UNIVERSITY OF SOUTH AFRICA
Department of Computer Science
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
COS2661: Formal Logic II
Assignment 2 — Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
COS2661
Module Code:
Formal Logic II
Module Name:
Propositional and Predicate Logic
Assignment Topic:
180808
Unique Number:
10 July 2026
Due Date:
Submitted in partial fulfilment of the requirements for COS2661 — UNISA 2026
,UNISA | COS2661 Formal Logic II — Assignment 2
Question 1
1.1.1
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
P → Q, Q → R, ¬R Conclusion: ¬P
An argument is valid when every row of the truth table that makes all premises true also
makes the conclusion true. The formula contains three atomic sentences, P , Q, and R, pro-
ducing eight rows.
P Q R P →Q Q→R ¬R ¬P All premises true?
T T T T T F F no
T T F T F T F no
T F T F T F F no
T F F F T T F no
F T T T T F T no
F T F T F T T no
F F T T T F T no
F F F T T T T yes
Row 8 is the only row where P → Q, Q → R, and ¬R are all true at once. In that row ¬P is
also true. No row produces true premises together with a false conclusion, so the argument is
valid.
1.1.2
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
A → B, B → C, C → D Conclusion: D → A
Four atomic sentences appear, giving sixteen rows.
Three rows, A = F, B = T, C = T, D = T and A = F, B = F, C = T, D = T and A = F, B =
F, C = F, D = T , make every premise true while the conclusion D → A comes out false. A
single counter-model is enough to settle the matter, so the argument is invalid.
Page 2 of 12
,UNISA | COS2661 Formal Logic II — Assignment 2
A B C D A→B B→C C→D D→A All premises true?
T T T T T T T T yes
T T T F T T F T no
T T F T T F T T no
T T F F T F T T no
T F T T F T T T no
T F T F F T F T no
T F F T F T T T no
T F F F F T T T no
F T T T T T T F yes, conclusion false
F T T F T T F T no
F T F T T F T F no
F T F F T F T T no
F F T T T T T F yes, conclusion false
F F T F T T F T no
F F F T T T T F yes, conclusion false
F F F F T T T T yes
1.1.3
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
A → B, B Conclusion: A
Two atomic sentences give four rows.
A B A→B B All premises true?
T T T T yes, conclusion true
T F F F no
F T T T yes, conclusion false
F F T F no
Row 3 gives A = F and B = T . Both premises are true there, yet the conclusion A is false.
This single row is a counter-model, so the argument is invalid. It matches the pattern known
as affirming the consequent.
1.2
For the following sentence determine whether the indicated pairs of sentences are tautologi-
cally equivalent. Use truth tables to show your answer.
P ↔ (Q ↔ (R ↔ S)) and ((P ↔ Q) ↔ R) ↔ S
Page 3 of 12
, UNISA | COS2661 Formal Logic II — Assignment 2
Two sentences are tautologically equivalent when they carry the same truth value on every
row of a joint truth table. Four atomic sentences give sixteen rows.
P Q R S R↔S Q ↔ (R ↔ S) Formula 1 P ↔Q (P ↔ Q) ↔ R
T T T T T T T T T
T T T F F F F T T
T T F T F F F T F
T T F F T T T T F
T F T T T F F F F
T F T F F T T F F
T F F T F T T F T
T F F F T F F F T
F T T T T T F F F
F T T F F F T F F
F T F T F F T F T
F T F F T T F F T
F F T T T F T T T
F F T F F T F T T
F F F T F T F T F
F F F F T F T T F
S Formula 2 = ((P ↔ Q) ↔ R) ↔ S Match with Formula 1?
T T yes
F F yes
T F yes
F T yes
T F yes
F T yes
T T yes
F F yes
T F yes
F T yes
T T yes
F F yes
T T yes
F F yes
T F yes
F T yes
Every row produces the same value for both formulas, so the two sentences are tautologically
equivalent. This follows a general property of the biconditional connective: ↔ is associative
and commutative in the same way that exclusive-or is, so any grouping of a chain of bicondi-
tionals over the same set of atomic sentences returns the same truth value regardless of where
the brackets are placed.
