Notation 1. For a formula $\varphi $ , $\text { FV} \left ({\varphi }\right )$ denotes the set of free variables in $\varphi $ . Quantifier-free formulas are denoted with subscript “qf” as ${\varphi }_{\text {qf}}$ . In addition, a list of variables is denoted with an over-line as $\overline {x}$ . In particular, a list of quantifiers of the same kind is denoted as $\exists \overline {x}$ and $\forall \overline {x}$ respectively.

Proof Straightforward by the fact that

(1) $$ \begin{align} \mathsf{HA} \vdash \xi \leftrightarrow \forall z \xi \leftrightarrow \exists z \xi \end{align} $$

for any $ z\notin \mathrm{FV} \left ({\xi }\right )$ .⊣

Proof We first claim that $\mathsf {HA} + \text {CT}_{0}$ proves $\varphi _0$ . For the sake of contradiction, assume

(2) $$ \begin{align} \forall x \left( \neg \exists u \left( \mathrm{T}(x,x,u) \land \mathrm{U}(u)=0 \right) \lor \neg \exists u \left( \mathrm{T}(x,x,u) \land \text{ U}(u)\neq 0 \right) \right) \end{align} $$

and reason in $\mathsf {HA} + \text {CT}_{0}$ . Since $\varphi _1\lor \varphi _2 \leftrightarrow \exists k\left ( (k=0 \to \varphi _1) \land (k\neq 0 \to \varphi _2) \right )$ (see [Reference Troelstra8, §1.3.7]), by $\text {CT}_{0}$ , there exists e such that

$$ \begin{align*} \forall x \exists v \left( \begin{array}{l} \mathrm{T}(e,x,v) \\ \land \left(\mathrm{U}(v)=0 \to \neg \exists u \left( \mathrm{T}(x,x,u) \land \mathrm{U}(u)=0 \right) \right) \\ \land \left(\mathrm{U}(v)\neq 0 \to \neg \exists u \left( \mathrm{T}(x,x,u) \land \mathrm{U}(u) \neq 0 \right) \right) \end{array} \right). \end{align*} $$

In particular, for that e, there exists $v_e$ such that $\text {T}(e,e,v_e)$ ,

$$ \begin{align*}\mathrm{U}(v_e)=0 \to \neg \exists u \left( \mathrm{T}(e,e,u) \land \mathrm{U}(u)=0 \right) \end{align*} $$

and

$$ \begin{align*}\mathrm{U}(v_e)\neq 0 \to \neg \exists u \left( \mathrm{T}(e,e,u) \land \mathrm{U}(u) \neq 0 \right). \end{align*} $$

Since $\mathrm{U}(v_e)=0 \lor \mathrm{U}(v_e) \neq 0$ , we obtain a contradiction straightforwardly.

If $\varphi _0 $ is equivalent to some sentence $\varphi _0 ' \in \Sigma _1 $ over $\mathsf {HA} +{\Sigma _1}\text {-}\mathrm{DNE}$ , we have $\mathsf {HA} + {\Sigma _1}\text {-}\mathrm{DNE} +\text {CT}_{0} \vdash \varphi _0 '$ from the above claim. Since $\varphi _0' \in \Sigma _1$ , by the soundness of Kleene realizability (see [Reference Troelstra8, §3.2.22]), we have that

$$ \begin{align*} \mathsf{HA} + {\Sigma_1}\text{-}\mathrm{DNE} \vdash \varphi_0 ', \end{align*} $$

and hence, $\mathsf {HA} + {\Sigma _1}\text {-}\mathrm{DNE} \vdash \varphi _0 $ . On the other hand, since

$$ \begin{align*}\forall x \neg \left( \exists u \left( \mathrm{T}(x,x,u) \land \mathrm{U}(u)=0 \right) \land \exists u \left( \mathrm{T}(x,x,u) \land \mathrm{U}(u)\neq 0 \right) \right) \end{align*} $$

is provable in $\mathsf {HA}$ , we have $\mathsf {HA} +{\Sigma _1}\text {-}\text {DML} \vdash (2)$ . Therefore we have

$$ \begin{align*} \mathsf{HA} + {\Sigma_1}\text{-}\mathrm{DNE} + {\Sigma_1}\text{-}\text{DML} \vdash \perp, \end{align*} $$

and hence, $\mathsf {PA} \vdash \perp $ , which is a contradiction.⊣

Proof Assume $\mathsf {HA} + \mathrm{P} \vdash \varphi $ . Then there exists finite instances $\psi _1, \dots , \psi _k$ of $\mathrm{P}$ such that $\mathsf {HA} + \psi _1 + \dots + \psi _k \vdash \varphi $ . Since $\mathsf {HA}$ satisfies the deduction theorem, we have that $\mathsf {HA}$ proves $\psi _1 \land \dots \land \psi _k \to \varphi $ , and hence, $\neg \neg {\left (\psi _1 \land \dots \land \psi _k \to \varphi \right )}$ , which is equivalent to $\neg \neg {\psi _1} \land \dots \land \neg \neg {\psi _k} \to \neg \neg {\varphi }$ . Then we have $\mathsf {HA} + \neg \neg \mathrm{P} \vdash \neg \neg {\varphi }$ .⊣

Proof Immediate from Lemma 4.1 and the fact that $\neg \neg {\left (\varphi _1 \leftrightarrow \varphi _2\right )}$ is intuitionistically equivalent to $\neg \neg {\varphi _1} \leftrightarrow \neg \neg {\varphi _2}$ .⊣

Proof Straightforward by simultaneous induction on k.⊣

Proof Note that $\varphi _1 \lor \varphi _2$ is equivalent to

$$ \begin{align*} \exists k \left(\left(k=0 \to \varphi_1 \right) \land \left( k\neq 0 \to \varphi_2 \right) \right) \end{align*} $$

over $\mathsf {HA}$ (see [Reference Troelstra8, §1.3.7]). Since ${\varphi }_{\text {qf}} \to \exists x \psi (x)$ and ${\varphi }_{\text {qf}} \to \forall x \psi (x)$ are equivalent to $\exists x\left ({\varphi }_{\text {qf}} \to \psi (x)\right )$ and $\forall x \left ( {\varphi }_{\text {qf}} \to \psi (x) \right )$ respectively over $\mathsf {HA}$ when $x \notin \mathrm{FV} \left ({{\varphi }_{\text {qf}}}\right )$ , our assertion follows from Lemma 4.3 straightforwardly.⊣

Proof (1): Assume $\varphi _1 \land \varphi _2 \in \mathrm{U}_k^+$ . Then $l(s) \leq k$ for all $s \in \mathit {Alt}( \varphi _1 \land \varphi _2) = \mathit {Alt}( \varphi _1) \cup \mathit {Alt}( \varphi _2)$ .

• • If $l(s) < k$ for all $s \in \mathit {Alt}( \varphi _1)$ , then $\varphi _1$ is in $\displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{U}_k^+$ .

• • Otherwise, there is $s_0 \in \mathit {Alt}( \varphi _1)$ such that $l(s_0)=k$ . Then, since $\varphi _1 \land \varphi _2 \notin \displaystyle \bigcup _{i<k} \text {F}_i$ , we have $\varphi _1 \land \varphi _2 \in \mathrm{U}_k$ . Then, for each $s \in \mathit {Alt}(\varphi _1)$ such that $l(s)=k$ , we have $i(s)\,{\equiv}\ \kern1pt{-}\, $ since $s \in \mathit {Alt}(\varphi _1 \land \varphi _2 )$ . Thus $\varphi _1 \in \mathrm{U}_k \subseteq \mathrm{U}_k^+$ .

We also have $\varphi _2 \in \mathrm{U}_k^+$ in the same manner.

For the converse direction, assume that $\varphi _1 $ and $\varphi _2$ are in $\mathrm{U}_k^+$ . Then, for all $s\in \mathit {Alt}(\varphi _1 \land \varphi _2)$ , since $s\in \mathit {Alt}(\varphi _1)$ or $s\in \mathit {Alt}(\varphi _2)$ , we have $l(s)\leq k$ , in particular, $i(s) \,{\equiv}\ \kern1pt{-}$ if $l(s)=k$ . Thus $\varphi _1 \land \varphi _2 $ is in $\mathrm{U}_k^+$ .

As for the case of $\mathrm{E}_k^+$ , an analogous proof works.

(2): Analogous to (1).

(3): Assume $\varphi _1 \to \varphi _2 \in \mathrm{U}_k^+$ . Let s be in $\mathit {Alt}(\varphi _1 )$ . By the definition of $\mathit {Alt}(\varphi _1 \to \varphi _2)$ , we have $s^{\perp } \in \mathit {Alt}(\varphi _1 \to \varphi _2)$ and $l(s) \leq k$ .

• • If $l(s) < k$ for all $s \in \mathit {Alt}( \varphi _1)$ , then $\varphi _1$ is in $\displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{E}_k^+$ .

• • Otherwise, there is $s_0 \in \mathit {Alt}( \varphi _1)$ such that $l(s_0)=k$ . Since ${s_0}^{\perp } \in \mathit {Alt}(\varphi _1 \to \varphi _2)$ , we have $\varphi _1 \to \varphi _2 \in \mathrm{U}_k$ . Then, for each $s \in \mathit {Alt}(\varphi _1)$ such that $l(s)=k$ , we have $i(s^{\perp })\,{\equiv}\ \kern1pt{-}\,$ , and hence, $i(s)\equiv +$ . Thus $\varphi _1 \in \mathrm{E}_k \subseteq \mathrm{E}_k^+$ .

We also have $\varphi _2 \in \mathrm{U}_k^+$ in the same manner.

For the converse direction, assume $\varphi _1 \in \mathrm{E}_k^+$ and $\varphi _2 \in \mathrm{U}_k^+$ . Since $\mathit {deg}(\varphi _1)\leq k$ and $\mathit {deg}(\varphi _2)\leq k$ , we have $\mathit {deg}(\varphi _1 \to \varphi _2)\leq k$ .

• • If $\mathit {deg}(\varphi _1 \to \varphi _2)<k$ , then $\varphi _1 \to \varphi _2\in \displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{U}_k^+$ .

• • If $\mathit {deg}(\varphi _1 \to \varphi _2)=k$ , for all $s\in \mathit {Alt}(\varphi _1 \to \varphi _2)$ such that $l(s)=k$ , we have $s\in \mathit {Alt}(\varphi _2)$ or $s\equiv {s_0}^{\perp }$ for some $s_0\in \mathit {Alt}(\varphi _1)$ . In the former case, we have $i(s) \,{\equiv}\ \kern1pt{-}\,$ by $\varphi _2 \in \mathrm{U}_k^+$ . In the latter case, we have $i(s_0) \equiv +$ by $\varphi _1 \in \mathrm{E}_k^+$ , and hence, $i(s)\,{\equiv}\ \kern1pt{-}\,$ .

