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IGNMENT
创建日期：2024-04-08 16:20:39
标签： IGNMENT
SOLUTIONS FOR ASSIGNMENT

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SOLUTIONS FOR ASSIGNMENT 1

(1) (a) (i) [? 4 points ?] Claim: P ⇒ Q, Q⇒ R ` P ⇒ R
Proof.
1 (1) P ⇒ Q A
2 (2) Q⇒ R A
3 (3) P A
1,3 (4) Q MP 1,3
1,2,3 (5) R MP 2,4
1,2 (6) P ⇒ R CP 3,5

(ii) [? 4 points ?]Claim: P, P ⇔ Q ` Q
Proof.
1 (1) P A
2 (2) P ⇔ Q A
2 (3) (P ⇒ Q) ∧ (Q⇒ P ) Def 2
2 (4) P ⇒ Q ∧ E 3
1,2 (5) Q MP 1,4

(iii) [? 4 points ?]Claim: P, ∼ (P ∧Q) ` ∼ Q
Proof.
1 (1) P A
2 (2) ∼ (P ∧Q) A
3 (3) Q A
1,3 (4) P ∧Q ∧ I 1,3
1,2,3 (5) (P ∧Q) ∧ (∼ (P ∧Q)) ∧ I 2,4
1,2 (6) ∼ Q RAA 3,5

(iv) [? 4 points ?]Claim: P ∧ (Q ∨R) ` (P ∧Q) ∨ (P ∧R)
Proof.
1 (1) P ∧ (Q ∨R) A
1 (2) P ∧E 1
1 (3) Q ∨R ∧E 1
4 (4) Q A
1,4 (5) P ∧Q ∧I 2,4
1,4 (6) (P ∧Q) ∨ (P ∧R) ∨I 5
7 (7) R A
1,7 (8) P ∧R ∧I 2,7
1,7 (9) (P ∧Q) ∨ (P ∧R) ∨I 8
1 (10) (P ∧Q) ∨ (P ∧R) ∨E 3,4,6,7,9

1
2 SOLUTIONS FOR ASSIGNMENT 1
(b) (i) [? 2 points ?]
P Q ∼ P P ⇒ Q (∼ P )⇒ (P ⇒ Q)
T T F T T
T F F F T
F T T T T
F F F T T
(ii) [? 2 points ?]
P Q R P ∧Q ∼ R P ∧ (∼ R) (P ∧Q) ∨ (P ∧ (∼ R))
T T T T F F T
T T F T T T T
T F T F F F F
T F F F T T T
F T T F F F F
F T F F T F F
F F T F F F F
F F F F T F F
(c) (i) [? 4 points ?]
(∼ Q) ⇒ (P ⇒ (Q ⇒ R))
T F F T F T F F
2 8 1 4 3 6 5 7
Comparing steps 6 and 8 we reach a contradiction, hence it’s a theorem.
(ii) [? 4 points ?]
(P ⇒ R) ⇒ ((Q ⇒ R) ⇒ ((P ∨Q) ⇒ R))
F T F F F T F F T F F
11 2 9 1 10 4 8 3 6 5 7
Comparing steps 6,10 and 11 we reach a contradiction, hence it’s a theorem.
(2) (a) (i) [? 3 points ?] Suppose the sequent (P ⇒ Q)⇒ (∼ P ), P ⇒ Q ` P is valid. Then
by the Soundness Theorem (P ⇒ Q) ⇒ (∼ P ), P ⇒ Q ( P is valid as well.
But suppose that V (P ) = F . Then we would have that V (P ⇒ Q) = T and
V ((P ⇒ Q) ⇒ (∼ P )) = T . This shows that (P ⇒ Q) ⇒ (∼ P ), P ⇒ Q * P .
This is a contradiction, and hence the sequent cannot be valid.
