COS3761 2026 LATEST EXAM QUESTIONS WITH
CORRECT VERIFIED ANSWERS|ALREADY GRADED
A|GUARANTEED EXAM|LATEST 2026.
PROPOSITIONAL LOGIC – Translation (English to Symbolic)
Q1: "The house is used as a guesthouse only if it has seven bathrooms." (s = house
used as guesthouse, r = house has seven bathrooms)
A1: s → r (Option 2)
Q2: "Unless Sam saves, he does not buy the house with seven bathrooms." (p =
Sam saves, q = Sam buys the house, r = house has seven bathrooms)
A2: ¬p → ¬(q ∧ r) or equivalently ¬p → (¬q ∨ ¬r). None of the given options
exactly match (Option 5)
Q3: Formula: q → ¬r. Translate to English. (q = Sam buys the house, r = house has
seven bathrooms)
A3: "Sam buys the house only if the house does not have seven bathrooms"
(Option 3)
Q4: Formula: t → (p ∧ q ∧ s). (t = Peter likes the house, p = Sam saves, q = Sam
buys the house, s = house used as guesthouse)
A4: "If Peter likes the house, then Sam saves and buys the house used as
guesthouse"
Q5: "He will come on the 8:15 or the 9:15 train; if the former, he will have time to
visit us." (p = He will come on the 8:15, q = He will come on the 9:15, r = He will
have time to visit us)
A5: (p ∨ q) ∧ (p → r) (Option E)
Q6: "If inflation is up and an election is approaching, then public borrowing goes
up."
,A6: (p ∧ q) → r (Option C)
Q7: "If the grass is wet, then either it is raining or the sprinklers are on." (p = It is
raining, q = Sprinklers are on, r = Grass is wet)
A7: r → (p ∨ q)
Q8: Formula: ¬(p → r) ∨ (q → r). Translate to English.
A8: "Either it is not the case that if it rains then the grass is wet, or if the sprinklers
are on then the grass is wet."
Q9: "There is a hail storm only if it is cloudy and the temperature is not above
20°C." (s = There is a hail storm, q = It is cloudy, t = Temperature is above 20°C)
A9: s → (q ∧ ¬t)
Q10: Formula: p ∧ (t → r). Translate to English. (p = It is windy, t = Temperature
is above 20°C, r = There is a dust storm)
A10: "It is windy and if the temperature is above 20°C then there is a dust storm."
PROPOSITIONAL LOGIC – Natural Deduction Proofs
Q11: Prove: ⊢ (p → q) → ((p ∧ r) → (q ∧ r))
A11: [Proof using natural deduction: Assume p → q. Assume p ∧ r. From p ∧ r get
p and r. From p and p → q get q by →-elimination. From q and r get q ∧ r by ∧-
introduction. Discharge assumptions to get (p ∧ r) → (q ∧ r), then (p → q) → ((p ∧
r) → (q ∧ r)).]
Q12: Prove: ⊢ ¬(p ∧ ¬p)
A12: [Proof by contradiction: Assume p ∧ ¬p. From this get p and ¬p, a
contradiction. Therefore ¬(p ∧ ¬p).]
Q13: Prove: q → r ⊢ (p → q) → (p → r)
A13: [Assume q → r. Assume p → q. Assume p. From p and p → q get q. From q
and q → r get r. Discharge assumptions to get p → r, then (p → q) → (p → r).]
, Q14: Prove: q → (p → r), ¬r, q ⊢ ¬p
A14: [Assume q → (p → r) and ¬r and q. From q and q → (p → r) get p → r.
Assume p. From p and p → r get r. But we have ¬r, a contradiction. Therefore ¬p.]
Q15: Prove: p → q ⊢ ¬p ∨ q
A15: [Assume p → q. By the law of excluded middle, either p or ¬p. If ¬p, then ¬p
∨ q. If p, then from p → q get q, so ¬p ∨ q. Therefore ¬p ∨ q.]
Q16: Prove: ⊢ (p → r) ∧ (q → r)
A16: [Proof using natural deduction with ∧-introduction after proving both
implications.]
Q17: Prove: p → (q → r), p ∧ q ⊢ r
A17: [Assume p → (q → r) and p ∧ q. From p ∧ q get p and q. From p and p → (q
→ r) get q → r. From q and q → r get r.]
Q18: Prove: (p → q) ∧ (q → r) ⊢ p → r
A18: [Assume (p → q) ∧ (q → r). From this get p → q and q → r. Assume p. From
p and p → q get q. From q and q → r get r. Therefore p → r.]
Q19: Prove: p → (q ∨ r), ¬q, ¬r ⊢ ¬p
A19: [Assume p → (q ∨ r), ¬q, ¬r. Assume p. From p and p → (q ∨ r) get q ∨ r.
From q ∨ r, ¬q, and ¬r get a contradiction. Therefore ¬p.]
Q20: Prove: (p ∨ q) → r ⊢ p → r
A20: [Assume (p ∨ q) → r. Assume p. From p get p ∨ q by ∨-introduction. From p
∨ q and (p ∨ q) → r get r. Therefore p → r.]