One can also show that $\varphi _1 \to \varphi _2$ is in $\mathrm{E}_k^+$ if and only if $\varphi _1 $ is in $\mathrm{U}_k^+$ and $\varphi _2$ is in $\mathrm{E}_k^+$ analogously.

(4): Assume $\forall x \varphi _1 \in \mathrm{U}_k^+$ .

• • If $\forall x \varphi _1 \notin \mathrm{U}_k$ , then $\forall x \varphi _1 \in \displaystyle \bigcup _{i<k} \text {F}_i$ . Since $\mathit {deg}(\varphi _1)\leq \mathit {deg}(\forall x \varphi _1) <k$ , we have $\varphi _1 \in \displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{U}_k^+$ .

• • Otherwise, $\mathit {deg}(\varphi _1)\leq \mathit {deg}(\forall x \varphi _1) =k$ . If $\mathit {deg}(\varphi _1)< k$ , then we have $\varphi _1 \in \displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{U}_k^+$ . Assume $\mathit {deg}(\varphi _1)= k$ . Let s be an alternation path of $\varphi _1$ such that $l(s)=k$ . If $i(s) \not\equiv\ -\,$ , by the definition of $\mathit {Alt}(\forall x \varphi _1)$ , we have ${-}s\in \mathit {Alt}(\forall x \varphi _1)$ , which contradicts $\mathit {deg}(\forall x \varphi _1) =k$ since $l({-}s)=k+1$ . Then we have $i(s)\,{\equiv}\ \kern1pt{-}\,$ . Thus we have $\varphi _1 \in \mathrm{U}_k \subseteq \mathrm{U}_k^+$ .

For the converse direction, assume $\varphi _1 \in \mathrm{U}_k^+$ .

• • If $\varphi _1 \notin \mathrm{U}_k$ , then $\varphi _1 \in \displaystyle \bigcup _{i<k} \text {F}_i$ . Thus $\mathit {deg}(\varphi _1)<k$ , and hence, $\mathit {deg}(\forall x \varphi _1)\leq k$ . If $\mathit {deg}(\forall x \varphi _1)< k$ , then $\forall x \varphi _1 \in \displaystyle \bigcup _{i<k} \text {F}_i \subseteq \mathrm{U}_k^+$ . If $\mathit {deg}(\forall x \varphi _1)= k$ , since $i(s)\,{\equiv}\ \kern1pt{-}\,$ for all $s\in \mathit {Alt}(\forall x \varphi _1)$ , we have $\forall x \varphi _1 \in \mathrm{U}_k \subseteq \mathrm{U}_k^+$ .

• • Otherwise, $\mathit {deg}(\varphi _1)=k$ and $i(s)\,{\equiv}\ \kern1pt{-}\,$ for all $s \in \mathit {Alt}(\varphi _1)$ such that $l(s) =k$ . By the definition of $\mathit {Alt}(\forall x\varphi _1)$ , for all $s \in \mathit {Alt}( \forall x \varphi _1)$ , we have $l(s)\leq k$ , and hence, $\mathit {deg}(\forall x\varphi _1)= k$ . In addition, again by the definition of $\mathit {Alt}(\forall x\varphi _1)$ , we have $i(s) \equiv\ \,-$ for all $s\in \mathit {Alt}(\forall x \varphi _1)$ such that $l(s) =k$ . Thus $\forall x \varphi _1 \in \mathrm{U}_k \subseteq \mathrm{U}_k^+$ .

(5): Analogous to (4).

(6): Assume $\forall x \varphi _1 \in \mathrm{E}_{k+1}^+$ . Since $i(s)\,{\equiv}\ {-}\,$ for all $s\in \mathit {Alt}(\forall x \varphi _1)$ , $\forall x \varphi _1$ is not in $\mathrm{E}_{k+1}$ . Then $\forall x \varphi _1 \in \displaystyle \bigcup _{i\leq k} \text {F}_i $ , and hence, $\mathit {deg}(\forall x \varphi _1) \leq k$ .

• • If $\mathit {deg}(\forall x \varphi _1) < k$ , then $\forall x \varphi _1 \in \displaystyle \bigcup _{i< k} \text {F}_i \subseteq \mathrm{U}_k^+$ .

• • If $\mathit {deg}(\forall x \varphi _1) = k$ , since $i(s) \,{\equiv}\ \kern1pt{-}\,$ for all $s\in \mathit {Alt}(\forall x \varphi _1)$ , we have $\forall x \varphi _1 \in \mathrm{U}_k\subseteq \mathrm{U}_k^+$ .

The converse direction is trivial since $ \mathrm{U}_k^+ \subseteq \displaystyle \bigcup _{i< k+1} \text {F}_i \subseteq \mathrm{E}_{k+1}^+$ .

(7): Analogous to (6).⊣

Proof By simultaneous induction on k. The base case is trivial.

For the induction step, assume that the assertion holds for k. We show the assertion for $k+1$ by induction on the structure of formulas.

The case of that $\varphi $ is prime: Suppose $\varphi \in \mathrm{U}_{k+1}^+ \cup \mathrm{E}_{k+1}^+$ . Since $\varphi $ is prime, the assertion for k ensures that there exist $\varphi ' \in \mathrm{U}_k$ and $\varphi ''\in \mathrm{E}_k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )=\mathrm{FV} \left ({\varphi ''}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \varphi '\leftrightarrow \varphi ''$ . Put $\psi ' :\equiv \exists x \varphi '$ and $\psi '' :\equiv \forall x \varphi ''$ where $x\notin \mathrm{FV} \left ({\varphi '}\right ) \cup \mathrm{FV} \left ({\varphi ''}\right )$ . By the definition, it is straightforward to show that $\psi ' \in \mathrm{E}_{k+1}$ , $\psi '' \in \mathrm{U}_{k+1}$ and $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\psi '}\right )=\mathrm{FV} \left ({\psi ''}\right )$ . In addition, by (1) in the proof of Lemma 2.3, we have $\mathsf {HA} \vdash \varphi \leftrightarrow \psi ' \leftrightarrow \psi ''$ .

The case of $\varphi : \equiv \varphi _1 \land \varphi _2$ : Suppose $\varphi \in \mathrm{U}_{k+1}^+$ . By Lemma 4.5.(1), we have $\varphi _1 \in \mathrm{U}_{k+1}^+$ and $\varphi _2 \in \mathrm{U}_{k+1}^+$ . By the induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \mathrm{U}_k$ such that $\mathrm{FV} \left ({\varphi _1}\right )=\mathrm{FV} \left ({\varphi _1'}\right )$ , $\mathrm{FV} \left ({\varphi _2}\right )=\mathrm{FV} \left ({\varphi _2'}\right )$ , $\mathsf {HA} \vdash \varphi _1 \leftrightarrow \varphi _1'$ and $\mathsf {HA} \vdash \varphi _2 \leftrightarrow \varphi _2'$ . Then it is straightforward to show that $\varphi _1' \land \varphi _2'\in \mathrm{U}_k$ , $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi _1' \land \varphi _2'}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \varphi _1'\land \varphi _2'$ . In the same manner, if $\varphi \in \mathrm{E}_{k+1}^+$ , there exists $\varphi '\in \mathrm{E}_{k+1}$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \varphi '$ .

The case of $\varphi : \equiv \varphi _1 \lor \varphi _2$ is similar to the case of $\varphi : \equiv \varphi _1 \land \varphi _2$ (use Lemma 4.5.(2) instead of Lemma 4.5.(1)).

The case of $\varphi : \equiv \varphi _1 \to \varphi _2$ : Suppose $\varphi \in \mathrm{U}_{k+1}^+$ . By Lemma 4.5.(3), we have $\varphi _1\in \mathrm{E}_{k+1}^+ $ and $\varphi _2\in \mathrm{U}_{k+1}^+$ . By the induction hypothesis, there exist $\varphi _1'\in \mathrm{E}_{k+1}$ and $\varphi _2'\in \mathrm{U}_{k+1}$ such that $\mathrm{FV} \left ({\varphi _1}\right )=\mathrm{FV} \left ({\varphi _1'}\right )$ , $\mathrm{FV} \left ({\varphi _2}\right )=\mathrm{FV} \left ({\varphi _2'}\right )$ , $\mathsf {HA} \vdash \varphi _1 \leftrightarrow \varphi _1'$ and $\mathsf {HA} \vdash \varphi _2 \leftrightarrow \varphi _2'$ . Then it is straightforward to show that $\varphi _1'\to \varphi _2'\in \mathrm{U}_{k+1}$ , $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi _1' \to \varphi _2'}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow (\varphi _1'\to \varphi _2')$ . In the same manner, if $\varphi \in \mathrm{E}_{k+1}^+$ , there exists $\varphi '\in \mathrm{E}_{k+1}$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \varphi '$ .

The case of $\varphi :\equiv \forall x \varphi _1$ : First, suppose $\varphi \in \mathrm{U}_{k+1}^+$ . By Lemma 4.5.(4), we have $\varphi _1\in \mathrm{U}_{k+1}^+$ . By the induction hypothesis, there exists $\varphi _1'\in \mathrm{U}_{k+1}$ such that $\mathrm{FV} \left ({\varphi _1}\right )=\mathrm{FV} \left ({\varphi _1'}\right )$ and $\mathsf {HA} \vdash \varphi _1 \leftrightarrow \varphi _1'$ . Then it is straightforward to show that $\forall x \varphi _1' \in \mathrm{U}_{k+1}$ , $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\forall x \varphi _1'}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \forall x \varphi _1'$ . Next, suppose $\varphi \in \mathrm{E}_{k+1}^+$ . By Lemma 4.5.(6), we have $\varphi \in \mathrm{U}_{k}^+$ . The assertion for k ensures that there exists $\varphi '\in \mathrm{U}_k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \varphi '$ . Put $\psi :\equiv \exists y\varphi '$ where $y\notin \mathrm{FV} \left ({\varphi '}\right )$ . By the definition, it is straightforward to show that $\psi \in \mathrm{E}_{k+1}$ , $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\psi }\right )$ and $\mathsf {HA} \vdash \varphi \leftrightarrow \psi $ .