(ii) [? 3 points ?] We apply the same idea as part (i) above. This time, we take
V (P ) = F and V (Q) = T . Then V ((P ⇒ Q) ⇒ (Q ⇒ P )) = F . This shows
that * (P ⇒ Q) ⇒ (Q ⇒ P ), so by the Soundness Theorem it follows that the
sequence ` (P ⇒ Q)⇒ (Q⇒ P ) is not valid.
(b) [? 4 points ?] Consider the sequent S : W1, ...,Wn ` W . We let N ≥ 1 and suppose
the Soundness Theorem holds for any sequent which has a proof in at most N lines.
Suppose S has a proof in N + 1 lines, and the last line is an application of DN . Then
SOLUTIONS FOR ASSIGNMENT 1 3
the proof looks like:
...
A (r) X · · ·
...
A (N + 1) W DN, r
Here X is a wff and A is a set of assumptions. Note that since this is a valid proof of
W we must have that A ⊂ {W1, ...,Wn}.
The first r lines of this proof prove that the sequent W1, ...,Wn ` X is valid. Since
r ≤ N the inductive hypothesis applies to this sequent, and we can deduce that
W1, ...,Wn ( X holds. Now since we are applying DN we must have that either
X =∼ (∼W ) or W =∼ (∼ X).
We want to show that W1, ...,Wn (W holds. To this end, suppose that V (W1) = · · · =
V (Wn) = T . Then V (X) = T since W1, ...,Wn ( X. In either case (X =∼ (∼ W ) or
W =∼ (∼ X)) this implies that V (W ) = T as well, completing this proof.
(c) Suppose W is built out of propositional variables P1, ..., Pn. Fix a choice of truth values
V (P1), ..., V (Pn) ∈ {T, F}. Recall that from this choice we define
Wi =
{
Pi if V (Pi) = T, and
∼ Pi if V (Pi) = F,
and the Key Lemma states: V (W ) = T ⇒ W1, ...,Wn ` W , and V (W ) = F ⇒
W1, ...,Wn ` ∼W .
(i) [? 6 points ?] Now suppose that W = X ∨ Y where X,Y are some wffs. Note
that necessarily X,Y are also built out of P1, ..., Pn and `(X), `(Y ) ≤ N . Hence
the Key Lemma holds for X and Y .
If V (W ) = T then V (X) = T or V (Y ) = T . In the case V (X) = T the
Key Lemma implies that S1 : W1, ...,Wn ` X is valid. Here is a proof that
W1, ...,Wn `W :
1 (1) W1 A
...
n (n) Wn A
1, ..., n (n + 1) X SI(S1) 1, ..., n
1, ..., n (n + 2) X ∨ Y ∨I n + 1
1, ..., n (n + 3) W Def n + 2
The case V (Y ) = T is the same, except the roles of X and Y are interchanged
in the above proof.
If V (W ) = F then V (X) = V (Y ) = F . Since `(X), `(Y ) ≤ N we can use the
Key Lemma to deduce that S2 : W1, ...,Wn `∼ X and S3 : W1, ...,Wn `∼ Y are
both valid. We’d like to prove that W1, ...,Wn ` ∼W is valid.
Before doing so, we need to a lemma, which is essentially a part of de Morgan’s
laws in syntactic form. Recall that in lecture we proved the from a contradiction
one can prove anything, i.e. that the sequent S4 : S ∧ (∼ S) ` T is valid, where
S, T are any wffs.
4 SOLUTIONS FOR ASSIGNMENT 1
Now we will prove the following sequent is valid
S5 : (∼ X) ∧ (∼ Y ) `∼ (X ∨ Y )
1 (1) (∼ X) ∧ (∼ Y ) A
1 (2) ∼ X ∧E 1
1 (3) ∼ Y ∧E 1
4 (4) X ∨ Y A
5 (5) X A
1,5 (6) X ∧ (∼ X) ∧I 1, 5
7 (7) Y A
1,7 (8) Y ∧ (∼ Y ) ∧I 3, 7
1,7 (9) X ∧ (∼ X) SI(S4) 8
1,4 (10) X ∧ (∼ X) ∨E 4, 5, 6, 7, 9
1 (11) ∼ (X ∨ Y ) RAA 4, 10
Note that in line (9), we are applying sequent introduction with substitution,
where we substitute S = Y and T = X ∧ (∼ X).
