Goal-Oriented Proof-Search in Natural Deduction for Intuitionistic Propositional Logic

Abstract

We address the problem of proof-search in the natural deduction calculus for Intuitionistic propositional logic. Our aim is to improve the usual proof-search procedure where introduction rules are applied upwards and elimination rules downwards. In particular, we introduce $\mathbf {Nbu}$, a variant of the usual natural deduction calculus for Intuitionistic Propositional Logic, and we show that it can be used as a base for a goal-oriented proof-search procedure. We also show that the implementation of our proof-search procedure is competitive with those based on sequent or tableaux calculi.

Appendices

A Completeness of $\mathbf {NI}$

We prove the completeness of $\mathbf {NI}$ (Theorem 2). We start from the calculus $\mathbf {NI}_0$ introduced in [5, 7, 27] to build derivations in normal form for $\mathrm {IPL}$. Rules of $\mathbf {NI}_0$ are the same as $\mathbf {NI}$ (see Fig. 2), except that $\downarrow \!\uparrow ^{}$, $\bot E$ and $\vee E$ are applied without restrictions (p, F and D are any formulas). By the Normalization Theorem of natural deduction calculus of $\mathrm {IPL}$ we get:

Fig. 9

The reduction $\rho$ on $\mathbf {NI}_0$

Theorem 8

(Completeness of $\mathbf {NI}_0$) $A\in \mathrm {IPL}$ iff . $\square$

In order to prove the completeness of $\mathbf {NI}$, we introduce the calculus $\mathbf {NI}_1$, obtained from $\mathbf {NI}_0$ by imposing on $\downarrow \!\uparrow ^{}$ and $\bot E$ the same restrictions as in $\mathbf {NI}$ (see Fig. 2). Note that $\mathbf {NI}_1$-derivations correspond to derivations in “long normal form”.

Let $\rho$ be the reduction on $\mathbf {NI}_0$-derivations defined in Fig. 9; given an $\mathbf {NI}_0$-derivation $\mathscr {D}$ of $\sigma$, $\rho (\mathscr {D})$ is an $\mathbf {NI}_0$-derivation of $\sigma$ obtained from $\mathscr {D}$ by replacing an application of $\mathscr {R}\in \{\!\downarrow \!\uparrow ^{},\bot E\}$ with one or more application of $\mathscr {R}$ to smaller formulas. By repeatedly applying $\rho$, we eventually get an $\mathbf {NI}_0$-derivation $\mathscr {E}$ of $\sigma$ which is an $\mathbf {NI}_1$-derivation. This proves that:

Lemma 17

$\mathbf {NI}_0 \vdash \sigma$ implies $\mathbf {NI}_1 \vdash \sigma$. $\square$

By the completeness of $\mathbf {NI}_0$ and Lemma 17, the completeness of $\mathbf {NI}_1$ follows:

Theorem 9

(Completeness of $\mathbf {NI}_1$) $A\in \mathrm {IPL}$ iff . $\square$

To complete the proof of the completeness of $\mathbf {NI}$, we have to map $\mathbf {NI}_1$ into $\mathbf {NI}$. Let $\mathscr {D}$ be an $\mathbf {NI}_1$-derivation and let $\mathscr {C}^{\vee E}(\mathscr {D})$ be the multiset collecting all the sequents which occur in $\mathscr {D}$ as the conclusion of an applications of $\vee E$. The $\vee$-rank of $\mathscr {D}$, denoted by $\mathrm {Rnk}^{\vee }(\mathscr {D})$, is defined as follows:

Note that, if $\mathrm {Rnk}^{\vee }(\mathscr {D})=0$, then every application of $\vee E$ in $\mathscr {D}$ has a conclusion either of the form , with $F\in {\mathscr {V}} \cup \{\bot \}$, or of the form , hence $\mathscr {D}$ is an $\mathbf {NI}$-derivation. In the next lemma we show how we can reduce the $\vee$-rank of an $\mathbf {NI}_1$-derivation.

Lemma 18

Let $\mathscr {D}$ be an $\mathbf {NI}_1$-derivation of $\sigma$ such that $\mathrm {Rnk}^{\vee }(\mathscr {D})=k$ and $k >0$. Then, there exists an $\mathbf {NI}_1$-derivation of $\sigma$ having $\vee$-rank bounded by $k-1$.

Proof

Since $\mathrm {Rnk}^{\vee }(\mathscr {D})>0$, $\mathscr {D}$ contains at least a subderivation $\mathscr {E}$ of the form

such that:

• $H=H_1\wedge H_2$ or $H= H_1\rightarrow H_2$

• $\mathrm {Rnk}^{\vee }(\mathscr {E}_0) \,=\, \mathrm {Rnk}^{\vee }(\mathscr {E}_1)\,=\, \mathrm {Rnk}^{\vee }(\mathscr {E}_2)\,=\,0$.