The case of $\varphi :\equiv \exists x \varphi _1$ is similar to the case of $\varphi :\equiv \forall x \varphi _1$ (use Lemma 4.5.(7) and Lemma 4.5.(5)).⊣

Proof Without loss of generality, assume $k>0$ , $\varphi _1 :\equiv \forall x \rho _1(x)$ and $\varphi _2 :\equiv \forall y \rho _2(y)$ where $\rho _1(x), \rho _2(y) \in \Sigma _{k-1}$ (see Remark 2.2). By Lemma 4.4, it suffices to show

$$ \begin{align*} \mathsf{HA} + \neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \neg \left( \forall x \rho_1(x) \lor \forall y \rho_2(y)\right) \leftrightarrow \neg \neg \forall x,y \left( \rho_1(x) \lor \rho_2 (y) \right). \end{align*} $$

The implication from the left to the right is straightforward. The converse implication is shown as follows:

$$ \begin{equation*}\begin{array}{cl} & \neg \neg \forall x,y \left( \rho_1(x) \lor \rho_2 (y) \right) \\ \underset{ \neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow}&\neg \neg \forall x,y \left( \neg \neg \rho_1(x) \lor \neg \neg \rho_2 (y) \right) \\ \longrightarrow & \forall x,y \neg \neg \left( \neg \neg \rho_1(x) \lor \neg \neg \rho_2 (y) \right) \\ \longleftrightarrow & \neg \exists x,y \neg \left( \neg \neg \rho_1(x) \lor \neg \neg \rho_2 (y) \right) \\ \longleftrightarrow & \neg \exists x,y \left( \neg \rho_1(x) \land \neg \rho_2 (y) \right)\end{array} \end{equation*} $$

$$ \begin{equation*}\begin{array}{cl}\longleftrightarrow & \neg \left(\neg \neg \exists x \neg \rho_1(x) \land \neg \neg \exists y \neg \rho_2 (y) \right) \\ \longleftrightarrow & \neg \left(\neg \forall x \neg \neg \rho_1(x) \land \neg \forall y \neg \neg \rho_2 (y) \right) \\ \longleftrightarrow& \neg \neg \left(\forall x \neg \neg \rho_1(x) \lor \forall y \neg \neg \rho_2 (y) \right)\\ \underset{ \neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow}& \neg \neg \left(\forall x \rho_1(x) \lor \forall y \rho_2 (y) \right). \end{array} \end{equation*} $$

⊣

Proof Let $\varphi \in \Pi _k$ . By Lemma 4.8.(1), there exists $\psi \in \Sigma _k$ such that $\mathrm{FV} \left ({\varphi }\right ) = \mathrm{FV} \left ({\psi }\right )$ and $\mathsf {HA} + {\Sigma _k}\text {-}\mathrm{DNE} \vdash \neg \varphi \leftrightarrow \psi $ . By Corollary 4.2, we have $\mathsf {HA} + \neg \neg {{\Sigma _k}\text {-}\mathrm{DNE}} \vdash \neg \varphi \leftrightarrow \neg \neg \psi $ .⊣

Proof Fix an instance of $\neg \neg {{\Sigma _{k-1}}\text {-}\mathrm{LEM}}$

$$ \begin{align*}\varphi :\equiv \neg \neg \forall x \left( \varphi_1(x) \lor \neg \varphi_1(x) \right), \end{align*} $$

where $\varphi _1(x) \in \Sigma _{k-1}$ . Note $\left ( \varphi _1(x) \lor \neg \varphi _1(x) \right ) \in \text {F}_{k-1} \subseteq \mathrm{U}_k^+$ . Since $\mathsf {HA} $ proves $\forall x \neg \neg \left ( \varphi _1(x) \kern1pt{\lor}\kern1pt \neg \varphi _1(x)\kern-0.5pt \right )$ , we have that $\mathsf {HA} \kern1.2pt{+}\kern1.2pt {\mathrm{U}_k}\text {-}\mathrm{DNS}$ proves $ \neg \neg \forall x\! \left ( \varphi _1(x)\kern1pt{\lor}\kern1pt \neg \varphi _1\kern-1pt(x) \kern-0.5pt\right )\kern-0.2pt$ , namely, $\varphi $ .⊣

Proof Immediate from Lemma 4.10 and the fact that ${\Sigma _{k-1}}\text {-}\mathrm{LEM}$ implies ${\Sigma _{k-1}}\text {-}\mathrm{DNE}$ .⊣

Proof By simultaneous induction on k, we show the following two statements (which are in fact equivalent):

1. 1. If $\varphi \in \mathrm{E}_k^+$ , then there exists $\varphi ' \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and

   $$ \begin{align*}\mathsf{HA} + {\mathrm{U}_k^+}\text{-}\mathrm{DNS} \vdash \neg \varphi \leftrightarrow \neg \neg \varphi'. \end{align*} $$

2. 2. If $\varphi \in \mathrm{U}_k^+$ , then there exists $\varphi ' \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and

   $$ \begin{align*}\mathsf{HA} + {\mathrm{U}_k^+}\text{-}\mathrm{DNS} \vdash \neg \neg \varphi \leftrightarrow \neg \neg \varphi'.\end{align*} $$

The base case is trivial (one can take $\varphi '$ as $\varphi $ itself). In what follows, we show the induction step.

For the induction step, assume the items 1 and 2 for $k-1$ . We show the items 1 and 2 for k simultaneously by induction on the structure of formulas. When $\varphi $ is a prime formula, by Lemma 2.3, we have $\varphi '$ which satisfies the requirement. For the induction step, assume that the items 1 and 2 hold for $\varphi _1$ and $\varphi _2$ . When it is clear from the context, we suppress the argument on free variables.

The case o $\varphi _1 \land \varphi _2$ : First, assume $\varphi _1 \land \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{E}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves $\neg \varphi _1 \leftrightarrow \neg \neg \varphi _1'$ and $ \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.7, there exists $\varphi ' \in \Pi _k$ such that

$$ \begin{align*}\mathsf{HA} +\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \neg (\varphi_1' \lor \varphi_2') \leftrightarrow \neg \neg \varphi'.\end{align*} $$

By Corollary 4.11, $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \neg (\varphi_1 \land \varphi_2) \\ \longleftrightarrow & \neg (\neg \neg \varphi_1 \land \neg \neg \varphi_2) \\ \underset{\text{[I.H.] } {\mathrm{U}_k^+}\text{-}\mathrm{DNS}}{\longleftrightarrow} & \neg (\neg \varphi_1' \land \neg \varphi_2') \\ \longleftrightarrow & \neg \neg (\varphi_1' \lor \neg \varphi_2') \\ \underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow} & \neg \neg \varphi'. \end{array} \end{align*} $$

Next, assume $\varphi _1 \land \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{U}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves $\neg \neg \varphi _1 \leftrightarrow \neg \neg \varphi _1' $ and $\neg \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.3, there exists $\varphi ' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi ' \leftrightarrow \varphi _1' \land \varphi _2'$ . Then $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\neg \neg (\varphi_1 \land \varphi_2) \leftrightarrow \neg \neg \varphi_1 \land \neg \neg \varphi_2 \underset{\text{[I.H.] }{\mathrm{U}_k^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} \neg \neg \varphi_1' \land \neg \neg \varphi_2' \leftrightarrow \neg \neg (\varphi_1' \land \varphi_2') \leftrightarrow \neg \neg \varphi'. \end{align*} $$

The case of $\varphi _1 \lor \varphi _2$ : First, assume $\varphi _1 \lor \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{E}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} $ proves $\neg \varphi _1 \leftrightarrow \neg \neg \varphi _1' $ and $\neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.3, there exists $\varphi ' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi ' \leftrightarrow \varphi _1' \land \varphi _2'$ . Then $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \neg (\varphi_1 \lor \varphi_2)\\ \longleftrightarrow & \neg \varphi_1 \land \neg \varphi_2\\ \underset{\text{[I.H.] }{\mathrm{U}_k^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} & \neg \neg \varphi_1' \land \neg \neg \varphi_2'\\ \longleftrightarrow & \neg \neg (\varphi_1' \land \varphi_2')\\ \longleftrightarrow & \neg \neg \varphi'. \end{array} \end{align*} $$

Next, assume $\varphi _1 \lor \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{U}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} $ proves $\neg \neg \varphi _1 \leftrightarrow \neg \neg \varphi _1' $ and $\neg \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.7, there exists $\varphi ' \in \Pi _k$ such that

$$ \begin{align*}\mathsf{HA} +\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \neg (\varphi_1' \lor \varphi_2') \leftrightarrow \neg \neg \varphi'.\end{align*} $$

By Corollary 4.11, $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \neg \neg (\varphi_1 \lor \varphi_2)\\ \longleftrightarrow& \neg \neg ( \neg \neg \varphi_1 \lor \neg \neg \varphi_2)\\ \underset{\text{[I.H.] }{\mathrm{U}_k^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} & \neg \neg ( \neg \neg \varphi_1' \lor \neg \neg \varphi_2')\\ \longleftrightarrow &\neg \neg ( \varphi_1' \lor \varphi_2')\\ \underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow}& \neg \neg \varphi'. \end{array} \end{align*} $$

The case of $\varphi _1 \to \varphi _2$ : First, assume $\varphi _1 \to \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1 \in \mathrm{U}_k^+ $ and $\varphi _2 \in \mathrm{E}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} $ proves $\neg \neg \varphi _1 \leftrightarrow \neg \neg \varphi _1' $ and $ \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.3, there exists $\varphi ' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi ' \leftrightarrow \varphi _1' \land \varphi _2'$ . Then $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \neg (\varphi_1 \to \varphi_2)\\ \longleftrightarrow & \neg \neg \varphi_1 \land \neg \varphi_2\\ \underset{\text{[I.H.] }{\mathrm{U}_k^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} & \neg \neg \varphi_1' \land \neg \neg \varphi_2'\\ \longleftrightarrow & \neg \neg (\varphi_1' \land \varphi_2')\\ \longleftrightarrow & \neg \neg \varphi'. \end{array} \end{align*} $$

Next, assume $\varphi _1 \to \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1 \in \mathrm{E}_k^+ $ and $\varphi _2 \in \mathrm{U}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves $\neg \varphi _1 \leftrightarrow \neg \neg \varphi _1' $ and $ \neg \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . By Lemma 4.7, there exists $\varphi ' \in \Pi _k$ such that

$$ \begin{align*}\mathsf{HA} +\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \neg (\varphi_1' \lor \varphi_2') \leftrightarrow \neg \neg \varphi'.\end{align*} $$

By Corollary 4.11, $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \neg \neg (\varphi_1 \to \varphi_2)\\ \longleftrightarrow & \neg (\neg \neg \varphi_1 \land \neg \varphi_2)\\ \underset{\text{[I.H.] }{\mathrm{U}_k^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} & \neg( \neg \varphi_1' \land \neg \varphi_2')\\ \longleftrightarrow & \neg \neg (\varphi_1' \lor \varphi_2')\\ \underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} }{\longleftrightarrow} & \neg \neg \varphi'. \end{array} \end{align*} $$

The case of $\forall x \varphi _1(x)$ : First, assume $\forall x \varphi _1(x) \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\forall x \varphi _1(x) \in \mathrm{U}_{k-1}^+$ . By the item 2 for $k-1$ , there exists $\varphi ' \in \Pi _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} \vdash \neg \neg \forall x \varphi_1(x) \leftrightarrow \neg \neg \varphi'. \end{align*} $$

By Lemma 4.9, there exists $\varphi ''\in \Sigma _{k-1} \subseteq \Pi _k$ (see Remark 2.5) such that

$$ \begin{align*}\mathsf{HA} +\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \varphi' \leftrightarrow \neg \neg \varphi''. \end{align*} $$

By Corollary 4.11, $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\neg \forall x \varphi_1(x) \underset{\text{[I.H.] }{\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} \neg \varphi' \underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} }{\longleftrightarrow} \neg \neg \varphi''. \end{align*} $$

