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Assume the conjunction and obtain a contradiction in each of the two cases supplied by the premise:

The two case assumptions disappear at line 9, while the conjunction assumption remains available until line 10. The proof uses only the primitive rules printed in the paper.

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.

Let $N(\varphi ,\psi )$ abbreviate $\neg \varphi \wedge \neg \psi$. Suitable rules areTheir derivations require no new axiom. For introduction, apply the conjunction introduction rule to the two premises, obtaining $\neg \varphi \wedge \neg \psi$, which is exactly the abbreviation. For left and right elimination, unfold the abbreviation and apply the corresponding conjunction elimination rule. These are therefore derived rules for the NOR connective.

An equivalent subproof presentation of introduction permits one subproof from assumption $\varphi$ to $\perp$ and another from assumption $\psi$ to $\perp$: use the negation introduction rule to discharge each assumption, then apply the conjunction introduction rule. If elimination is desired in contradiction form, combine the left projection with $\varphi$, or the right projection with $\psi$, using the negation elimination rule. All these presentations reduce directly to the permitted rules.

Choose arbitrary fresh variables $y,x,z$ and assume $H=((x=y\vee y=z)\wedge P(y))$ under label $a$. The conjunction elimination rule gives $x=y\vee y=z$ and $P(y)$. Apply the disjunction elimination rule to the equality disjunction.

In the case $x=y$, the supplied symmetry rule gives $y=x$, and the equality elimination rule applied to $P(y)$ gives $P(x)$. Disjunction introduction rule gives $P(x)\vee P(z)$. In the case $y=z$, equality elimination rule applied directly to $P(y)$ gives $P(z)$; the other disjunction introduction rule gives the same conclusion $P(x)\vee P(z)$. Discharge the two equality case assumptions by disjunction elimination rule.

Discharge $a$ with the implication introduction rule. There are now no undischarged assumptions. Use the universal quantifier introduction rule successively for $z$, then $x$, then $y$, yieldingFreshness holds at every generalization because the sole temporary assumption has already been discharged; the arbitrary variables are not tied to any premise.