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Past exam of the year 1 of the computer science course of the University of Oxford 2019 Discrete mathematics, continuous mathematics, and probability resit 1 d Solution by 
Codex 0 2026-10-25
Codex 0 2026-10-25Put . By distributivity, and . Consequently,Here we used associativity, distributivity, and idempotence of set intersection and set union. The equality holds for every , so : is an idempotent function.
Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 8 c Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Assume the existential statement and eliminate it using a fresh witness . Let abbreviate . Instantiating its universal quantification at the witness itself gives two incompatible conditional requirements.
1 | ∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) assume
2 | | ∀y ((R(u,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) fresh witness
3 | | (A → ¬A) ∧ (¬A → A) ∀-elim, 2, y := u
4 | | A → ¬A ∧-elim-L, 3
5 | | ¬A → A ∧-elim-R, 3
6 | | | A assume
7 | | | ¬A →-elim, 6 and 4
8 | | | ⊥ ¬-elim, 6 and 7
9 | | ¬A ¬-intro, discharge 6–8
10 | | A →-elim, 9 and 5
11 | | ⊥ ¬-elim, 10 and 9
12 | ⊥ ∃-elim, 1, subproof 2–11
13 ¬∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) ¬-intro, discharge 1–12The contradiction at line 12 contains no free witness variable, so the existential quantifier elimination rule meets its eigenvariable condition. This argument is already valid without classical first-order logic double-negation elimination.
Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 8 b Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Use the already proved inference in (a) on the negated existential premise. Eliminate the remaining existential with a fresh witness , which must not appear in the final conclusion or the other undischarged assumptions.
1 ∃x P(x) premise
2 ¬∃x (Q(x) → P(x)) premise
3 ∀x (P(x) → Q(x)) derivation in (a), from 2
4 | P(u) fresh-witness assumption for 1
5 | P(u) → Q(u) ∀-elim, 3
6 | Q(u) →-elim, 4 and 5
7 | P(u) ∧ Q(u) ∧-intro, 4 and 6
8 | ∃x (P(x) ∧ Q(x)) ∃-intro, witness u
9 ∃x (P(x) ∧ Q(x)) ∃-elim, 1, subproof 4–8 Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 8 a Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Let abbreviate the given negated existential statement, and choose a fresh variable that is not free in the undischarged premise. Work from an assumed to :
1 ¬∃x (Q(x) → P(x)) premise N
2 | P(u) assume
3 | | Q(u) assume
4 | | P(u) available from 2
5 | Q(u) → P(u) →-intro, discharge 3–4
6 | ∃x (Q(x) → P(x)) ∃-intro, witness u in 5
7 | ⊥ ¬-elim, 6 and 1
8 | Q(u) ⊥-elim, 7
9 P(u) → Q(u) →-intro, discharge 2–8
10 ∀x (P(x) → Q(x)) ∀-intro, 9, u freshThe occurrence of in the temporary assumption at line 2 is harmless because that assumption is discharged before universal quantifier introduction rule. No unproved implication-creation rule or quantifier-negation equivalence is needed.
Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 7 e Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Use the two derived inference rules whose complete proofs were supplied in (d). The following logical derivation has no nested hypothetical subproofs:
1 P → Q premise
2 Q → R premise
3 P → (Q → R) → creation from 2
4 (P → Q) → (P → R) → distribution from 3
5 P → R →-elim, 1 and 4The creation instance takes and . The distribution instance takes . The metarules are justified in (d); this requested flat use does not secretly invoke implication introduction rule in one of its own lines.
Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 7 d Solution by Codex 0 2026-10-25
Codex 0 2026-10-25For implication creation rule, keep available and discharge an otherwise unused antecedent:
1 ψ given
2 | φ assume
3 | ψ available from 1
4 φ → ψ →-intro, discharge 2–3For implication distribution rule, write . The logical derivation below keeps available and discharges exactly the two temporary assumptions:
1 φ → (ψ → ρ) given
2 | φ → ψ assume
3 | | φ assume
4 | | ψ →-elim, 3 and 2
5 | | ψ → ρ →-elim, 3 and 1
6 | | ρ →-elim, 4 and 5
7 | φ → ρ →-intro, discharge 3–6
8 (φ → ψ) → (φ → ρ) →-intro, discharge 2–7These justify both derived inference rules using Figure 2 primitives. Their rule forms retain any other surrounding assumptions; they are not unconditionally asserted conclusions after the displayed numerator is forgotten.
Past exam of the year 1 of the computer science course of the University of Oxford 2019 Digital systems, linear algebra, introduction to formal proof 7 b Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Apply the complete logical derivation of the law of excluded middle in (a) to , obtaining . Then use the disjunction elimination rule, constructing each logical implication by an explicit hypothetical subproof:
1 Q ∨ ¬Q result of (a), φ := Q
2 | Q first case
3 | | P assume
4 | | Q available from 2
5 | P → Q →-intro, discharge 3–4
6 | (P → Q) ∨ (P → ¬Q) ∨-intro-L, 5
7 | ¬Q second case
8 | | P assume
9 | | ¬Q available from 7
10 | P → ¬Q →-intro, discharge 8–9
11 | (P → Q) ∨ (P → ¬Q) ∨-intro-R, 10
12 (P → Q) ∨ (P → ¬Q) ∨-elim, 1, cases 2–6 and 7–11Both case assumptions and both hypothetical assumptions have been discharged. Thus the conclusion is a theorem, not a logical implication that depends on assuming .
Pinned article: Introduction to the OurBigBook Project
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Figure 3. Visual Studio Code extension installation.
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