Next, assume $\forall x \varphi _1(x) \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1(x) \in \mathrm{U}_{k}^+$ . By induction hypothesis, there exists $\varphi _1'(x) \in \Pi _{k}$ such that

$$ \begin{align*}\mathsf{HA} + {\mathrm{U}_{k}^+}\text{-}\mathrm{DNS} \vdash \neg \neg \varphi_1(x) \leftrightarrow \neg \neg \varphi_1' (x). \end{align*} $$

Then $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\neg \neg \forall x \varphi_1(x) \underset{{\mathrm{U}_{k}^+}\text{-}\mathrm{DNS}}{\longleftrightarrow} \forall x \neg \neg \varphi_1(x) \underset{\text{[I.H.] }{\mathrm{U}_{k}^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} \forall x \neg \neg \varphi_1'(x) \underset{{\mathrm{U}_{k}^+}\text{-}\mathrm{DNS}}{\longleftrightarrow} \neg \neg \forall x \varphi_1'(x). \end{align*} $$

The case of $\exists x \varphi _1(x)$ : First, assume $\exists x \varphi _1(x) \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1(x) \in \mathrm{E}_{k}^+$ . By induction hypothesis, there exists $\varphi _1'(x) \in \Pi _{k}$ such that

$$ \begin{align*}\mathsf{HA} + {\mathrm{U}_{k}^+}\text{-}\mathrm{DNS} \vdash \neg \varphi_1(x) \leftrightarrow \neg \neg \varphi_1'(x). \end{align*} $$

Then $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\neg \exists x \varphi_1(x) \leftrightarrow \forall x \neg \varphi_1(x) \underset{\text{[I.H.] }{\mathrm{U}_{k}^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} \forall x \neg \neg \varphi_1'(x) \underset{{\mathrm{U}_{k}^+}\text{-}\mathrm{DNS}}{\longleftrightarrow} \neg \neg \forall x \varphi_1'(x). \end{align*} $$

Next, assume that $\exists x \varphi _1(x) \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\exists x \varphi _1(x) \in \mathrm{E}_{k-1}^+$ . By the item 1 for $k-1$ , there exists $\varphi ' \in \Pi _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} \vdash \neg \exists x \varphi_1(x) \leftrightarrow \neg \neg \varphi'. \end{align*} $$

By Lemma 4.9, there exists $\varphi ''\in \Sigma _{k-1} \subseteq \Pi _k$ (see Remark 2.5) such that

$$ \begin{align*}\mathsf{HA} +\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} \vdash \neg \varphi' \leftrightarrow \neg \neg \varphi''. \end{align*} $$

By Corollary 4.11, $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\neg \neg \exists x \varphi_1(x) \underset{\text{[I.H.] }{\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} \neg \varphi'\underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}} }{\longleftrightarrow} \neg \neg \varphi''. \end{align*} $$

⊣

Proof (1): For formulas $\varphi _1$ and $\varphi _2$ in $\Pi _{k}$ , $\varphi _1 \lor \varphi _2$ is equivalent (over $\mathsf {HA}$ ) to

$$ \begin{align*} \exists k \left(\left(k=0 \to \varphi_1 \right) \land \left( k\neq 0 \to \varphi_2 \right) \right), \end{align*} $$

which is equivalent to some $\varphi \in \Sigma _{k+1}$ such that $\mathrm{FV} \left ({\varphi }\right )= \mathrm{FV} \left ({\varphi _1 \lor \varphi _2}\right )$ over $\mathsf {HA}$ by Lemma 4.3.(2). Therefore any instance of ${(\Pi _{k} \lor \Pi _{k})}\text {-}\mathrm{DNE}$ is derived from some instance of ${\Sigma _{k+1}}\text {-}\mathrm{DNE}$ .

(2): Any instance of ${\Sigma _{k}}\text {-}\mathrm{DNE}$ is derived from some instance of ${(\Pi _{k+1} \lor \Pi _{k+1})}\text {-}\mathrm{DNE}$ since $\varphi \in \Sigma _k$ is equivalent to $\forall y \varphi \in \Pi _{k+1}$ with a variable y not occurring freely in $\varphi $ (cf. Lemma 2.3).

(3): Immediate from (2) and Corollary 4.2.

(4): Note that $\neg \neg \forall x \varphi (x)$ implies $\neg \neg \forall x \neg \neg \varphi (x)$ , which is intuitionistically equivalent to $\forall x \neg \neg \varphi (x)$ . Then any instance of ${\Pi _{k+1}}\text {-}\mathrm{DNE}$ is derived from some instance of ${\Sigma _{k}}\text {-}\mathrm{DNE}$ .⊣

Proof For the proof, we prepare the following auxiliary assertion (which is in fact a consequence of the item 2):

1. 3. If $\varphi \in \mathrm{E}_k^+$ , then there exists $\varphi ' \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and

   $$ \begin{align*}\mathsf{HA} + \neg \neg {{(\Pi_k\lor \Pi_k)}\text{-}\mathrm{DNE}} \vdash \neg \varphi \leftrightarrow \neg \neg \varphi'. \end{align*} $$

We show the items 1–3 by induction on k simultaneously. The base case is trivial (one can take $\varphi '$ as $\varphi $ itself). In what follows, we show the induction step.

Assume the items 1–3 for $k-1$ . Since $\mathsf {HA} + {\Pi _{k-1}}\text {-}\mathrm{DNE} \vdash {\Pi _{k-1}}\text {-}\mathrm{DNS}$ , by the item 2 for $k-1$ , we have

(3) $$ \begin{align} \mathsf{HA} + {\left(\Pi_{k-1}\lor \Pi_{k-1}\right)}\text{-}\mathrm{DNE} \vdash {\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS}. \end{align} $$

We show the items 1–3 simultaneously by induction on the structure of formulas. When $\varphi $ is a prime formula, by Lemma 2.3, we have $\varphi '$ which satisfies the requirement. For the induction step, assume that the items 1–3 hold for $\varphi _1$ and $\varphi _2$ . When it is clear from the context, we suppress the argument on free variables.

The case of $\varphi _1 \land \varphi _2$ : For the second item, assume $\varphi _1 \land \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{U}_k^+$ . By induction hypothesis, there exist $\varphi _1' ,\varphi _2' \in \Pi _k$ such that $\mathsf {HA} +{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE} $ proves $\varphi _1 \leftrightarrow \varphi _1'$ and $\varphi _2 \leftrightarrow \varphi _2'$ . By Lemma 4.3, there exists $\varphi ' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi ' \leftrightarrow \varphi _1'\land \varphi _2'$ . Then $\mathsf {HA} +{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}\varphi_1 \land \varphi_2 \underset{\text{[I.H.] }{(\Pi_k\lor \Pi_k)}\text{-}\mathrm{DNE} }{\longleftrightarrow} \varphi_1' \land \varphi_2' \leftrightarrow \varphi'.\end{align*} $$

For the first and third items, assume $\varphi _1 \land \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{E}_k^+$ . Then we have $\varphi '\in \Sigma _k$ such that $ \mathsf {HA} +{\Sigma _k}\text {-}\mathrm{DNE} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} \vdash \varphi _1 \land \varphi _2 \leftrightarrow \varphi '$ as in the second item. On the other hand, by induction hypothesis, there exist $\varphi _1'' ,\varphi _2'' \in \Pi _k$ such that $\mathsf {HA} + \neg \neg {{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE}} $ proves $\neg \varphi _1 \leftrightarrow \neg \neg \varphi _1'' $ and $ \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2''$ . In addition, by Lemma 4.7, there exists $\varphi '' \in \Pi _k$ such that $\mathsf {HA} + \neg \neg {{\Sigma _{k-1}}\text {-}\mathrm{DNE}} \vdash \neg \neg \varphi '' \leftrightarrow \neg \neg (\varphi _1'' \lor \varphi _2'')$ . Then, by Lemma 5.2, we have that $\mathsf {HA} + \neg \neg {{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE}}$ proves

$$ \begin{align*}\begin{array}{cl} &\neg (\varphi_1 \land \varphi_2) \\ \longleftrightarrow& \neg (\neg \neg \varphi_1 \land \neg \neg \varphi_2)\\ \longleftrightarrow& \neg \neg (\neg \varphi_1 \lor \neg \varphi_2)\\ \underset{\text{[I.H.] }\neg \neg {{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} & \neg \neg (\neg \neg \varphi_1'' \lor \neg \neg \varphi_2'' )\\ \longleftrightarrow& \neg (\neg \varphi_1'' \land \neg \varphi_2'')\\ \longleftrightarrow&\neg \neg (\varphi_1'' \lor \varphi_2'')\\ \underset{\neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow}&\neg \neg \varphi''. \end{array} \end{align*} $$

The case of $\varphi _1 \lor \varphi _2$ : For the second item, assume $\varphi _1 \lor \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{U}_k^+$ . Then, by induction hypothesis, there exist $\rho _1(x_1), \rho _2(x_2) \in \Sigma _{k-1}$ such that $\mathsf {HA} + {(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE} $ proves $\varphi _1 \leftrightarrow \forall x_1 \rho _1(x_1) $ and $\varphi _2 \leftrightarrow \forall x_2 \rho _2(x_2) $ . By Lemma 5.2, $\mathsf {HA} + {(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE} $ proves

$$ \begin{align*}\begin{array}{cl} & \forall x_1 \rho_1(x_1) \lor \forall x_2 \rho_2(x_2)\\ \longrightarrow& \forall x_1, x_2 \left( \rho_1(x_1) \lor \rho_2(x_2)\right)\\ \longrightarrow& \neg \left( \exists x_1 \neg \rho_1 (x_1) \land \exists x_2 \neg \rho_2 (x_2) \right) \\ \longleftrightarrow& \neg \left( \neg \neg \exists x_1 \neg \rho_1 (x_1) \land \neg \neg \exists x_2 \neg \rho_2 (x_2) \right) \\ \underset{{\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow} &\neg \left( \neg \forall x_1 \rho_1 (x_1) \land \neg \forall x_2 \rho_2 (x_2) \right) \\ \longleftrightarrow & \neg \neg \left( \forall x_1 \rho_1 (x_1) \lor \forall x_2 \rho_2 (x_2) \right) \\ \underset{{(\Pi_k\lor \Pi_k)}\text{-}\mathrm{DNE}}{\longrightarrow} &\forall x_1 \rho_1 (x_1) \lor \forall x_2 \rho_2 (x_2). \end{array} \end{align*} $$

