Warning: this is a 50-minute test.
The Midterm Exam will be longer and cover more material,
including the topics covered in class on Tuesday, October 18.
Math 2000 Midterm Practice Exam (Fall 2011)
50 minute time limit.
No books, notes, calculators, or collaboration.
1) Provide a justification (rule and line numbers) for each line of this proof:
1
(H & I) ⇒ J
2
¬J ⇒ H
3
I
4
H ∨ ¬H
5
H
6
H &I
7
J
8
H⇒J
10
¬H ⇒ ¬¬J
11
¬H ⇒ J
12
J
hypothesis
hypothesis
hypothesis
Law of Excluded Middle
assumption (want J)
&-intro (lines 5 and 3)
⇒-elim (lines 1 and 6)
⇒-intro (lines 5–7)
contrapositive of line 2
Rules of Negation applied to line 10
Proof by cases (lines 4, 8, and 11)
2) Assume h, c, x ∈ Z. Show that if hc | x, then h | cx.
PROOF. Since hc | x, we know there exists k ∈ Z, such that (hc)k = x. Let
k 0 = c2 k. Then
hk 0 = h(c2 k) = c (hc)k = c x = cx,
so h | cx.
2
3) Explain how you know that the following deduction is not valid.
Hypotheses:
1. ∀t ∈ T, (t ∈ E)
2. ∃t ∈ T, (t ∈ F )
Conclusion: ∀t ∈ T, (t ∈ F )
Let T = {1, 2}, E = {1, 2}, and F = {1}. Then:
• T = E, so every element of T is also an element of E. That is, Hypothesis 1
is true.
• 1 ∈ T , and we also have 1 ∈ F , so (letting t = 1) ∃t ∈ T, (t ∈ F ) is true.
That is, Hypothesis 2 is true.
So both hypotheses are true.
On the other hand, 2 ∈ T , but 2 ∈
/ F , so the assertion ∀t ∈ T, (t ∈ F ) is
false. That is, the conclusion is false.
Since we have a counterexample in which both hypotheses are true, but the
conclusion is false, the deduction is not valid.
4) Assume
(a) G ⊂ R,
(b) H ⊂ R,
(c) for every s ∈ R, if s2 = 2s, then s + 1 ∈ G, and
(d) for every g ∈ G, we have g 2 ∈ H.
Show H 6= ∅.
PROOF. Let s = 2 ∈ R. Then s2 = 22 = 4 = 2(2) = 2s,
so, from Assumption (c), we know s + 1 ∈ G.
Since s + 1 = 2 + 1 = 3, this means 3 ∈ G.
Therefore Assumption (d) tells us 32 ∈ H.
So H has at least one element (namely, 32 ), so H 6= ∅.
3
5) Using the symbolization key
U : the set of all university classes
F : the set of all classes that are fun
W : the set of all classes that are worthwhile
M : the set of all math classes
L: the set of all classes in Lethbridge
translate the following assertion into the notation of First-Order Logic:
If some math class in Lethbridge is worthwhile, then every fun class is a math class.
∃m ∈ (M ∩ L), m ∈ W ⇒ ∀f ∈ F, f ∈ M
6) Give a two-column proof of the following deduction:
(A ∨ B) ⇒ (C ∨ D),
B ⇒ ¬D,
.˙. B ⇒ C
1
(A ∨ B) ⇒ (C ∨ D)
hypothesis
2
B ⇒ ¬D
hypothesis
3
B
4
A∨B
∨-intro (line 3)
5
C ∨D
⇒-elim (lines 1 and 4)
6
¬D
⇒-elim (lines 2 and 3)
7
C
∨-elim (lines 5 and 6)
8
assumption (want C)
B⇒C
⇒-intro (lines 3–7)
4
7) Short answer.
(You do not need to show your work and you do not need to answer in complete sentences.)
(a) What does it mean to say that X is disjoint from Y ?
X ∩Y =∅
(b) What is an assertion?
A sentence that is either true or false.
(c) What is the power set of A?
It is the set of all subsets of A.
(d) What is P ∩ Q?
P ∩ Q = { x | (x ∈ P ) & (x ∈ Q) }
(e) Assume L = {1, 2, 3, 4, 5}. Specify the following set by listing its elements:
{ ` ∈ L | `2 > 7` − 11 } = {1,2,5}
5
8) Simplify the following assertion (so that ¬ is applied only to atomic predicates).
You do not need to show your work.
!
¬ ∀a ∈ A, (a R a) ⇒ ∃p ∈ P, (p ∈ Z) & (p 6= a) ∨ (a R p)
!
∃a ∈ A, (a R a) & ∀p ∈ P, (p ∈
/ Z) ∨ (p = a) & ¬(a R p)
9) Assume I, J, and K are sets. Show that if I r J ⊂ K, then I ⊂ J ∪ K.
PROOF. Given i ∈ I, we wish to show i ∈ J ∪ K. From the Law of Excluded Middle, we know that either i ∈ J or i ∈
/ J, and we consider these two
possibilities as separate cases.
Case 1. Assume i ∈ J. Then it is obviously true that either i ∈ J or i ∈ K, so
we have i ∈ J ∪ K, as desired.
Case 2. Assume i ∈
/ J. Since i ∈ I and i ∈
/ J, we have i ∈ I r J. Since
I r J ⊂ K, this implies i ∈ K. Then it is obviously true that either i ∈ J or
i ∈ K, so we have i ∈ J ∪ K, as desired.