Now we are ready to prove that W1, ...,Wn ` ∼W is valid:
1 (1) W1 A
...
n (n) Wn A
1, ..., n (n + 1) ∼ X SI(S2) 1, ..., n
1, ..., n (n + 2) ∼ Y SI(S3) 1, ..., n
1, ..., n (n + 3) (∼ X) ∧ (∼ Y ) ∧I n + 1, n + 2
1, ..., n (n + 4) ∼ (X ∨ Y ) SI(S5) n + 3
1, ..., n (n + 5) ∼W Def n + 4
(ii) [? 6 points ?] In preparation for this case we need two sequents. The first is:
S6 : (∼ X) ∨ Y ` X ⇒ Y
1 (1) (∼ X) ∨ Y A
2 (2) X A
3 (3) ∼ X A
2,3 (4) X ∧ (∼ X) ∧I 2,3
2,3 (5) Y SI(S4) 4
6 (6) Y A
1,2 (7) Y ∨E 1,3,5,6,6
1 (8) X ⇒ Y CP 2, 7
Before we prove the second sequent, we note an easy theorem. Suppose S, T are
wffs. Then
S ` T =⇒ ∼ T ` ∼ S (0.1)
Indeed, suppose S7 : S ` T is valid. There here’s a proof of ∼ T ` ∼ S:
1 (1) ∼ T A
2 (2) S A
2 (3) T SI(S7) 2
1,2 (4) T ∧ (∼ T ) ∧I 1,3
2,4 (5) ∼ S RAA 2, 4
SOLUTIONS FOR ASSIGNMENT 1 5
Now, the second sequent we want to prove is S8 : ∼ ((∼ X) ∨ Y ) ` ∼ (X ⇒ Y ).
By (0.1) this will follow if we prove S9 : X ⇒ Y ` (∼ X) ∨ Y . Here’s a proof of
S9:
1 (1) X ⇒ Y A
(2) X ∨ (∼ X) TI
3 (3) X A
1,3 (4) Y MP 1, 3
1,3 (5) (∼ X) ∨ Y ∨I 4
6 (6) ∼ X A
6 (7) (∼ X) ∨ Y ∨I 6
1 (8) (∼ X) ∨ Y ∨E 2, 3, 5, 6, 7
This proves S9 and hence also S8.
Now we suppose that W = X ⇒ Y . Since `(X), `(Y ) ≤ N the Key Lemma
applies to X,Y . If V (W ) = T we have that either V (X) = F or V (X) =
V (Y ) = T .
If V (X) = F then we get that S2 is valid. Here’s a proof that W1, ...,Wn `W is
valid:
1 (1) W1 A
...
n (n) Wn A
1, ..., n (n + 1) ∼ X SI(S2) 1, ..., n
1, ..., n (n + 2) (∼ X) ∨ Y ∨I n + 1
1, ..., n (n + 3) X ⇒ Y SI(S6) n + 2
1, ..., n (n + 4) W Def n + 3
If V (X) = V (Y ) = T then in particular we get that S10 : W1, ...,Wn ` Y is
valid. Then the proof of W1, ...,Wn ` W is the same as the above one, except
we replace the n + 1-th line by
1, ..., n (n + 1) Y SI(S10) 1, ..., n
If V (W ) = F then V (X) = T and V (Y ) = F . Hence we get that S1 and S3 are
valid. Here’s a proof that W1, ...,Wn ` ∼W is valid:
1 (1) W1 A
...
