Math 307, Homework #3
Due Wednesday, October 19
The point of learning symbolic logic proofs is to slow your brain down and force you to think about what
exactly you are doing in your reasoning. When we start doing more advance proofs, like in problems 8–12
on this worksheet, we will ease up on this and let your brain go a little faster. But when doing symbolic
logic proofs like in problems 1–3, you need to still go slow and justify all of your steps. This goes for every
symbolic logic proof you do this quarter.
1. Show that [U ∧ P ] ⇒ [Q ∧ R], P ⇔ [S ∨ T ], R ∧ T ` U ⇒ Q.
2. Show that ∼ P ⇒ Q, P ⇒∼ Q, P ⇔ R ` ∼ Q ⇔ R. (Hint: Recall that X ⇔ Y is an abbreviation for
something.)
3. Show that R ∨ S, ∼ P , Q∨ ∼ R, P ⇔ Q ` S.
4. Translate each of the following into a normal English sentence. Also, identify the statement as True or
False. For example, the proposition
(∀x)[(x ∈ Z ∧ x|24) ⇒ x|48]
says that “Every divisor of 24 is a divisor of 48”. It is True.
(a) (∀x)[(x ∈ N ∧ x|13) ⇒ x ∈ {1, 13}]
(b) (∃a)[a ∈ N ∧ (a > 51 ∧ a < 52)]
(c) (∀x)[(x ∈ N ∧ x2 = 2) ⇒ x = 5]
(d) (∀x)[x ∈ N ⇒ (∃y)[y ∈ N ∧ y|x]]
(e) (∀n)[(n ∈ N ∧ (4|n ∧ 6|n)) ⇒ 24|n]
5. The phrase “the integer x is a perfect square” means x ∈ Z ∧ (∃y ∈ Z)[x = y 2 ]. Keeping this in mind,
write out mathematical statements—using only quantifiers and other math symbols, no English words—
which say the same thing as the following sentences. Do not worry about whether the statements are
true or false!
(a) For every integer a, if a is even then a2 is a multiple of 5.
(b) 5 is smaller than every integer.
(c) Every integer which is a perfect square is also a perfect cube.
(d) Every nonzero integer is either a multiple of six or a multiple of seven.
(e) There exists an integer which is larger than every other integer.
(f) There is an element of the set A having the property that every element of the set B divides it.
(g) Every element of the set A is either even or a multiple of 13.
(h) There exists an integer which is not divisible by any divisor of 240.
(i) Every nonzero element of Z5 has a multiplicative inverse.
(j) For all integers x, if x is congruent to three mod eight then there exists an integer y such that x · y
is congruent to one mod eight.
(k) There exists an integer p with the following properties: p is even and for every pair of integers a
and b, if p divides ab then it must divide either a or b.
6. In each part, identify the given set by listing all of its elements. For example:
Given the set {x ∈ Z | x ≡4 3 ∧ 5 ≤ x ∧ x ≤ 20} you would answer {7, 11, 15, 19}, as these two sets are equal.
(a) {x ∈ Z4 | (∃y)[y ∈ Z4 ∧ y 6= 0 ∧ xy = 0]}
(b) {x3 + 1 | x ∈ N} ∩ {y | y ∈ N ∧ 1 ≤ y ≤ 30}
(c) {x ∈ N | (∃a)(∃b)[a ∈ N ∧ b ∈ N ∧ x = a2 + b2 ]} ∩ {x | x ∈ N ∧ x ≤ 20}
(d) {z ∈ Z30 | (∃u)[u ∈ Z30 ∧ zu = 1]}
(e) {a | a ∈ Z ∧ a ≡4 1} ∩ {b | b ∈ Z ∧ b ≡2 0}
(f) {(a, b) | a ∈ Z2 ∧ b ∈ Z2 ∧ a + b = 1}
(g) {x ∈ N | x ≡5 1 ∧ x ≤ 40} ∩ {x ∈ N | x ≡6 4}
SIDE DISCUSSION
In class we learned that (∀x)[P (x)] is an abbreviation for an infinite number of AND statements, whereas
(∃x)[Q(x)] is an abbreviation for an infinite number of OR statements. You also know DeMorgan’s Laws,
saying
∼ (P1 ∧ P2 ∧ P3 ∧ · · · ) ⇔ (∼ P1 ∨ ∼ P2 ∨ ∼ P3 ∨ · · · )
and
∼ (P1 ∨ P2 ∨ P3 ∨ · · · ) ⇔ (∼ P1 ∧ ∼ P2 ∧ ∼ P3 ∧ · · · ).
