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Derive each negated component in its own subproof, then combine them. Assumption lines below begin local scopes; the listed discharge ranges close those scopes.

Only the original premise remains open. No classical principle is needed for this direction of De Morgan's laws.

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.

The supplied double-negation elimination rule provides the classical step needed to recover each positive component:

Both local negated assumptions have been discharged. The proof establishes the positive conjunction without assuming an unproved derived version of De Morgan's laws.

Write $H$ for $(\varphi \to \psi )\to \varphi$. The following derives Peirce's law directly from the printed primitive rules, including double-negation elimination:

Every assumption is discharged, so the last formula is a theorem. In particular, no unproved appeal to the derived RAA rule is hidden in the argument.

Let $K$ abbreviate $Q(c_1)\vee Q(c_2)$, and group the three equality alternatives on the left. The proof uses the binary disjunction elimination rule twice, so no unproved three-way derived rule is needed.

The witness name is absent from $K$ and from every assumption remaining after line 19, so the existential elimination is legal. Each equality is used in the forward direction supplied by the printed equality rule. The three named constants need not denote distinct objects: the case proof requires only that they exhaust the domain. This is existential quantification over a finitely named domain followed by exclusion of the first case.

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.

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 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.

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.

Temporarily assume $Q(c)$ under label $a$. The implication premise and the modus ponens rule yield $\forall x(P(x)\leftrightarrow P(c))$. Instantiate it at $c^{\prime }$ using the universal quantifier elimination rule, obtaining $P(c^{\prime })\leftrightarrow P(c)$. Its right biconditional elimination rule gives $P(c)\to P(c^{\prime })$.

Apply the modus ponens rule with the premise $P(c)$ to obtain $P(c^{\prime })$. Together with $\neg P(c^{\prime })$, negation elimination rule gives $\perp$. Discharge $a$ using the negation introduction rule to conclude $\neg Q(c)$. Nothing assumes that the two constant symbols denote different objects; if they denoted the same object, the stated premises would themselves already be inconsistent.

Let $u$ be fresh, not free in any of the three universal premises. Universal quantifier elimination rule gives $\neg P(u)\vee R(c,u)$, $Q(u)\to R(c,u)$ and $P(u)\vee Q(u)$. Apply disjunction elimination rule to the last formula.

In the $P(u)$ branch, apply disjunction elimination rule to $\neg P(u)\vee R(c,u)$. Its $\neg P(u)$ case contradicts $P(u)$ by negation elimination rule, and logical falsity elimination rule then gives $R(c,u)$. Its $R(c,u)$ case already gives that conclusion. Discharge those two inner case assumptions to obtain $R(c,u)$ under the outer $P(u)$ assumption. In the outer $Q(u)$ branch, the modus ponens rule applied to $Q(u)\to R(c,u)$ gives $R(c,u)$ directly.

Discharge the two outer case assumptions by disjunction elimination rule. The conclusion $R(c,u)$ now depends only on the universal premises, in which $u$ is not free. Universal quantifier introduction rule therefore gives $\forall xR(c,x)$. All case assumptions have been discharged, and the argument requires no classical reductio ad absurdum.

Keep the premise $P\to Q$ open. Assume $Q\to \neg P$, then temporarily assume $P$. Two applications of the implication elimination rule produce $Q$ and then $\neg P$, contradicting the temporary $P$. Discharge $P$ using the negation introduction rule, then discharge $Q\to \neg P$ using the implication introduction rule.This natural deduction proof uses only the supplied introduction and elimination rules. Line 6 is an application of the negation elimination rule; no classical contradiction principle is needed.