n (n) Wn A
1, ..., n (n + 1) X SI(S1) 1, ..., n
1, ..., n (n + 2) ∼ Y SI(S3) 1, ..., n
1, ..., n (n + 3) ∼ (∼ X) DN n + 1
1, ..., n (n + 4) ∼ (∼ X) ∧ (∼ Y ) ∧I n + 2, n + 3
1, ..., n (n + 5) ∼ ((∼ X) ∨ Y ) SI(S5) n + 4
1, ..., n (n + 6) ∼ (X ⇒ Y ) SI(S8) n + 5
1, ..., n (n + 7) ∼W Def n + 6
(3) (a) [? 5 points ?] First we note the following two computations:
25 ≡ 4 (mod 7)
42 ≡ 2 (mod 7) =⇒ 44 ≡ 4 (mod 7)
6 SOLUTIONS FOR ASSIGNMENT 1
Using these computations we proceed as follows:
2525 ≡ 425 (mod 7)
≡ 4244 (mod 7)
≡ (44)64 (mod 7)
≡ 464 (mod 7)
≡ 4443 (mod 7)
≡ 4(43) (mod 7)
≡ 44 (mod 7)
≡ 4 (mod 7)
Hence if today is Monday in 2525 days it will be Friday.
(b) (i) [? 5 points ?]First we note that we have x = dk(10
k−1) + dk−1(10k−2) + · · · +
d2(10) + d1. Observe that:
10 ≡ 1 (mod 9)
10 ≡ 1 (mod 3)
10 ≡ −1 (mod 11)
Therefore we have that
x ≡ dk(10k−1) + dk−1(10k−2) + · · ·+ d2(10) + d1 (mod 9)
≡ dk(1k−1) + dk−1(1k−2) + · · ·+ d2(1) + d1 (mod 9)
≡ dk + dk−1 + · · ·+ d2 + d1 (mod 9)
Replacing 9 by 3 above we also get that x ≡ dk + dk−1 + · · ·+ d2 + d1 (mod 3).
We also have that
x ≡ dk(10k−1) + dk−1(10k−2) + · · ·+ d2(10) + d1 (mod 11)
≡ dk(−1)k−1 + dk−1(−1)k−2 + · · ·+ d2(−1) + d1 (mod 11)
≡ d1 − d2 + d3 − · · ·+ (−1)k−1dk (mod 11)
(ii) [? 4 points. Some people may have interpreted this question as asking to apply
the divisibility test on four separate 3 digit numbers. That’s ok - no penalty for
my ambiguously phrased question. ?]We can use to define easy divisibility tests
for any number. For instance, to check if a number is divisible by 9, we just add
all its digits and check if the result is divisible by 9. The same would hold if 9
is replaced by 3. To check if a number is divisible by 11 we take the alternating
sum of all its digits and check if that is divisible by 11.
For instance, the sum of the digits of x = 911, 219, 186, 429 is 53 which is not
divisible by 3 nor 9. Hence x is not divisible by 3 nor 9. The alternating sum of
the digits of x is 9− 2 + 4− 6 + 8− 1 + 9− 1 + 2− 1 + 1− 9 = 13 which is not
divisible by 11. Hence x is also not divisible by 11.
SOLUTIONS FOR ASSIGNMENT 1 7
(4) (a) [? 6 points ?] As sets we have R = {I, 1 + I, x + I, x + 1 + I} and S = {J, 1 + J, x +
J, x + 1 + J}. Here is the addition table for R:
I 1 + I x + I x + 1 + I
I I 1 + I x + I x + 1 + I
1 + I 1 + I I x + 1 + I x + I
x + I x + I x + 1 + I I 1 + I
x + 1 + I x + 1 + I x + I 1 + I I
The addition table for S is the same, except every instance of I should be replaced by
a J .