By Lemma 4.4, there exists $\xi (x_1, x_2) \in \Sigma _{k-1}$ such that $\mathsf {HA} \vdash \xi (x_1, x_2) \leftrightarrow \rho _1(x_1) \lor \rho _2(x_2)$ . Then we have that $\mathsf {HA} + {(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE} $ proves

$$ \begin{align*}\begin{array}{cl} &\varphi_1 \lor \varphi_2\\ \underset{\text{[I.H.] }{(\Pi_k\lor \Pi_k)}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \forall x_1 \rho_1 (x_1) \lor \forall x_2 \rho_2 (x_2)\\ \underset{{(\Pi_k\lor \Pi_k)}\text{-}\mathrm{DNE}}{\longleftrightarrow} & \forall x_1, x_2 \left( \rho_1(x_1) \lor \rho_2(x_2)\right)\\ \longleftrightarrow& \forall x_1, x_2\, \xi(x_1, x_2) \in \Pi_k. \end{array}\end{align*} $$

For the first and third items, assume $\varphi _1 \lor \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1,\varphi _2 \in \mathrm{E}_k^+$ . By induction hypothesis, there exist $\rho _1(x_1), \rho _2(x_2) \in \Pi _{k-1}$ such that $\mathsf {HA} + {\Sigma _k}\text {-}\mathrm{DNE} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} $ proves $\varphi _1 \leftrightarrow \exists x_1 \rho _1(x_1) $ and $ \varphi _2 \leftrightarrow \exists x_2 \rho _2(x_2) $ . By the item 2 for $k-1$ , there exists $\xi (x_1, x_2) \in \Pi _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE} \vdash \xi(x_1,x_2) \leftrightarrow \rho_1(x_1) \lor \rho_2(x_2).\end{align*} $$

By Lemma 5.2, we have that $\mathsf {HA} + {\Sigma _k}\text {-}\mathrm{DNE} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} $ proves

$$ \begin{align*}\begin{array}{cl} &\varphi_1 \lor \varphi_2 \\ \underset{\text{[I.H.] }{\Sigma_k}\text{-}\mathrm{DNE},\, {\mathrm{U}_k^+}\text{-}\mathrm{DNS}}{\longleftrightarrow} & \exists x_1 \rho_1 (x_1) \lor \exists x_2 \rho_2 (x_2)\\ \longleftrightarrow &\exists x_1, x_2 (\rho_1(x_1) \lor \rho_2(x_2)) \\ \underset{\text{[I.H.] }{(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE}}{\longleftrightarrow} &\exists x_1, x_2\, \xi(x_1, x_2). \end{array} \end{align*} $$

Thus we are done for the first item. For the third item, by induction hypothesis, there exist $\varphi _1'' ,\varphi _2'' \in \Pi _k$ such that $\mathsf {HA} + \neg \neg {{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE}}$ proves $\neg \varphi _1 \leftrightarrow \neg \neg \varphi _1''$ and $ \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2''$ . In addition, by Lemma 4.3, there exists $\varphi '' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi '' \leftrightarrow \varphi _1'' \land \varphi _2''$ . Then $\mathsf {HA} + \neg \neg {{(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE}}$ proves

$$ \begin{align*}\begin{array}{r}\neg (\varphi_1 \lor \varphi_2 ) \leftrightarrow \neg \varphi_1 \land \neg \varphi_2 \underset{\text{[I.H.] }\neg \neg {{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} \neg \neg \varphi_1'' \land \neg \neg \varphi_2''\\ \leftrightarrow \neg \neg (\varphi_1'' \land \varphi_2'') \leftrightarrow \neg \neg \varphi''.\end{array} \end{align*} $$

The case of $\varphi _1 \to \varphi _2$ : For the second item, assume $\varphi _1 \to \varphi _2 \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1 \in \mathrm{E}_k^+$ and $\varphi _2 \in \mathrm{U}_k^+$ . By induction hypothesis, there exist $ \rho _1(x_1), \rho _2(x_2)\in \Sigma _{k-1}$ such that $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}} \vdash \neg \varphi _1 \leftrightarrow \neg \neg \forall x_1 \rho _1(x_1)$ and $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} \vdash \varphi _2 \leftrightarrow \forall x_2 \rho _2(x_2)$ . By Lemma 4.5, we have that $\neg \rho _1(x_1) \to \rho _2(x_2)$ is in $\mathrm{E}_{k-1}^+$ . Then, by the item 1 for $k-1$ , there exists $\xi (x_1, x_2) \in \Sigma _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {\Sigma_{k-1}}\text{-}\mathrm{DNE} + {\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} \vdash \xi(x_1, x_2) \leftrightarrow \left( \neg \rho_1(x_1) \to \rho_2(x_2) \right).\end{align*} $$

Then, using Lemma 5.2 and (3), we have that $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} $ proves

$$ \begin{align*}\begin{array}{cl} &\varphi_1 \to \varphi_2\\ \underset{\text{[I.H.] }{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}{\longleftrightarrow} & \varphi_1 \to \forall x_2 \rho_2 (x_2)\\ \underset{{\Pi_k}\text{-}\mathrm{DNE}}{\longleftrightarrow} & \varphi_1 \to \neg \neg \forall x_2 \rho_2 (x_2)\\ \longleftrightarrow & \neg \neg \varphi_1 \to \neg \neg \forall x_2 \rho_2 (x_2) \\ \underset{\text{[I.H.] }\neg \neg {{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} & \neg \forall x_1 \rho_1(x_1) \to \neg \neg \forall x_2 \rho_2 (x_2)\\ \underset{{\Sigma_{k-1}}\text{-}\mathrm{DNE} }{\longleftrightarrow} & \neg \neg \exists x_1 \neg \rho_1(x_1) \to \neg \neg \forall x_2 \rho_2 (x_2)\\ \longleftrightarrow & \neg \neg \forall x_1, x_2 \left( \neg \rho_1(x_1) \to \rho_2 (x_2) \right)\\ \underset{\text{[I.H.] }{\Sigma_{k-1}}\text{-}\mathrm{DNE} , \, {\mathrm{U}_{k-1}^+}\text{-}\mathrm{DNS} }{\longleftrightarrow} &\neg \neg \forall x_1, x_2 \, \xi( x_1, x_2)\\ \underset{{\Pi_k}\text{-}\mathrm{DNE}}{\longleftrightarrow} &\forall x_1, x_2 \, \xi( x_1, x_2) \in \Pi_k.\\ \end{array} \end{align*} $$

For the first and third items, assume $\varphi _1 \to \varphi _2 \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1 \in \mathrm{U}_k^+$ and $\varphi _2 \in \mathrm{E}_k^+$ . By induction hypothesis, there exists $ \rho _2(x_2) \in \Pi _{k-1}$ such that $\mathsf {HA} + {\Sigma _{k}}\text {-}\mathrm{DNE} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} \vdash \varphi _2 \leftrightarrow \exists x_2 \rho _2(x_2) $ . In addition, by Lemma 5.1, there exists $ \rho _1(x_1)\in \Sigma _{k-1}$ such that $\mathsf {HA} + {\mathrm{U}_k^+}\text {-}\mathrm{DNS} \vdash \neg \neg \varphi _1 \leftrightarrow \neg \neg \forall x_1 \rho _1(x_1)$ . By Lemma 4.5, we have $\neg \rho _2(x_2) \to \neg \rho _1(x_1)$ is in $\mathrm{U}_{k-1}^+$ . Then, by the item 2 for $k-1$ , there exists $\xi (x_1, x_2) \in \Pi _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE} \vdash \xi(x_1, x_2) \leftrightarrow \left(\neg \rho_2(x_2) \to \neg \rho_1(x_1) \right).\end{align*} $$

Then, using Lemma 5.2, we have that $\mathsf {HA} + {\Sigma _{k}}\text {-}\mathrm{DNE}+ {\mathrm{U}_k^+}\text {-}\mathrm{DNS}$ proves

$$ \begin{align*}\begin{array}{cl} & \varphi_1 \to \varphi_2 \\ \underset{\text{[I.H.] }{\Sigma_{k}}\text{-}\mathrm{DNE}, \, {\mathrm{U}_k^+}\text{-}\mathrm{DNS}}{\longleftrightarrow}& \varphi_1 \to \exists x_2 \rho_2(x_2)\\ \underset{{\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \varphi_1 \to \neg \neg \exists x_2 \rho_2(x_2)\\ \longleftrightarrow& \neg \neg \varphi_1 \to \neg \neg \exists x_2 \rho_2(x_2)\\ \underset{ {\mathrm{U}_k^+}\text{-}\mathrm{DNS}}{\longleftrightarrow}&\neg \neg \forall x_1 \rho_1(x_1) \to \neg \neg \exists x_2 \rho_2(x_2)\\ \longleftrightarrow&\neg \neg \exists x_2 \left( \forall x_1 \rho_1(x_1) \to \rho_2(x_2) \right) \\ \underset{{\Pi_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_2 \left(\neg \rho_2(x_2) \to \neg \forall x_1 \rho_1(x_1) \right) \\ \underset{{\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_2 \left(\neg \rho_2(x_2) \to \neg \neg \exists x_1 \neg \rho_1(x_1) \right) \\ \longleftrightarrow &\neg \neg \exists x_2 \neg \neg \exists x_1 \left(\neg \rho_2(x_2) \to \neg \rho_1(x_1) \right) \\ \underset{\text{[I.H.] } {(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_1, x_2 \, \xi (x_1, x_2)\\ \underset{ {\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \exists x_1, x_2 \, \xi (x_1, x_2) \in \Sigma_k .\\ \end{array} \end{align*} $$

Thus we are done for the first item. For the third item, by induction hypothesis, there exist $\varphi _1', \varphi _2' \in \Pi _k$ such that $\mathsf {HA} + {(\Pi _k \lor \Pi _k)}\text {-}\mathrm{DNE} \vdash \varphi _1 \leftrightarrow \varphi _1'$ and $\mathsf {HA} + \neg \neg {{(\Pi _k \lor \Pi _k)}\text {-}\mathrm{DNE}} \vdash \neg \varphi _2 \leftrightarrow \neg \neg \varphi _2'$ . On the other hand, by Lemma 4.3, there exists $\xi ' \in \Pi _k$ such that $\mathsf {HA} \vdash \varphi _1' \land \varphi _2' \leftrightarrow \xi '$ . Then $\mathsf {HA} +\neg \neg {{(\Pi _k \lor \Pi _k)}\text {-}\mathrm{DNE}} $ proves

$$ \begin{align*}\begin{array}{r} \neg( \varphi_1 \to \varphi_2) \leftrightarrow (\neg \neg \varphi_1 \land \neg \varphi_2)\underset{\text{[I.H.] } \neg \neg {{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} \neg \neg \varphi_1' \land \neg \neg \varphi_2'\\ \leftrightarrow \neg \neg (\varphi_1' \land \varphi_2') \leftrightarrow \neg \neg \xi'. \end{array}\end{align*} $$

The case of $\forall x \varphi _1 (x )$ : For the second item, assume $\forall x \varphi _1 (x ) \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $\varphi _1 (x ) \in \mathrm{U}_k^+$ . By induction hypothesis, there exists $\varphi _1'(x ) \in \Pi _{k}$ such that $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} \vdash \varphi _1(x ) \leftrightarrow \varphi _1'(x )$ . Then $\forall x \varphi _1 (x )$ is equivalent to $\forall x \varphi _1'(x ) \in \Pi _k$ over $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} $ .