PROPOSITIONAL LOGIC – Invalidity / Counterexamples
Q21: Show that the following sequent is not valid: p → q ⊢ (p ∧ q) → r
CORRECT VERIFIED ANSWERS|ALREADY GRADED
A|GUARANTEED EXAM|LATEST 2026.
PROPOSITIONAL LOGIC – Translation (English to Symbolic)
Q1: "The house is used as a guesthouse only if it has seven bathrooms." (s = house
used as guesthouse, r = house has seven bathrooms)
A1: s → r (Option 2)
Q2: "Unless Sam saves, he does not buy the house with seven bathrooms." (p =
Sam saves, q = Sam buys the house, r = house has seven bathrooms)
A2: ¬p → ¬(q ∧ r) or equivalently ¬p → (¬q ∨ ¬r). None of the given options
exactly match (Option 5)
Q3: Formula: q → ¬r. Translate to English. (q = Sam buys the house, r = house has
seven bathrooms)
A3: "Sam buys the house only if the house does not have seven bathrooms"
(Option 3)
Q4: Formula: t → (p ∧ q ∧ s). (t = Peter likes the house, p = Sam saves, q = Sam
buys the house, s = house used as guesthouse)
A4: "If Peter likes the house, then Sam saves and buys the house used as
guesthouse"
Q5: "He will come on the 8:15 or the 9:15 train; if the former, he will have time to
visit us." (p = He will come on the 8:15, q = He will come on the 9:15, r = He will
have time to visit us)
A5: (p ∨ q) ∧ (p → r) (Option E)
Q6: "If inflation is up and an election is approaching, then public borrowing goes
up."
,A6: (p ∧ q) → r (Option C)
Q7: "If the grass is wet, then either it is raining or the sprinklers are on." (p = It is
raining, q = Sprinklers are on, r = Grass is wet)
A7: r → (p ∨ q)
Q8: Formula: ¬(p → r) ∨ (q → r). Translate to English.
A8: "Either it is not the case that if it rains then the grass is wet, or if the sprinklers
are on then the grass is wet."
Q9: "There is a hail storm only if it is cloudy and the temperature is not above
20°C." (s = There is a hail storm, q = It is cloudy, t = Temperature is above 20°C)
A9: s → (q ∧ ¬t)
Q10: Formula: p ∧ (t → r). Translate to English. (p = It is windy, t = Temperature
is above 20°C, r = There is a dust storm)
A10: "It is windy and if the temperature is above 20°C then there is a dust storm."
PROPOSITIONAL LOGIC – Natural Deduction Proofs
Q11: Prove: ⊢ (p → q) → ((p ∧ r) → (q ∧ r))
A11: [Proof using natural deduction: Assume p → q. Assume p ∧ r. From p ∧ r get
p and r. From p and p → q get q by →-elimination. From q and r get q ∧ r by ∧-
introduction. Discharge assumptions to get (p ∧ r) → (q ∧ r), then (p → q) → ((p ∧
r) → (q ∧ r)).]
Q12: Prove: ⊢ ¬(p ∧ ¬p)
A12: [Proof by contradiction: Assume p ∧ ¬p. From this get p and ¬p, a
contradiction. Therefore ¬(p ∧ ¬p).]
Q13: Prove: q → r ⊢ (p → q) → (p → r)
A13: [Assume q → r. Assume p → q. Assume p. From p and p → q get q. From q
and q → r get r. Discharge assumptions to get p → r, then (p → q) → (p → r).]
, Q14: Prove: q → (p → r), ¬r, q ⊢ ¬p
A14: [Assume q → (p → r) and ¬r and q. From q and q → (p → r) get p → r.
Assume p. From p and p → r get r. But we have ¬r, a contradiction. Therefore ¬p.]
Q15: Prove: p → q ⊢ ¬p ∨ q
A15: [Assume p → q. By the law of excluded middle, either p or ¬p. If ¬p, then ¬p
∨ q. If p, then from p → q get q, so ¬p ∨ q. Therefore ¬p ∨ q.]
Q16: Prove: ⊢ (p → r) ∧ (q → r)
A16: [Proof using natural deduction with ∧-introduction after proving both
implications.]
Q17: Prove: p → (q → r), p ∧ q ⊢ r
A17: [Assume p → (q → r) and p ∧ q. From p ∧ q get p and q. From p and p → (q
→ r) get q → r. From q and q → r get r.]
Q18: Prove: (p → q) ∧ (q → r) ⊢ p → r
A18: [Assume (p → q) ∧ (q → r). From this get p → q and q → r. Assume p. From
p and p → q get q. From q and q → r get r. Therefore p → r.]
Q19: Prove: p → (q ∨ r), ¬q, ¬r ⊢ ¬p
A19: [Assume p → (q ∨ r), ¬q, ¬r. Assume p. From p and p → (q ∨ r) get q ∨ r.
From q ∨ r, ¬q, and ¬r get a contradiction. Therefore ¬p.]
Q20: Prove: (p ∨ q) → r ⊢ p → r
A20: [Assume (p ∨ q) → r. Assume p. From p get p ∨ q by ∨-introduction. From p
∨ q and (p ∨ q) → r get r. Therefore p → r.]
PROPOSITIONAL LOGIC – Invalidity / Counterexamples
Q21: Show that the following sequent is not valid: p → q ⊢ (p ∧ q) → r