Here is the multiplication table for R:
R I 1 + I x + I x + 1 + I
I I I I I
1 + I I 1 + I x + I x + 1 + I
x + I I x + I 1 + I x + 1 + I
x + 1 + I I x + 1 + I x + 1 + I I
Here is the multiplication table for S:
S J 1 + J x + J x + 1 + J
J J J J J
1 + J J 1 + J x + J x + 1 + J
x + J J x + J x + J J
x + 1 + J J x + 1 + J J x + 1 + J
(b) [? 6 points ?] We claim that R and S are not isomorphic. Indeed, suppose we have an
isomorphism f : R→ S. Since we must have that f(I) = J , i.e. f maps the zero in R
to the zero in S, it follows that f induces a bijection between the nonzero elements of
R to the nonzero elements of S. Therefore
f((1 + I)(x + I)(x + 1 + I)) = f(1 + I)f(x + I)f(x + 1 + I) (0.2)
= (1 + J)(x + J)(x + 1 + J) (0.3)
This implies that f(x + 1 + I) = J , a contradiction since the isomorphism can’t map
a nonzero element of R to the zero element of S.
Note that in (0.3) we are not claiming that f(1+I) = 1+J, f(x+I) = x+J, f(x+1+I) =
x + 1 + J . We are using the fact that
{f(1 + I), f(x + I), f(x + 1 + I)} = {1 + J, x + J, x + 1 + J}
(5) (a) [? 6 points ?]First we show that ∼ is an equivalence relation. For x, y, z ∈ X we have:
x ∼ x since f(x) = f(x) (reflexivity), x ∼ y =⇒ f(x) = f(y) =⇒ f(y) = f(x) =⇒ y ∼
x (symmetry), and (x ∼ y) ∧ (y ∼ z) =⇒ (f(x) = f(y)) ∧ (f(y) = f(z)) =⇒ f(x) =
f(z) =⇒ x ∼ z (transitivity).
Define ϕ : X/∼ → im(f) by ϕ([x]) = f(x). We claim that ϕ is a well-defined function.
Suppose that [x] = [y]; we need to show that ϕ([x]) = ϕ([y]). Indeed,
[x] = [y] =⇒ x ∼ y =⇒ f(x) = f(y) =⇒ ϕ([x]) = ϕ([y])
Now we’ll show that ϕ is a bijection. To prove injectivity, suppose that ϕ([x]) = ϕ([y])
for [x], [y] ∈ X/∼. Then f(x) = f(y), so x ∼ y, and hence [x] = [y]. To prove
surjectivity, let z ∈ im(f). Then z = f(x) for some x ∈ X, and hence ϕ([x]) = z.
8 SOLUTIONS FOR ASSIGNMENT 1
(b) [? 6 points ?]We define ϕ : Z[x]→ Z[i] by ϕ(p(x)) = p(i). This is a ring homomorphism
since if p(x), q(x) ∈ Z[x] then
ϕ(p(x)) + ϕ(q(x)) = p(i) + q(i) = ϕ(p(x) + q(x))
ϕ(p(x))ϕ(q(x)) = p(i)q(i) = ϕ(p(x)q(x)).
Given a + bi ∈ Z[x] we have that ϕ(a + bx) = a + bi, proving that ϕ is surjective, i.e.
im(ϕ) = Z[i]. Next, we claim that ker(ϕ) = I. Indeed, I ⊂ ker(ϕ) since if p(x) ∈ I
then p(x) = (x2 + 1)q(x) for some q(x) ∈ Z[x]. Hence p(i) = (i2 + 1)q(i) = 0, so
p(x) ∈ ker(ϕ). Conversely, if p(x) ∈ ker(ϕ) then p(i) = 0. By the fact in the statement
of the problem, this implies that x2 + 1 divides p(x), i.e. p(x) ∈ I. This shows that
ker(ϕ) = I. By the First Isomorphism Theorem we conclude that Z[x]/I ∼= Z[i].
(6) (a) [? 6 points ?] Let x + ay, x′ + ay′ ∈ J . Then
(x + ay) + (x′ + ay′) = (x + x′) = a(y + y′) ∈ J
since x+ x′ ∈ I. Moreover, if z ∈ R then (x+ ay)z = xz + ayz ∈ J since xz ∈ I. This
shows that J is an ideal. We have that I ⊂ J since for any x ∈ I, x = x + a0 ∈ J .
