6.1. For sg, we use primitive recursion. let f be 0 (the constant
function of 0 arguments with value 0), and let g(a, b) = 1. Clearly sg(0) = f
and, for all n, sg(S(n) = g(n, sg(n).
MATHEMATICS 114L
SPRING 2014
Solution to Exercise 5.7
(a) a is a prime number if and only if a 2 and
b < a(b > 1 b does not divide a).
(b) (a, b) is a pair of adjacent primes if and only if a is prime and b is prime
and a < b and
c < b(a < c c is not prim
MATHEMATICS 114L
SPRING 2014
Solution to 3rd Exercise of Homework 3
Statement of Problem. For terms t and t and variable x, let t (x; t) be
the result of replacing the occurrences of x in t by occurrences of t.
Let A be a model and let s be a variable ass
MATHEMATICS 114L
SPRING 2014
Solutions to Exercises 5.3, 5.4, 5.5, and 5.6
Exercise 5.3. Assume that t is u1 u2 . Let j1 = (u1 )N and let j2 = (u2 )N .
By the fact that u1 and u2 are shorter than t,
(a) Q |= u1 = Sj1 0
(b) Q |= u2 = Sj2 0.
By mathematical