These tell us that
∼ (∀x)[P (x)] ⇔ (∃x)[∼ P (x)]
and
∼ (∃x)[Q(x)] ⇔ (∀x)[∼ Q(x)].
Here is a specific example. The negation of
(∀x)[2|x]
is equivalent to
(∃x)[26 |x].
The first statement is false, and the second is true.
Usually we want to apply DeMorgan’s Laws a few times at once. For example, the negation of
(1)
(∃x)[x ∈ Z ∧ 2|x]
is
(2)
(∀x)[∼ (x ∈ Z ∧ 2|x)],
but we prefer to write that as
(3)
(∀x)[x ∈
/ Z ∨ 26 |x].
Notice that we used DeMorgan’s Laws to negate the ∃ quantifer, but then we used DeMorgan’s Laws again
to negate the AND.
Now, for various reasons we like it best when ∃ applies to AND statements and ∀ applies to IMPLIES
statements. So I’m not completely satisfied with statement (3) above, although it is technically correct. But
you know that every OR statement is equivalent to an IMPLIES statement: (X ∨ Y ) ⇔ (∼ X ⇒ Y ). So I
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can use this tautology to transform (3) into the equivalent form
(∀x)[x ∈ Z ⇒ 26 |x].
(4)
This is the best form of the negation of (1).
I can summarize the passage from (1) to (4) by saying that we used the tautology
∼ (P ∧ Q) ⇔ (P ⇒∼ Q).
The other useful tautology in this kind of problem is
∼ (P ⇒ Q) ⇔ (P ∧ ∼ Q).
As one more example, the negation of
(∀y)[(y ∈ Z ∧ 4|y) ⇒ 2|y]
is
(∃y)[(y ∈ Z ∧ 4|y) ∧ 26 |y].
Here the first statement happens to be true, and the second statement is false.
Back to homework problems...
7. In each part below, use mathematical notation to write the negation of the given statement in such a
way that no quantifier is immediately preceded by a negation sign, every universal quantifer is applied
to a conditional, and every existential quantifier is applied to a conjunction. In each part, decide which
statement is true: the given statement or its negation.
(a) (∀y)[y ∈ Z ⇒ y > 0]
(b) (∀x)[(x ∈ Z ∧ 2|x) ⇒ 4|x]
(c) (∀x)[x ∈ Z6 ⇒ (∃y)[y ∈ Z6 ∧ x +6 y = 0]]
(d) (∃x)[x ∈ Z6 ∧ (∀y)[y ∈ Z6 ⇒ x +6 y = 0]]
(e) (∀x)[x ∈ Z ⇒ (∃y)[y ∈ Z ∧ x = y 2 ]]
(f) (∃a)[a ∈ N ∧ (∀b)[[b ∈ N ∧ a 6= b] ⇒ b > a]]
(g) (∃m)(∃n)[m, n ∈ N ∧ m > n]
(h) (∀m)(∀n)[m, n ∈ N ⇒ m > n]
(i) (∃m)[m ∈ N ∧ (∀n)[n ∈ N ⇒ m > n]
(j) (∀m)[m ∈ N ⇒ (∃n)[n ∈ N ∧ m > n]
(k) (∀a)(∀b)[(a, b ∈ R ∧ a < b) ⇒ (∃c)[c ∈ R ∧ [a < c ∧ c < b]]]
Homework continues on next page.
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In class we have started doing what I will call “line proofs”. These are a little less formal than logic proofs,
in that you may condense several logical steps into one when you think it is clear enough; yet the proofs still
keep their numbered line-by-line structure. Here is an example we discussed in class:
(∀n)[(n ∈ N ∧ 4|n) ⇒ 16|n2 + 12n]
Proof:
1. Assume n ∈ N and 4|n.
2. (∃k)[k ∈ N ∧ n = 4 · k]
3. n = 4 · k, for some k ∈ N.
