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THE UNIVERSITY OF SYDNEY
MATH3066 ALGEBRA AND LOGIC
Semester 1
First Assignment
2014
This assignment comprises a total of 60 marks, and is worth 15% of the overall
assessment. It should be completed, accompanied by a signed cover sheet, and handed
in at the lecture on Thursday 17 April. Acknowledge any sources or assistance.
1. Construct truth tables for each of the following ws:
(a)
(P ∨ Q) ∧ R
(b)
(P ∧ R ) ∨ Q
Use your tables to explain briey why
(P ∨ Q) ∧ R
|=
(P ∧ R ) ∨ Q ,
(P ∧ R ) ∨ Q
|=
(P ∨ Q) ∧ R .
but
(6 marks)
2. Use truth values to determine which one of the following ws is a theorem (in
the sense of always being true).
(a)
(b)
P ⇒ Q⇒R
⇒
P ⇒Q ⇒R
P ⇒Q ⇒R ⇒ P ⇒ Q⇒R
For the one that isn’t a theorem, produce all counterexamples. For the one
that is a theorem, provide a formal proof also using rules of deduction in the
Propositional Calculus (but avoiding derived rules of deduction).
(8 marks)
3. Use the rules of deduction in the Propositional Calculus (but avoiding derived
rules) to nd formal proofs for the following sequents:
(a)
P ⇒ (Q ⇒ R ) , ∼ R
⊢
(b)
(P ∨ Q) ∧ (P ∨ R )
P ∨ (Q ∧ R )
(c)
P ∨ (Q ∧ R ) ⊢ (P ∨ Q) ∧ (P ∨ R )
⊢
P ⇒∼Q
(12 marks)
4. Let W = W (P1 , . . . , Pn ) be a proposition built from variables P1 , . . . , Pn . Say
that W is even if
W ≡ W ( ∼ P1 , ∼ P2 , . . . , ∼ Pn ) .
Say that W is odd if
W ≡ ∼ W ( ∼ P1 , ∼ P2 , . . . , ∼ Pn ) .
(a) Use truth tables to decide which of the following are even or odd:
(i) W = (P1 ⇔ P2 )
(ii) W = (P1 ⇔ P2 ) ⇔ P3
(b) Use De Morgan’s laws and logical equivalences to explain why the following
proposition is odd:
W=
P1 ∨ P2 ∧ P3 ∨ P1 ∧ P2
(c) Explain why the number of truth tables that correspond to propositions
n
n −1
in variables P1 , . . . , Pn is 22 , and, of those, 22
tables correspond to
2 n −1
tables correspond to odd propositions.
even propositions, and 2
(16 marks)
5. Evaluate each of
in Z11
3
9
10
1
,
,
,
,
5
7
10
9
and Z14 , or explain briey why the given fraction does not exist.
(8 marks)
6. Prove that the only integer solution to the equation
x2 + y 2 = 3 z 2
is x = y = z = 0.
[Hint: rst interpret this equation in Zn for an appropriate n.]