For the first and third items, assume $\forall x \varphi _1 (x ) \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\forall x \varphi _1 (x ) \in \mathrm{U}_{k-1}^+$ . Then, by the item 2 for $k-1$ , there exists $\xi \in \Pi _{k-1} \subseteq \Sigma _k$ (see Remark 2.5) such that

(4) $$ \begin{align} \mathsf{HA} + {(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE} \vdash \forall x \varphi_1(x ) \leftrightarrow \xi. \end{align} $$

By Lemma 5.2, we are done for the first item. For the third item, by Lemma 4.8, there exists $\xi '\in \Sigma _{k-1} \subseteq \Pi _k$ (see Remark 2.5) such that $\mathsf {HA} + {\Sigma _{k-1}}\text {-}\mathrm{DNE} \vdash \neg \xi \leftrightarrow \xi '$ . By Corollary 4.2, we have $\mathsf {HA} + \neg \neg {{\Sigma _{k-1}}\text {-}\mathrm{DNE}} \vdash \neg \xi \leftrightarrow \neg \neg \xi '$ . In addition,

$$ \begin{align*}\mathsf{HA} + \neg \neg {{(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE}} \vdash \neg \neg \forall x \varphi_1(x ) \leftrightarrow \neg \neg \xi\end{align*} $$

follows from (4). Then, by Lemma 5.2, we have that $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}} $ proves

$$ \begin{align*}\neg \forall x \varphi_1(x ) \underset{\text{[I.H.] } \neg \neg {{(\Pi_{k-1}\lor \Pi_{k-1})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} \neg \xi \underset{ \neg \neg {{\Sigma_{k-1}}\text{-}\mathrm{DNE}}}{\longleftrightarrow} \neg \neg \xi'. \end{align*} $$

Thus we have shown the third item.

The case of $\exists x \varphi _1 (x )$ : For the second item, assume $\exists x \varphi _1 (x ) \in \mathrm{U}_k^+$ . By Lemma 4.5, we have $ \exists x \varphi _1 (x ) \in \mathrm{E}_{k-1}^+$ . Then, by the item 1 for $k-1$ , there exists $\xi \in \Sigma _{k-1}\subseteq \Pi _k$ (see Remark 2.5) such that $\mathsf {HA} +{\Sigma _{k-1}}\text {-}\mathrm{DNE} + {\mathrm{U}_{k-1}^+}\text {-}\mathrm{DNS} \vdash \exists x \varphi _1(x ) \leftrightarrow \xi $ . By Lemma 5.2 and (3), we have $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} \vdash \exists x \varphi _1(x ) \leftrightarrow \xi $ .

For the first and third items, assume $\exists x \varphi _1 (x ) \in \mathrm{E}_k^+$ . By Lemma 4.5, we have $\varphi _1 (x ) \in \mathrm{E}_k^+$ . By induction hypothesis, there exist $\varphi _1'(x ) \in \Sigma _{k}$ and $\varphi _1''(x ) \in \Pi _k$ such that $\mathsf {HA} +{\Sigma _{k}}\text {-}\mathrm{DNE} +{\mathrm{U}_{k}^+}\text {-}\mathrm{DNS} \vdash \varphi _1(x ) \leftrightarrow \varphi _1'(x )$ and $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}} \vdash \neg \varphi _1 (x) \leftrightarrow \neg \neg \varphi _1'' (x)$ . Then $\exists x \varphi _1 (x )$ is equivalent to $\exists x \varphi _1'(x ) \in \Sigma _k$ over $\mathsf {HA} +{\Sigma _{k}}\text {-}\mathrm{DNE} +{\mathrm{U}_{k}^+}\text {-}\mathrm{DNS} $ . Thus we are done for the first item. For the third item, since $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}}$ proves

$$ \begin{align*}\neg \exists x \varphi_1(x )\leftrightarrow \forall x \neg \varphi_1(x ) \underset{ \text{[I.H.] }\neg \neg {{(\Pi_{k}\lor \Pi_{k})}\text{-}\mathrm{DNE}}}{\longleftrightarrow} \forall x \neg \neg \varphi_1''(x ) , \end{align*} $$

we have that $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}}$ proves

$$ \begin{align*}\neg \exists x \varphi_1(x )\leftrightarrow \neg \neg \forall x \neg \neg \varphi_1''(x ). \end{align*} $$

On the other hand, the latter is equivalent to $ \neg \neg \forall x \varphi _1''(x ) $ in the presence of $ \neg \neg {{\Pi _{k}}\text {-}\mathrm{DNE}}$ . Thus we have $\mathsf {HA} + \neg \neg {{(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}} \vdash \neg \exists x \varphi _1(x )\leftrightarrow \neg \neg \forall x \varphi _1''(x )$ .⊣

Proof Since ${\mathrm{U}_k}\text {-}\mathrm{DNS} $ is intuitionistically equivalent to $\neg \neg {{\mathrm{U}_k}\text {-}\mathrm{DNS} }$ (see Remark 2.8), it suffices to show $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} \vdash {\mathrm{U}_k}\text {-}\mathrm{DNS} $ . By Theorem 5.3.(2), any formula $\varphi \in \mathrm{U}_k$ is equivalent to some $\varphi ' \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi }\right ) = \mathrm{FV} \left ({\varphi '}\right )$ over $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}$ . Since $\mathsf {HA} + {\Pi _k}\text {-}\mathrm{DNE} \vdash {\Pi _k}\text {-}\mathrm{DNS}$ , we have $\mathsf {HA} + {(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE} \vdash {\mathrm{U}_k}\text {-}\mathrm{DNS} $ .⊣

Proof We mimic the proof of Theorem 5.3. Thus we first prepare the following auxiliary assertion (which is in fact a consequence of the item 2):

1. 3. If $\varphi \in \mathrm{E}_k^+$ , then there exists $\varphi ' \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and

   $$ \begin{align*}\mathsf{HA} + \neg \neg {\Sigma_{k-1}}\text{-}\mathrm{DNE} \vdash \neg \varphi \leftrightarrow \neg \neg \varphi'. \end{align*} $$

Then we show the items 1–3 by induction on k simultaneously. The base case is trivial. Most of the parts for the induction step is the same as those for Theorem 5.3. The same proof works since each of the logical principles in the items 1 and 2 implies both of them for $k-1$ and the logical principle in the item 3 is the double negation of the logical principle in the item 2 as in Theorem 5.3.

Only the difference with the proof of Theorem 5.3 is in proving the item 1 for $\varphi : \equiv \varphi _1 \to \varphi _2\in \mathrm{E}_k^+$ , where we use Lemma 5.1. Here one can use the item 2 instead of Lemma 5.1. This is because ${\Sigma _k}\text {-}\mathrm{DNE}$ includes ${\Sigma _{k-1}}\text {-}\mathrm{DNE}$ while the verification theory of the item 1 in Theorem 5.3 contains the verification theory of Lemma 5.1. To be absolutely clear, we present the proof of this part: Let $ \varphi _1 \to \varphi _2\in \mathrm{E}_k^+$ . By Lemma 4.5, we have $ \varphi _1 \in \mathrm{U}_k^+$ and $\varphi _2\in \mathrm{E}_k^+$ . By induction hypothesis, there exists $\rho _1(x_1) \in \Sigma _{k-1}$ and $ \rho _2(x_2) \in \Pi _{k-1}$ such that $\mathsf {HA} + {\Sigma _{k-1}}\text {-}\mathrm{DNE} \vdash \varphi _1 \leftrightarrow \forall x_1 \rho _1(x_1)$ and $\mathsf {HA} + {\Sigma _{k}}\text {-}\mathrm{DNE} \vdash \varphi _2 \leftrightarrow \exists x_2 \rho _2(x_2) $ . By Lemma 4.5, we have $\neg \rho _2(x_2) \to \neg \rho _1(x_1)$ is in $\mathrm{U}_{k-1}^+$ . Then, by the item 2 for $k-1$ , there exists $\xi (x_1, x_2) \in \Pi _{k-1}$ such that

$$ \begin{align*}\mathsf{HA} + {\Sigma_{k-2}}\text{-}\mathrm{DNE} \vdash \xi(x_1, x_2) \leftrightarrow \left(\neg \rho_2(x_2) \to \neg \rho_1(x_1) \right).\end{align*} $$

Then $\mathsf {HA} + {\Sigma _{k}}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}\begin{array}{cl} & \varphi_1 \to \varphi_2 \\ \underset{\text{[I.H.] }{\Sigma_{k}}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \varphi_1 \to \exists x_2 \rho_2(x_2)\\ \underset{{\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \varphi_1 \to \neg \neg \exists x_2 \rho_2(x_2)\\ \underset{\text{[I.H.] }{\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \forall x_1 \rho_1(x_1) \to \neg \neg \exists x_2 \rho_2(x_2)\\ \longleftrightarrow&\neg \neg \exists x_2 \left( \forall x_1 \rho_1(x_1) \to \rho_2(x_2) \right) \\ \underset{{\Pi_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_2 \left(\neg \rho_2(x_2) \to \neg \forall x_1 \rho_1(x_1) \right) \\ \underset{{\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_2 \left(\neg \rho_2(x_2) \to \neg \neg \exists x_1 \neg \rho_1(x_1) \right) \\ \longleftrightarrow &\neg \neg \exists x_2 \neg \neg \exists x_1 \left(\neg \rho_2(x_2) \to \neg \rho_1(x_1) \right) \\ \underset{\text{[I.H.] }{\Sigma_{k-2}}\text{-}\mathrm{DNE}}{\longleftrightarrow}&\neg \neg \exists x_1, x_2 \, \xi (x_1, x_2)\\ \underset{ {\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \exists x_1, x_2 \, \xi (x_1, x_2) \in \Sigma_k .\\ \end{array} \end{align*} $$

⊣

Proof By induction on the structure of formulas in prenex normal form.⊣

Proof By induction on the length of the derivations (see the proof of [Reference Kohlenbach6, Proposition 10.3]).⊣

Proof By simultaneous induction on k. The base case is trivial. For the induction step, assume the items 1 and 2 for k to show those for $k+1$ . For the first item, let $\exists x \varphi _1 \in \Sigma _{k+1}$ where $\varphi _1 \in \Pi _k$ . We have that $\mathsf {HA} + {\Sigma _{k+1}}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}\left( \exists x \varphi_1 \right)^N\equiv \neg \neg \exists x {\left( \varphi_1 \right)}_{N} \leftrightarrow \neg \neg \exists x \neg \neg {\left( \varphi_1 \right)}_{N} \underset{\text{[I.H.] }{\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow} \neg \neg \exists x \varphi_1 \underset{{\Sigma_{k+1}}\text{-}\mathrm{DNE}}{\longleftrightarrow} \exists x \varphi_1. \end{align*} $$

For the second item, let $\forall x \varphi _1 \in \Pi _{k+1}$ where $\varphi _1 \in \Sigma _k$ . Since ${\Pi _{k+1}}\text {-}\mathrm{DNE}$ is derived from ${\Sigma _k}\text {-}\mathrm{DNE}$ (see Lemma 5.2.(4)), we have that $\mathsf {HA} + {\Sigma _k}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}\left( \forall x \varphi_1 \right)^N\equiv \neg \neg \forall x \neg \neg {\left( \varphi_1 \right)}_{N} \underset{\text{[I.H.] }{\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow} \neg \neg \forall x \varphi_1 \underset{{\Pi_{k+1}}\text{-}\mathrm{DNE}}{\longleftrightarrow} \forall x \varphi_1. \end{align*} $$

⊣

Proof By induction on the length of the derivations.⊣

Proof Fix a set X of $\mathsf {HA}$ -sentences. By induction on k, one can show straightforwardly that for any k and any $\mathsf {HA}^{*} $ -formula $\varphi $ , if $\mathsf {HA}^{*} + X \vdash \varphi $ with the proof of length k, then $\mathsf {HA} + X \vdash \varphi [\psi /{*} ]$ for any $\mathsf {HA}$ -formula $\psi $ such that the free variables of $\psi $ is not bounded in $\varphi $ . The variable condition is used to verify the case of the axioms and rules for quantifiers.⊣

Proof We show the items 1 and 2 simultaneously by induction on k.