4. n2 + 12n = (4k)2 + 12(4k) = 16k 2 + 48k = 16(k 2 + 3k)
5. (∃r)[r ∈ Z ∧ n2 + 12n = 16 · r]
6. 16|n2 + 12n
7. (n ∈ N ∧ 4|n) ⇒ 16|n2 + 12n
8. (∀n)[n ∈ N ∧ 4|n) ⇒ 16|n2 + 12n
IE
EI
DT, discharge For 1
IU
The following exercises will help you start writing proofs like this yourself. For each “fill in the blanks”
exercise, copy the whole proof onto a sheet of your homework.
You need to know two definitions: (1) a|b ⇔ (∃k)(k ∈ Z ∧ b = ka), and (2) a ≡n b ⇔ n|a − b.
8. Fill in the blanks in the outline below to prove that (∀a, b, k)[[a, b, k ∈ Z ∧ (k ≥ 1 ∧ a|b)] ⇒ ak |bk ].
Proof:
1. Assume a, b, k ∈ Z and k ≥ 1 and a|b.
(y ∈ Z ∧ b = y · a)
2.
3.
b = y · a for some
4.
bk =
5.
(∃u)(u ∈ Z∧
6.
k
a |b
)
k
DT, discharge For 1
7.
8.
k
k
(∀a, b, k)[[a, b, k ∈ Z ∧ (k ≥ 1 ∧ a|b)] ⇒ a |b ].
Answer the following questions:
(i) The above proof used one IE step, one EI step, and one IU step. Label them in column 2 of your
proof (the same column with DT in it).
(ii) The definition of a|b used k for the variable in the existential statement. In step 2, why did we
change and use y instead?
(iii) In step 5 we introduced the variable u in the existential statement. Could we have used k here
instead? What about y?
HW continues on next page
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9. Fill in the blanks in the outline below to prove that (∀a, b, c)[[a, b, c ∈ Z ∧ (a|b ∧ b|c)] ⇒ a|c]. [Hint:
Notice that P and Q, from step 6, will have to appear in steps 1–5 somewhere.]
Proof:
1.
Assume
2.
(∃k)[k ∈ Z ∧ b = k · a]
for some
3.
4.
(∃k)[k ∈ Z ∧ c = k · b]
5.
for some
6.
c = Q · b = P · (Qa) = P Q · a
7.
(∃r)[r ∈ Z ∧
8.
a|c
]
9.
DT, discharge For 1
10.
Answer the following questions:
(i) The above proof used two IE steps, one EI step, and one IU step. Label them in column 2 of your
proof (the same column with DT in it). You do not have to give reasons for any of the other steps.
(ii) It is okay that we used k in both step 2 and step 4. Why? (This might be hard to explain in words,
but at least try to come to some kind of understanding for yourself).
HW continues on next page
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10. Fill in the blanks in the outline below to prove that (∀n)[(n ∈ N ∧ 36 |n) ⇒ n2 ≡3 1]. Also:
(i) There are three uses of the Deduction Theorem in this proof. Label the appropriate DT steps.
(ii) There are two steps at the end of the proof with the phrase “Logical rule?” in bold. For these,
label the appropriate rule from symbolic logic that is being used.
1.
Assume
2.
3.
4.
n 6≡3 0
So either n ≡3 1 or n ≡3 2
Assume n ≡3 1.
5.
Then 3|
6.
So n − 1 = 3 · P for some
7.
n=
8.
n2 =
9.
n2 − 1 =
10.
(∃y)[y ∈ Z ∧
11.
3|
12.
13.
14.
n 2 ≡3 1
So n ≡3 1 ⇒ n2 ≡3 1.
Now assume n ≡3 2.
=3·
+1
]
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
So n ≡3 2 ⇒ n2 ≡3 1.
We therefore have (n ≡3 1 ∨ n ≡3 2) ⇒ n2 ≡3 1
So n2 ≡3 1
(n ∈ N ∧ 36 |n) ⇒ n2 ≡3 1
Logical rule?
Logical rule?
29.
Now try it yourself. Give line proofs for the following statements:
11. (∀n)[(n ∈ N ∧ n ≡6 3) ⇒ n2 + 2n + 10 ≡12 1]
12. (∀n)[(n ∈ N ∧ n ≡5 2) ⇒ (26 |n ∨ n2 ≡20 4)]
[Hint: Remember that (X ∨ Y ) ⇔ (∼ X ⇒ Y ). Use DT twice in this proof.]
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