(b) [? 6 points ?] Let a + I ∈ R/I be a nonzero element. To show that R/I is a field we
need to show that a + I has a multiplicative inverse. Since a + I is nonzero we have
that a /∈ I. Consider J = I + aR as in part (a). By (a), J is an ideal of R containing
I. Moreover, since a ∈ J and a /∈ I we have that I 6= J . Since I is large we must have
that J = R. In particular, this implies that 1 ∈ J , i.e. 1 = x + ay for some elements
y ∈ R, x ∈ I. Hence we get that
(a + I)(y + I) = ay + I
= ay + x + I
= 1 + I
Note that the second equality follows since (ay + x)− (ay) ∈ I. This shows that y + I
is the multiplicative inverse of a + I, and hence we conclude that R/I is a field.
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EARN 5 FE 825 IERG 5130 STAT 8178 EE161 ADS2 G54FUZ CS326 113B ECN 226 QBUS6860 ELEC3202 COMP9334 MGT11B MET CS 777 STA238 00099F Python SMM306 CMT206 EC116 SP MAST90084 CS480/680 BCPM0091 CMPT 300 ECON2209 FY20 STAT 432 CVEN9625 MATH1712 CMP9771M CMP3750M CIS 325 AI506 MAT344H1S MGT7180 PSTAT 160B CSYS5010 1D STAT 453/558 ME152B STA304H1F CSCI-128 ME428 CO3002 FINE 7640 COMP3234 ECSE 223 EE497 – 3D MATH5905 ECS648U ENG5009 MATH10242 EC665 A CPI221 ISOM5220 MACE40402 MGT B456F COMP9315 AFM 113 STAT 778 CS 564 MA231 CS 230 CS 124 COMP5511 CSC317 ITEC1010 ELEC207 INFT 3019 7CCMFM06 EE 161 5G ECLT5810 MATH 7349 CS526 O2 COMP90054 ELE00063M ACCT5034 ARC1015 CSCI 40 CSCI E40 FIN 524B CS 462 389COM CMT309 MFIN703 COMP4250 STA304 STA304H1S STA 1003 FIN 448 BUFN 732 CS520 CVEN3701 ECMT6006 PHAS0100 CSIT314 MDIA 5028 CSCI 2134 MAT 451 CMPT 459 STAT 457 INFO20003 COMP4141 IELE8066 F540 ECM9 CSE 6321 FIN 430 MATH3911 BFC5916 COMP1011 EEE 6209 STATS 240 CW3 COMP206 COMP1003 4331W ISE 530 CMPT 295 MATH 239 QF5203 INFR08010 IT114105 FIT2093 ECON3124 ECON1195 MATH 337 STAT 3006 COMP202 COMP3130 MATH3609 COMP 250 CS6140 IEEE ELE00041I FIN40090 CIV 5899 MATH3066 IGNMENT MATH 2009 STAT3405 HIST 101 MATH 3281 CEGE0042 ICT532 LAB 7 LAB 5 MCO455 CSIT214 Q4 IFB104 A1 ECO220Y5Y DA5020 CMPSC473 FIN9007 DSCI553 MSCI534 COMP3308 CSE-112 CMPT 310 EE104 F 2C P20 ALY2010 BUSM060 STA258 ECO00030M R BIOSTAT 274 COS6012-B FIT3155 CMPUT 340 EC421 INFS4205 ECOM135 NATS 1740 COMP809 MBE 6005 STA465 CSSE2310 CIVL2700 DATA4700 CMPSC 473 ISE 529 CS5044 MAST30020 COMP2207 ENGG1001 COMP0046 FIT3165 STAT GR5241 MECH0007 vATHK1001 ANALYTIC ATHK1001 MATH352001 HW3 S1 BMAN 24662 COMP 636 Stat-461 STAT 302 STAT 2132 STAR534 ENG4137 CMP3752M CS3103 BISM3222 STA 135 ME155A IIMT3601 ECON3371 ECON 3371 HW4 