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Year 1 of the computer science course of the University of Oxford

Past exam of the year 1 of the computer science course of the University of Oxford

2014

Digital systems, linear algebra, introduction to formal proof

Assuming $\neg P$ lets the <disjunction introduction rule> produce the <logical disjunction> excluded by the premise. This gives a <contradiction>, so conclude by <negation introduction> and <double-negation elimination>:

|| Line
|| Context
|| Formula
|| Justification

| 1
| $\{1\}$
| $\neg(\neg P\lor\neg Q)$
| Premise

| 2
| $\{1,2\}$
| $\neg P$
| Temporary assumption

| 3
| $\{1,2\}$
| $\neg P\lor\neg Q$
| <disjunction introduction rule> $\lor\text{-intro-L}$ on 2

| 4
| $\{1,2\}$
| $\bot$
| <negation elimination rule> $\neg\text{-elim}$ on 3, 1

| 5
| $\{1\}$
| $\neg\neg P$
| <negation introduction rule> $\neg\text{-intro}$ on 4; discharge 2

| 6
| $\{1\}$
| $P$
| <double-negation elimination> $\neg\neg\text{-elim}$ on 5

Thus
$$
\boxed{\neg(\neg P\lor\neg Q)\vdash P.}
$$
No separate result about $Q$ or use of a derived De Morgan rule is needed.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Digital systems, linear algebra, introduction to formal proof / 7 / c / iv / Solution

Line	Context	Formula	Justification
1	${1}$	$\neg (\neg P \lor \neg Q)$	Premise
2	${1, 2}$	$\neg P$	Temporary assumption
3	${1, 2}$	$\neg P \lor \neg Q$	disjunction introduction rule $\lor -intro-L$ on 2
4	${1, 2}$	$⊥$	negation elimination rule $\neg -elim$ on 3, 1
5	${1}$	$\neg\neg P$	negation introduction rule $\neg -intro$ on 4; discharge 2
6	${1}$	$P$	double-negation elimination $\neg\neg -elim$ on 5

Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Digital systems, linear algebra, introduction to formal proof / 7 / c / iv / Solution

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