The base case: Since every quantifier-free formula ${\varphi }_{\text {qf}}$ such that $\mathrm{FV} \left ({{\varphi }_{\text {qf}}}\right )=\{\overline {x} \} $ is equivalent to a prime formula $t(\overline {x})=0$ for some closed term t (see e.g., [Reference Kohlenbach6, Proposition 3.8]), by Proposition 6.8, it suffices to show the assertions only for prime formulas. Since $\left ({\left ( \varphi _{\text {p}} \right )}_{N}\right )^{*} \equiv {\varphi _{\text {p}}}^{*} \equiv \varphi _{\text {p}} \lor {*} \equiv {\left ( \varphi _{\text {p}} \right )}_{N} \lor {*}$ , we are done.

The induction step: Assume that the items 1 and 2 hold for k. We first show the item 1 for $k+1$ . Let $\varphi _1\in \Pi _k$ . Since

$$ \begin{align*}\left({\left( \exists x \varphi_1 \right)}_{N}\right)^{*} \equiv \left( \exists x\, {\left( \varphi_1 \right)}_{N}\right)^{*} \equiv \exists x \left({\left( \varphi_1 \right)}_{N}\right)^{*},\end{align*} $$

by induction hypothesis, we have

$$ \begin{align*}\mathsf{HA}^{*} + {\Sigma_k}\text{-}\text{LEM} \vdash\left({\left( \exists x \varphi_1 \right)}_{N}\right)^{*} \leftrightarrow \exists x \left({\left( \varphi_1 \right)}_{N} \lor {*} \right).\end{align*} $$

Since $\mathsf {HA}^{*} $ proves $\exists x \left ({\left ( \varphi _1 \right )}_{N} \lor {*} \right ) \leftrightarrow \left (\exists x \, {\left ( \varphi _1 \right )}_{N} \lor {*} \right ) \equiv \left ( {\left ( \exists x \varphi _1 \right )}_{N} \lor {*} \right )$ , we have

$$ \begin{align*}\mathsf{HA}^{*} + {\Sigma_k}\text{-}\text{LEM} \vdash \left({\left( \exists x \varphi_1 \right)}_{N}\right)^{*} \leftrightarrow \left( {\left( \exists x \varphi_1 \right)}_{N} \lor {*} \right).\end{align*} $$

Thus we have shown the item 1 for $k+1$ .

Next, we show the item 2 for $k+1$ . Let $\varphi _2\in \Sigma _k$ . We shall show that $\mathsf {HA}^{*} + {\Sigma _k}\text {-}\mathrm{DNE}$ (and hence, $\mathsf {HA}^{*} + {\Sigma _{k+1}}\text {-}\mathrm{LEM}$ ) proves ${\left ( \forall x \varphi _2 \right )}_{N} \lor {*} \to \left ( {\left ( \forall x \varphi _2 \right )}_{N} \right )^{*} $ . By Lemma 6.5.(1), we have

(5) $$ \begin{align} \mathsf{HA}+{\Sigma_k}\text{-}\mathrm{DNE} \vdash \varphi_2 \leftrightarrow \left(\varphi_2 \right)^N \equiv \neg \neg {\left( \varphi_2 \right)}_{N}. \end{align} $$

Then we have that $\mathsf {HA}^{*} + {\Sigma _k}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}{\left( \forall x \varphi_2 \right)}_{N}\lor {*} \equiv \left( \forall x \neg \neg {\left( \varphi_2 \right)}_{N} \lor {*} \right) \leftrightarrow \forall x \varphi_2 \lor {*}. \end{align*} $$

By Lemma 6.3, $\mathsf {HA}$ proves $\varphi _2 \to {\left ( \varphi _2 \right )}_{N}$ . Then, using induction hypothesis and the fact that ${\Sigma _{k}}\text {-}\mathrm{DNE}$ derives ${\Sigma _{k-1}}\text {-}\mathrm{LEM}$ , we have that $\mathsf {HA}^{*} + {\Sigma _k}\text {-}\mathrm{DNE}$ proves

$$ \begin{align*}\begin{array}{rcl} {\left( \forall x \varphi_2 \right)}_{N} \lor {*} &\underset{{\Sigma_k}\text{-}\mathrm{DNE}}{\longleftrightarrow}& \forall x \varphi_2 \lor {*}\\ &\longrightarrow & \forall x {\left( \varphi_2 \right)}_{N} \lor {*}\\[2pt] & \longrightarrow & \forall x ({\left( \varphi_2 \right)}_{N} \lor {*}) \\[2pt] & \underset{\text{[I.H.] }{\Sigma_{k-1}}\text{-}\text{LEM}}{\longleftrightarrow} &\forall x \left({\left( \varphi_2 \right)}_{N} \right)^{*}\\ & \longrightarrow & \forall x \left(\left( \left({\left( \varphi_2 \right)}_{N} \right)^{*} \to {*} \right) \to {*} \right)\\ &\longleftrightarrow & \left( {\left( \forall x \varphi_2 \right)}_{N}\right)^{*}. \end{array} \end{align*} $$

In the following, we show the converse direction:

(6) $$ \begin{align} \mathsf{HA}^{*} + {\Sigma_{k+1}}\text{-}\text{LEM} \vdash \left( {\left( \forall x \varphi_2 \right)}_{N} \right)^{*} \to {\left( \forall x \varphi_2 \right)}_{N} \lor {*}. \end{align} $$

Reason in $\mathsf {HA}^{*} + {\Sigma _{k+1}}\text {-}\mathrm{LEM} $ . Suppose $\left ( {\left ( \forall x \varphi _2 \right )}_{N} \right )^{*} $ , equivalently,

(7) $$ \begin{align} \forall x \left(\left( \left({\left( \varphi_2 \right)}_{N} \right)^{*} \to {*} \right) \to {*} \right). \end{align} $$

By induction hypothesis, (7) is equivalent to $\forall x \left (\left ({\left ( \varphi _2 \right )}_{N} \lor {*} \to {*} \right ) \to {*} \right )$ , which is intuitionistically equivalent to

$$ \begin{align*} \forall x \left(\left({\left( \varphi_2 \right)}_{N} \to {*} \right)\to {*} \right). \end{align*} $$

Then we have

(8) $$ \begin{align} \exists x \neg {\left( \varphi_2 \right)}_{N} \to {*}. \end{align} $$

By Lemma 4.8.(2), there exists $\psi _2 \in \Pi _k$ such that $\mathrm{FV} \left ({\varphi _2}\right )=\mathrm{FV} \left ({\psi _2}\right )$ and $\neg \varphi _2$ is equivalent to $\psi _2$ . Since $\exists x \psi _2\in \Sigma _{k+1}$ , by ${\Sigma _{k+1}}\text {-}\mathrm{LEM}$ , we have $\exists x \psi _2 \lor \neg \exists x \psi _2$ , and hence,

$$ \begin{align*} \exists x \neg \varphi_2 \lor \forall x \neg \neg \varphi_2. \end{align*} $$

Then, by (5), we obtain

$$ \begin{align*} \exists x \neg {\left( \varphi_2 \right)}_{N} \lor \forall x \neg \neg {\left( \varphi_2 \right)}_{N}. \end{align*} $$

In the former case, we have ${*}$ by (8). In the latter case, we have ${\left ( \forall x {\varphi _2} \right )}_{N}$ . Thus we have shown (6).⊣

Proof (1)–(3) are immediate from the definition of $\neg _{*} $ (see Definition 6.6). (4) follows from (1)–(3).⊣

Proof Since there exists a closed term t such that $\mathsf {HA} \vdash {\varphi }_{\text {qf}}(x_1, \dots , x_k) \leftrightarrow t(x_1, \dots , x_k) =0 $ for each quantifier-free formula ${\varphi }_{\text {qf}}$ such that $\mathrm{FV} \left ({{\varphi }_{\text {qf}}}\right )=\{x_1, \dots , x_k \}$ (see e.g., [Reference Kohlenbach6, Proposition 3.8]), by Proposition 6.4 and Proposition 6.8, one can assume that formulas in prenex normal form consist of the formulas of form $Q_{1}x_1 \dots Q_{k}x_k \, \varphi _{\text {p}} $ where $Q_i$ s are quantifiers and $\varphi _{\text {p}}$ is prime. We show our assertion by induction on the structure of formulas of this form.

For a prime formula $\varphi _{\text {p}}$ , it is trivial to see that $\mathsf {HA}^{*} $ proves

$$ \begin{align*}\varphi_{\text{p}} \to \varphi_{\text{p}} \lor {*} \to \neg_{*} \neg_{*} \left( \varphi_{\text{p}} \lor {*} \right) \leftrightarrow \left(\left(\varphi_{\text{p}}\right)^N \right)^{*}. \end{align*} $$

Assume the assertion for $\varphi $ . Then, using Lemma 6.12, $\mathsf {HA}^{*}$ proves

$$ \begin{align*}\begin{array}{r} \exists x \varphi \underset{\text{[I.H.]}}{\longrightarrow} \exists x \left(\varphi^N \right)^{*} \equiv \exists x \left(\neg \neg {\varphi}_{N} \right)^{*} \leftrightarrow \exists x \neg_{*} \neg_{*} \left({\varphi}_{N} \right)^{*} \to \neg_{*} \neg_{*} \exists x \left({\varphi}_{N} \right)^{*} \\ \leftrightarrow \left(\left( \exists x \varphi \right)^N \right)^{*} \end{array} \end{align*} $$

and

$$ \begin{align*}\begin{array}{r} \forall x \varphi \underset{\text{[I.H.]}}{\longrightarrow} \forall x \left(\varphi^N \right)^{*} \equiv \forall x \left(\neg \neg {\varphi}_{N} \right)^{*} \leftrightarrow \forall x \neg_{*} \neg_{*} \left({\varphi}_{N} \right)^{*} \to \neg_{*} \neg_{*} \forall x \neg_{*} \neg_{*} \left({\varphi}_{N} \right)^{*} \\ \leftrightarrow \left(\left( \forall x \varphi \right)^N \right)^{*}. \end{array} \end{align*} $$

⊣

Proof Let $\varphi :\equiv \forall x \exists y \varphi _1$ where $\varphi _1\in \Pi _k$ . Since one can freely replace the bound variables, assume that the free variables of $ \exists y \varphi _1 $ are not bounded in $\psi $ and x does not occur in $\psi $ without loss of generality.

