Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 7 / a

Use natural deduction with labelled temporary assumptions. Assume $\neg P\to \neg Q$ under label $a$, then $\neg P\to Q$ under label $b$, then $\neg P$ under label $c$. Two applications of the modus ponens rule give $\neg Q$ and $Q$. The negation elimination rule gives $\perp$, and the supplied reductio ad absurdum rule discharges $c$ to conclude $P$.

Discharge $b$ using the implication introduction rule, obtaining $(\neg P\to Q)\to P$ under $a$. Discharge $a$ with the implication introduction rule, obtainingAll three temporary assumptions have now been discharged; no premise remains.

Keep the four given premises in context and assume $P$ under label $a$. The modus ponens rule applied to the first two premises gives $Q\vee R$ and $\neg S$. Apply the disjunction elimination rule to $Q\vee R$ as follows.

Under a case assumption $Q$, the premise $Q\to S$ gives $S$ by the modus ponens rule. Together with $\neg S$, the negation elimination rule gives $\perp$, and the supplied logical falsity elimination rule gives $R$. Under the other case assumption $R$, the required conclusion $R$ is already available. Discharge both case assumptions with the disjunction elimination rule to conclude $R$.

Finally discharge $a$ with the implication introduction rule, giving $P\to R$. The premise $S\to Q$ is not needed: a derivation is permitted to leave a premise unused.

Use the disjunction elimination rule on the outer disjunction. In its left branch, assume $P\vee Q$ and apply the disjunction elimination rule again: from a case assumption $P$, the disjunction introduction rule gives $P\vee R$; from a case assumption $Q$, the premise $Q\to R$ and the modus ponens rule give $R$, then the disjunction introduction rule gives $P\vee R$. Discharging those two inner case assumptions yields $P\vee R$ under the left outer branch.

In the right outer branch, assume $R\vee S$. A case assumption $R$ gives $P\vee R$ by the disjunction introduction rule. A case assumption $S$, together with the premise $\neg S$, gives $\perp$ by the negation elimination rule, and then $P\vee R$ by logical falsity elimination rule. Discharge these inner case assumptions with the disjunction elimination rule.

Both outer branches now have the same conclusion. Discharge the two outer case assumptions with the disjunction elimination rule to obtain $P\vee R$ under exactly the stated premises.

Assume $E\to G$ under label $a$, $F\to G$ under label $b$, and $E\vee F$ under label $c$. A temporary case assumption $E$ gives $G$ by the modus ponens rule using $a$. A temporary case assumption $F$ gives $G$ by the modus ponens rule using $b$. The disjunction elimination rule discharges these case assumptions and concludes $G$ under $a,b,c$.

Apply the implication introduction rule three times, discharging $c$, then $b$, then $a$. The resulting theorem isThe displayed parentheses make explicit the right association of the source formula.