COS3761 Assignment 2 &3
solutions 2026
UNISA
Formal Logic III
THIS DOCUMENT CONTAIN TWO ASSIGNMENTS 2 and 3
All the questions are answered well.
If need help reach me I am your tutor iQ Level
,Assignment 2 — Formal Logic III (COS3761)
Predicate symbols used throughout:
Symbol Meaning
S(x) x is a student
C(x) x is a Computing module
G(x) x is a Maths module
T(x, y) x takes y
L(x, y) x loves y
K Kevin
b Bill
Question 1 — English → Predicate Logic
1.1 No students love Kevin.
¬∃𝑥 (𝑆(𝑥) ∧ 𝐿(𝑥, 𝐾))
Equivalently: ∀𝑥 (𝑆(𝑥) → ¬𝐿(𝑥, 𝐾))
1.2 Every student loves some student.
∀𝑥 (𝑆(𝑥) → ∃𝑦 (𝑆(𝑦) ∧ 𝐿(𝑥, 𝑦)))
1.3 Every student who takes a Computing module also takes a Maths module.
∀𝑥 (𝑆(𝑥) ∧ ∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝑥, 𝑚)) → ∃𝑛 (𝐺(𝑛) ∧ 𝑇(𝑥, 𝑛)))
1.4 Kevin takes a Computing module while Bill does not.
∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝐾, 𝑚)) ∧ ¬∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝑏, 𝑚))
1.5 There is at least one student who takes a Maths module.
∃𝑥 (𝑆(𝑥) ∧ ∃𝑚 (𝐺(𝑚) ∧ 𝑇(𝑥, 𝑚)))
, Question 2 — Predicate Logic → English
2.1 ∃𝑥 ∀𝑧 (𝑆(𝑥) ∧ 𝑇(𝑥, 𝐺(𝑧)) ∧ ∀𝑦(𝑆(𝑦) ∧ 𝑇(𝑦, 𝐺(𝑧)) → 𝑥 = 𝑦))
There exists a student x such that for every Maths module z, x takes that Maths
module, and x is the only student who takes it.
In plain English: There is a student who is the unique student taking every
Maths module.
2.2 ∃𝑥 (𝐶(𝑥) ∧ ∀𝑦 (𝑆(𝑦) → ¬𝑇(𝑦, 𝑥)))
There exists a Computing module that no student takes.
2.3 ∀𝑥 ∃𝑦 𝑇(𝑥, 𝑦) ∧ 𝑆(𝑥)
Note: By standard precedence, the ∧ binds less tightly than the quantifiers, so this
reads as: (∀𝑥 ∃𝑦 𝑇(𝑥, 𝑦)) ∧ 𝑆(𝑥), where the second conjunct has x free. More likely
the intended reading (treating it as a scope issue) is:
Every entity takes something, and x is a student.
If parsed as written (with x free in S(x)): Everything takes some module, and x is a
student. (x is a free variable here — the formula is not a sentence.)
If the intended reading is ∀𝑥(∃𝑦 𝑇(𝑥, 𝑦) ∧ 𝑆(𝑥)): Every student takes at least one
module.
2.4 ∀𝑥 [𝑆(𝑥) → 𝐿(𝑘, 𝑥)]
Kevin loves every student. (Using k as the constant for Kevin.)
solutions 2026
UNISA
Formal Logic III
THIS DOCUMENT CONTAIN TWO ASSIGNMENTS 2 and 3
All the questions are answered well.
If need help reach me I am your tutor iQ Level
,Assignment 2 — Formal Logic III (COS3761)
Predicate symbols used throughout:
Symbol Meaning
S(x) x is a student
C(x) x is a Computing module
G(x) x is a Maths module
T(x, y) x takes y
L(x, y) x loves y
K Kevin
b Bill
Question 1 — English → Predicate Logic
1.1 No students love Kevin.
¬∃𝑥 (𝑆(𝑥) ∧ 𝐿(𝑥, 𝐾))
Equivalently: ∀𝑥 (𝑆(𝑥) → ¬𝐿(𝑥, 𝐾))
1.2 Every student loves some student.
∀𝑥 (𝑆(𝑥) → ∃𝑦 (𝑆(𝑦) ∧ 𝐿(𝑥, 𝑦)))
1.3 Every student who takes a Computing module also takes a Maths module.
∀𝑥 (𝑆(𝑥) ∧ ∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝑥, 𝑚)) → ∃𝑛 (𝐺(𝑛) ∧ 𝑇(𝑥, 𝑛)))
1.4 Kevin takes a Computing module while Bill does not.
∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝐾, 𝑚)) ∧ ¬∃𝑚 (𝐶(𝑚) ∧ 𝑇(𝑏, 𝑚))
1.5 There is at least one student who takes a Maths module.
∃𝑥 (𝑆(𝑥) ∧ ∃𝑚 (𝐺(𝑚) ∧ 𝑇(𝑥, 𝑚)))
, Question 2 — Predicate Logic → English
2.1 ∃𝑥 ∀𝑧 (𝑆(𝑥) ∧ 𝑇(𝑥, 𝐺(𝑧)) ∧ ∀𝑦(𝑆(𝑦) ∧ 𝑇(𝑦, 𝐺(𝑧)) → 𝑥 = 𝑦))
There exists a student x such that for every Maths module z, x takes that Maths
module, and x is the only student who takes it.
In plain English: There is a student who is the unique student taking every
Maths module.
2.2 ∃𝑥 (𝐶(𝑥) ∧ ∀𝑦 (𝑆(𝑦) → ¬𝑇(𝑦, 𝑥)))
There exists a Computing module that no student takes.
2.3 ∀𝑥 ∃𝑦 𝑇(𝑥, 𝑦) ∧ 𝑆(𝑥)
Note: By standard precedence, the ∧ binds less tightly than the quantifiers, so this
reads as: (∀𝑥 ∃𝑦 𝑇(𝑥, 𝑦)) ∧ 𝑆(𝑥), where the second conjunct has x free. More likely
the intended reading (treating it as a scope issue) is:
Every entity takes something, and x is a student.
If parsed as written (with x free in S(x)): Everything takes some module, and x is a
student. (x is a free variable here — the formula is not a sentence.)
If the intended reading is ∀𝑥(∃𝑦 𝑇(𝑥, 𝑦) ∧ 𝑆(𝑥)): Every student takes at least one
module.
2.4 ∀𝑥 [𝑆(𝑥) → 𝐿(𝑘, 𝑥)]
Kevin loves every student. (Using k as the constant for Kevin.)