Suppose $\mathsf {PA} \vdash \psi \to \forall x \exists y \varphi _1$ . By Proposition 6.4, we have that $\mathsf {HA}$ proves $\neg \neg ({\psi }_{N} \to \forall x \neg \neg \exists y\, {\left ( \varphi _1 \right )}_{N})$ , which is intuitionistically equivalent to $\neg \neg {\psi }_{N} \to \forall x \neg \neg \exists y\, {\left ( \varphi _1 \right )}_{N}$ , namely, $\psi ^N \to \forall x \neg \neg \exists y \, {\left ( \varphi _1 \right )}_{N}$ . Then we have

$$ \begin{align*} \mathsf{HA} \vdash \psi^N \to \neg \neg \exists y \, {\left( \varphi_1 \right)}_{N}. \end{align*} $$

By Proposition 6.8, we have

$$ \begin{align*} \mathsf{HA}^{*} \vdash \left(\psi^N\right)^{*} \to \neg_{*} \neg_{*} \exists y \left( {\left( \varphi_1 \right)}_{N}\right)^{*} , \end{align*} $$

and hence,

$$ \begin{align*} \mathsf{HA}^{*} \vdash \psi \to \neg_{*} \neg_{*} \exists y \left( {\left( \varphi_1 \right)}_{N}\right)^{*} \end{align*} $$

by Lemma 6.13. Then, by Lemma 6.11.(2), we have that $\mathsf {HA}^{*} +{\Sigma _k}\text {-}\mathrm{LEM} $ proves

$$ \begin{align*} \psi \to \neg_{*} \neg_{*} \exists y \left( {\left( \varphi_1 \right)}_{N} \lor {*} \right), \end{align*} $$

which is intuitionistically equivalent to

$$ \begin{align*} \psi \to \neg_{*} \neg_{*} \exists y \, {\left( \varphi_1 \right)}_{N}. \end{align*} $$

Since the free variables of $ \exists y \varphi _1 $ are not bounded in $\psi $ , using Lemma 6.10 with Remark 6.2, we have

(9) $$ \begin{align} \mathsf{HA} +{\Sigma_k}\text{-}\text{LEM} \vdash \psi \to \left( (\exists y \, {\left( \varphi_1 \right)}_{N} \to \exists y \varphi_1) \to \exists y \varphi_1\right). \end{align} $$

On the other hand, by Lemma 6.5.(2) and the fact that $ {\Sigma _k}\text {-}\mathrm{LEM}$ derives ${\Sigma _k}\text {-}\mathrm{DNE}$ , we have that $\mathsf {HA} +{\Sigma _k}\text {-}\mathrm{LEM}$ proves

$$ \begin{align*} {\left( \varphi_1 \right)}_{N} \to \neg \neg {\left( \varphi_1 \right)}_{N} \equiv \left(\varphi_1\right)^N \underset{ {\Sigma_{k-1}}\text{-}\mathrm{DNE}}{\longleftrightarrow} \varphi_1. \end{align*} $$

and hence, $\exists y \, {\left ( \varphi _1 \right )}_{N} \to \exists y \varphi _1$ . Then, by (9), we have $\mathsf {HA} +{\Sigma _k}\text {-}\mathrm{LEM} \vdash \psi \to \exists y \varphi _1$ . By our assumption, x does not occur in $\psi $ , and hence, $\mathsf {HA} +{\Sigma _k}\text {-}\mathrm{LEM} \vdash \psi \to \forall x \exists y \varphi _1 $ follows.⊣

Notation 2. Let ${T} $ be an extension of $\mathsf {HA}$ . Let $\Gamma $ and $\Gamma '$ be classes of $\mathsf {HA}$ -formulas. Then $\mathrm{PNFT}_{T} \left ( {\Gamma }, {\Gamma '} \right ) $ denotes the following statement: for any $\varphi \in \Gamma $ , there exists $\varphi ' \in \Gamma '$ such that $\mathrm{FV} \left ({\varphi }\right )=\mathrm{FV} \left ({\varphi '}\right )$ and ${T} \vdash \varphi \leftrightarrow \varphi ' $ .

Proof Fix an instance of ${(\Pi _{k}\lor \Pi _{k})}\text {-}\mathrm{DNE}$

$$ \begin{align*}\varphi :\equiv \forall x \left(\neg\neg \left(\varphi_1(x) \lor \varphi_2(x) \right) \to \varphi_1(x) \lor \varphi_2(x) \right), \end{align*} $$

where $\varphi _1(x), \varphi _2(x) \in \Pi _{k}(x)$ . Since $\neg \neg \left (\varphi _1(x) \lor \varphi _2(x) \right ) $ and $ \varphi _1(x) \lor \varphi _2(x)$ are in $\mathrm{U}_{k}$ , by our assumption, there exist $\rho (x)$ and $\rho '(x)$ in $\Pi _{k}(x)$ such that ${T} $ proves $ \rho (x) \leftrightarrow \varphi _1(x) \lor \varphi _2(x) $ and $\rho '(x) \leftrightarrow \neg \neg \left (\varphi _1(x) \lor \varphi _2(x)\right ) $ . Since $\mathsf {PA} \vdash \varphi $ and $\mathsf {PA}$ is an extension of ${T}$ , we have $\mathsf {PA} \vdash \rho '(x) \to \rho (x) $ . By Theorem 6.14, we have that $\mathsf {HA} +{\Sigma _{k-2}}\text {-}\mathrm{LEM}$ proves $\rho '(x) \to \rho (x)$ , and hence, $\forall x\left ( \rho '(x) \to \rho (x) \right )$ . Since ${T}$ is an extension of $\mathsf {HA} + {\Sigma _{k-2}}\text {-}\mathrm{LEM}$ , we have ${T} \vdash \forall x\left (\rho '(x) \to \rho (x) \right )$ , and hence, ${T} \vdash \varphi $ .⊣

Proof By induction on k. The base case is trivial. For the induction step, assume $\mathrm{PNFT}_{T} \left ( {\mathrm{U}_{k'}}, {\Pi _{k'}} \right ) $ for all $k' \leq k+1$ . Then, by induction hypothesis, we have ${T} \vdash {\Sigma _{k-1}}\text {-}\mathrm{LEM}$ . Fix an instance of ${\Sigma _{k}}\text {-}\mathrm{LEM}$

$$ \begin{align*}\varphi :\equiv \forall x (\varphi_1(x) \lor \neg \varphi_1(x)), \end{align*} $$

where $\varphi _1(x) \in \Sigma _{k}(x)$ . Since $\varphi \in \mathrm{U}_{k+1}$ , by our assumption, there exists a sentence $\varphi ' \in \Pi _{k+1}$ such that ${T} \vdash \varphi \leftrightarrow \varphi '$ . Since $\mathsf {PA} \vdash \varphi $ , we have $\mathsf {PA} \vdash \varphi '$ . Then, by Theorem 6.14, we have $\mathsf {HA} + {\Sigma _{k-1}}\text {-}\mathrm{LEM} \vdash \varphi '$ , and hence, ${T} \vdash \varphi '$ . Thus we have ${T} \vdash \varphi $ .⊣

Proof The “only if” direction is immediate from Theorem 5.3.(2). We show the converse direction. Assume $\mathrm{PNFT}_{T} \left ( {\mathrm{U}_{k'}}, {\Pi _{k'}} \right ) $ for all $k' \leq k$ . Let $k>0$ without loss of generality. By Lemma 7.2, ${T} \vdash {\Sigma _{k-1}}\text {-}\mathrm{LEM}$ . Then, by Lemma 7.1, we have ${T} \vdash {(\Pi _k\lor \Pi _k)}\text {-}\mathrm{DNE} $ .⊣

Proof By induction on k. The base case is trivial. For the induction step, assume $\mathrm{PNFT}_{T} \left ( {{\mathrm{U}_{k'}}^{\mathrm{dn}}}, {{\Pi _{k'}}^{\mathrm{dn}}} \right ) $ for all $k' \leq k+1$ . Then, by induction hypothesis, ${T}$ proves $\neg \neg {{\Sigma _{k-1}}\text {-}\mathrm{LEM}}$ . Fix an instance of $\neg \neg {{\Sigma _{k}}\text {-}\mathrm{LEM}}$

$$ \begin{align*} \varphi :\equiv \neg \neg \forall x (\varphi_1(x) \lor \neg \varphi_1(x)), \end{align*} $$

where $\varphi _1(x) \in \Sigma _{k}(x)$ . Since $\varphi \in {\mathrm{U}_{k+1}}^{\mathrm{dn}}$ , by our assumption, there exists a sentence $\varphi ' \in \Pi _{k+1}$ such that ${T} \vdash \varphi \leftrightarrow \neg \neg \varphi '$ . Since $\mathsf {PA} \vdash \forall x (\varphi _1(x) \lor \neg \varphi _1(x))$ , we have $\mathsf {PA} \vdash \varphi '$ . Then, by Theorem 6.14, we have $\mathsf {HA} + {\Sigma _{k-1}}\text {-}\mathrm{LEM} \vdash \varphi '$ , and hence, $\mathsf {HA} + \neg \neg {{\Sigma _{k-1}}\text {-}\mathrm{LEM}} \vdash \neg \neg \varphi '$ by Lemma 4.1. Then ${T} \vdash \varphi $ .⊣

Proof The equivalence of (1) and (2) is trivial (cf. the proof of Lemma 5.1). In addition, (3 $\to $ 2) is immediate from Lemma 5.1 and Proposition 4.6. In what follows, we show (2 $\to $ 3). Assume $\mathrm{PNFT}_{T} \left ( {{\mathrm{U}_{k'}}^{\mathrm{dn}}}, {{\Pi _{k'}}^{\mathrm{dn}}} \right ) $ for all $k' \leq k$ . Since ${\mathrm{U}_{k}}\text {-}\mathrm{DNS}$ is intuitionistically equivalent to $\neg \neg {{\mathrm{U}_{k}}\text {-}\mathrm{DNS}}$ (See Remark 2.8), it suffices to show ${T} \vdash \neg \neg {{\mathrm{U}_{k}}\text {-}\mathrm{DNS}}$ . Let $k>0$ without loss of generality. Fix an instance of $\neg \neg {{\mathrm{U}_k}\text {-}\mathrm{DNS}}$