Page 4 of 12
Department of Computer Science
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
COS2661: Formal Logic II
Assignment 2 — Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
COS2661
Module Code:
Formal Logic II
Module Name:
Propositional and Predicate Logic
Assignment Topic:
180808
Unique Number:
10 July 2026
Due Date:
Submitted in partial fulfilment of the requirements for COS2661 — UNISA 2026
,UNISA | COS2661 Formal Logic II — Assignment 2
Question 1
1.1.1
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
P → Q, Q → R, ¬R Conclusion: ¬P
An argument is valid when every row of the truth table that makes all premises true also
makes the conclusion true. The formula contains three atomic sentences, P , Q, and R, pro-
ducing eight rows.
P Q R P →Q Q→R ¬R ¬P All premises true?
T T T T T F F no
T T F T F T F no
T F T F T F F no
T F F F T T F no
F T T T T F T no
F T F T F T T no
F F T T T F T no
F F F T T T T yes
Row 8 is the only row where P → Q, Q → R, and ¬R are all true at once. In that row ¬P is
also true. No row produces true premises together with a false conclusion, so the argument is
valid.
1.1.2
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
A → B, B → C, C → D Conclusion: D → A
Four atomic sentences appear, giving sixteen rows.
Three rows, A = F, B = T, C = T, D = T and A = F, B = F, C = T, D = T and A = F, B =
F, C = F, D = T , make every premise true while the conclusion D → A comes out false. A
single counter-model is enough to settle the matter, so the argument is invalid.
Page 2 of 12
,UNISA | COS2661 Formal Logic II — Assignment 2
A B C D A→B B→C C→D D→A All premises true?
T T T T T T T T yes
T T T F T T F T no
T T F T T F T T no
T T F F T F T T no
T F T T F T T T no
T F T F F T F T no
T F F T F T T T no
T F F F F T T T no
F T T T T T T F yes, conclusion false
F T T F T T F T no
F T F T T F T F no
F T F F T F T T no
F F T T T T T F yes, conclusion false
F F T F T T F T no
F F F T T T T F yes, conclusion false
F F F F T T T T yes
1.1.3
For the following sentence determine whether the indicated pairs of sentences are a valid argu-
ment. Use truth tables to show your answer.
A → B, B Conclusion: A
Two atomic sentences give four rows.
A B A→B B All premises true?
T T T T yes, conclusion true
T F F F no
F T T T yes, conclusion false
F F T F no
Row 3 gives A = F and B = T . Both premises are true there, yet the conclusion A is false.
This single row is a counter-model, so the argument is invalid. It matches the pattern known
as affirming the consequent.
1.2
For the following sentence determine whether the indicated pairs of sentences are tautologi-
cally equivalent. Use truth tables to show your answer.
P ↔ (Q ↔ (R ↔ S)) and ((P ↔ Q) ↔ R) ↔ S
Page 3 of 12
, UNISA | COS2661 Formal Logic II — Assignment 2
Two sentences are tautologically equivalent when they carry the same truth value on every
row of a joint truth table. Four atomic sentences give sixteen rows.
P Q R S R↔S Q ↔ (R ↔ S) Formula 1 P ↔Q (P ↔ Q) ↔ R
T T T T T T T T T
T T T F F F F T T
T T F T F F F T F
T T F F T T T T F
T F T T T F F F F
T F T F F T T F F
T F F T F T T F T
T F F F T F F F T
F T T T T T F F F
F T T F F F T F F
F T F T F F T F T
F T F F T T F F T
F F T T T F T T T
F F T F F T F T T
F F F T F T F T F
F F F F T F T T F
S Formula 2 = ((P ↔ Q) ↔ R) ↔ S Match with Formula 1?
T T yes
F F yes
T F yes
F T yes
T F yes
F T yes
T T yes
F F yes
T F yes
F T yes
T T yes
F F yes
T T yes
F F yes
T F yes
F T yes
Every row produces the same value for both formulas, so the two sentences are tautologically
equivalent. This follows a general property of the biconditional connective: ↔ is associative
and commutative in the same way that exclusive-or is, so any grouping of a chain of bicondi-
tionals over the same set of atomic sentences returns the same truth value regardless of where
the brackets are placed.
Page 4 of 12
