 Solution

Write $D(\varphi ,\psi )=\neg (\neg \varphi \wedge \neg \psi )$. We derive the two disjunction introduction rules and the disjunction elimination rule without using any disjunction rule inside the derivations.

For left introduction, start with $\varphi$ and temporarily assume $\neg \varphi \wedge \neg \psi$. The conjunction elimination rule gives $\neg \varphi$, and the negation elimination rule gives $\perp$. Discharge the temporary conjunction by the negation introduction rule, concluding $D(\varphi ,\psi )$. For right introduction, start with $\psi$ and use the right conjunction elimination rule to obtain $\neg \psi$ instead. Thus either input derives the required encoded disjunction.

For elimination, suppose $D(\varphi ,\psi )$ is available, together with a derivation of $\kappa$ from an additional assumption $\varphi$ and a derivation of $\kappa$ from an additional assumption $\psi$. Keep their common external context fixed. Temporarily assume $\neg \kappa$ under label $a$. Temporarily assume $\varphi$ under label $b$ and insert the supplied first branch derivation to obtain $\kappa$. Negation elimination rule gives $\perp$, so negation introduction rule discharges $b$ and gives $\neg \varphi$. Repeat with a temporary assumption $\psi$ under label $c$ and the supplied second branch derivation, discharging $c$ to obtain $\neg \psi$.

Conjunction introduction rule now gives $\neg \varphi \wedge \neg \psi$. Its negation $D(\varphi ,\psi )$ gives $\perp$ by negation elimination rule. The supplied reductio ad absurdum rule discharges $a$ and yields $\kappa$. This discharges the two branch assumptions as required and proves the encoded elimination rule. Its last step uses classical propositional logic; the same encoded elimination principle is not generally available in Intuitionistic propositional logic.