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 / iii

Table of contents
- Solution iii

Solution

The premise turns an assumed

\neg P

into a contradiction. Use the negation introduction rule to obtain

\neg\neg P

, then the supplied classical double-negation elimination rule:

Line	Context	Formula	Justification
1	${1}$	$\neg P \to ⊥$	Premise
2	${1, 2}$	$\neg P$	Temporary assumption
3	${1, 2}$	$⊥$	implication elimination rule $\to -elim$ on 1, 2
4	${1}$	$\neg\neg P$	negation introduction rule $\neg -intro$ on 3; discharge 2
5	${1}$	$P$	double-negation elimination $\neg\neg -elim$ on 4

Hence

(\neg P \to ⊥) ⊢ P

. The final classical rule is essential: deriving

\neg\neg P

alone is not yet a proof of

P

in Intuitionistic propositional logic.

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