FINA 5840 DT211C CTEC3902 MAT005 MATH2148 S261F Econ 332 MATH3734 MIS41270 7CCMFM13 SYST 664 CIS434 CENG10014 MT4111 MT4527 ECMM424 CMT219 Math 136 MATH 3120 COMP10002 MAST20009 MATH0058 BEEM117 ISYE 6420 MGMTMFE 405 25579 1ECE 273 M122 COMP4434 1ELEC ELEC5620M STAT0010 BFF2401 MXB361 MATH3811 CS330 CMT307 FIT1033 CptS 583 UP 431 MATH6005 AF5343 CS2435 ECOM30002 ES50061 CS 2506 ECMT6002 CODE 00099F MT4539 ESSAY ICM317 MATH1231 MFIT5010 BMA547 MATH334 CIVL2201 MATH39032 T-04 CIVE97119 EMESTER CAB302 CAB301 59PM STAT802 ISYS3412 RM3 COSC 2123 AcF 311 MATH2022 1FINAL 4CCS1PHC ICM 142 ENVS3023 Math 4023 I1 ACSE8 FIN B385F CS-275 STA 4109-01 A31021 EE140 MT1 EE124 ECO-6004B Inf2-IADS CS126 IS2184 UL18 MATH5092 ECOM009 COMP 6211 ELEC202 MKT3019 IAB230 FIT5046 584-S21 COMP3210 CO250 A29401 MATH 136 MGT6174 ECMT6007 DATA2001 MAT 443 MATH10222 PHDE0071 ELE00123M COSC1295 MAT00045H MATH32032 1004GRC MAS223 COMP10001 DATA3404 CIVL2110 STAT2008 FNCE10002 EEE6431 COM6515 COMP9120 NBS8185 ECTE942 ECON 405 MATH314301 STAT 2011 COMPSCI 350 COMP20007 CMPSC-132 ECON5002 STA 35 MTH6108 FIT5147 MTH6150 100C FINA 5260 DESC9115 ECON4998 NBS8257 GMM ECON 6070 BSA2 ECE599 STATS 102B STA120 INFS601 BFC3241 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MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A 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COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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210FIN316 MGMT10002 COMP0050 EC306 FIT 5124 59PM ET S2098367 CAD MMGT6001 COVID-19 COMP5940M ARC181H1S SCC 312 ECO102 MATH4061 ELEC 3848 DATA3888 MECH0024 STA302H1 CSE 510 LAWS8214 WORK6108 E2 ECON 7310 COMP3211 FIT5216 A4 MATH 110 IBUS 1101 CS355 SWEN90004 MATH1023 BEEM012 COMP3311 ACCT90013 EES A06 EC 504 ISE 580 ENG4701 CAS CS538 ELE3040 6013B0506Y BENV0115 ECON3104 CSCI-1200 COMP27112 IEOR4703 ELEC271 ECE-GY Math 412 BUSS1000 PP208 BCPM000028 7QQMM316 BISM7209 DASC7606 HW 4 STAT 4620 ECON8026 INFS3202 COMMGMT 3506 COMU1002 ECON 20003 MATH 348 BUSN7008 BENV0113 STATISTICS Function STAT2001 ECE6483 MATH1115 MGMT20002 CONSUMPTION ECO00003I LAWS6997 APMA 4300 ECON 7030 STAT 413 802G5 GEOG0051 LINA02 ECON 3208 GOVT 1621 ECOS2001 ECON 7520 MGTS7301 ECMT 6002 MATH4321 EIBS7300 MARK5813 BH50A0220 OFFICE280 ECON1012 CIS273 CIS 273 ELEC3106 ACCT3013 EC831 MMAN3400 EC365 MAST20006 ECON7520 ENGM030 CSC148H1S COMP5125M CMPSC 497 ST5188 MAST 90082 CSSE3010 ECON7040 GEOM3000 GEOM30009 ACCT5910 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