Note that $\mathrm {Rnk}^{\vee }(\mathscr {E})=|H|$. We build an $\mathbf {NI}_1$-derivation $\mathscr {F}$ of such that $\mathrm {Rnk}^{\vee }(\mathscr {F}) \le |H|-1$. By replacing in $\mathscr {D}$ the subderivation $\mathscr {E}$ with $\mathscr {F}$, we get an $\mathbf {NI}_1$-derivation of $\sigma$ having $\vee$-rank at most $k-1$.

Let $H=H_1\wedge H_2$. Since $\mathrm {Rnk}^{\vee }(\mathscr {E}_1)=0$ and $\mathrm {Rnk}^{\vee }(\mathscr {E}_2)=0$, the root rule of $\mathscr {E}_1$ and $\mathscr {E}_2$ must be an application of $\wedge I$. Thus, $\mathscr {E}$ has the form

We build $\mathscr {F}$ as follows:

The case $H=H_1\rightarrow H_2$ is similar. In this case $\mathscr {E}$ has the form

Fig. 10

Definition of $\psi (\mathscr {D})$, with $\mathscr {D}$ an $\mathbf {Nbu}$-derivation

Fig. 11

Definition of $\psi '(\mathscr {T})$, with $\mathscr {T}$ an $\uparrow$-reduced $\mathbf {Nbu}$-tree with the open assumption $\sigma$ in the main branch

Fig. 12

Definition of $\theta (\mathscr {E})$, with $\mathscr {E}$ an $\mathbf {LJTbu}$-derivation of

Fig. 13

Definition of $\theta '(\mathscr {E})$, with $\mathscr {E}$ an $\mathbf {LJTbu}$-derivation of ; $\sigma$ is the open assumption in the main branch of $\theta '(\mathscr {E})$

and we define $\mathscr {F}$ as follows

$\square$

By repeatedly applying Lemma 18, we can transform an $\mathbf {NI}_1$-derivation of $\sigma$ into an $\mathbf {NI}_1$-derivation of $\sigma$ having $\vee$-rank 0, namely into an $\mathbf {NI}$-derivation of $\sigma$. By the completeness of $\mathbf {NI}_1$ (Theorem 9), it follows that $A\in \mathrm {IPL}$ iff , and this proves the completeness of $\mathbf {NI}$ (Theorem 2).

B The Isomorphism Between $\mathbf {Nbu}$ and $\mathbf {LJTbu}$

We define a 1-1 translation from $\mathbf {Nbu}$-derivations to $\mathbf {LJTbu}$-derivations, where $\mathbf {LJTbu}$ is the sequent calculus obtained by adding labels b, u to Herbelin calculus $\mathbf {LJT}$ (see Sect. 7). Following [6,7,8, 20], we introduce two maps $\psi$ and $\psi '$, where I-rules are translated bottom-up and E-rules top-down, and we show that they admit inverse maps $\theta$ and $\theta '$ respectively.

An $\mathbf {Nbu}$-tree $\mathscr {T}$ is $\uparrow$-reduced if the root rule of $\mathscr {T}$ is one among $\downarrow \!\uparrow ^{}$, $\bot E$, $\vee E$. Let $\mathscr {T}$ be an $\uparrow$-reduced $\mathbf {Nbu}$-tree with root sequent $\sigma _r$; then, $\sigma _r$ has the form , where $D\in {\mathscr {V}} \cup \{\bot \}$ or $D=D_0\vee D_1$. The main branch of an $\uparrow$-reduced $\mathbf {Nbu}$-tree $\mathscr {T}$ is its left-most branch, namely the branch containing the major premises of rule applications. Note that, except for the root sequent $\sigma _r$, every sequent $\sigma$ in the main branch of $\mathscr {T}$ has the form and it is either a leaf of $\mathscr {T}$ or the conclusion of one of the rules $\mathrm {Id}$, $\wedge E_k$, $\rightarrow E$. By the notation

we indicate an $\uparrow$-reduced $\mathbf {Nbu}$-tree $\mathscr {T}$ such that is the root sequent of $\mathscr {T}$, is an open assumption on the main branch of $\mathscr {T}$ and $\mathscr {T}$ has no other open assumption. To translate $\mathbf {Nbu}$ into $\mathbf {LJTbu}$, we define the maps $\psi$ and $\psi '$ by mutual induction, so that the following properties hold:

• (P1) Let $\mathscr {D}$ be an $\mathbf {Nbu}$-derivation of . Then, $\psi (\mathscr {D})$ is an $\mathbf {LJTbu}$-derivation of .

• (P2) Let $\mathscr {T}$ be the $\uparrow$-reduced $\mathbf {Nbu}$-tree

  

  Then, $\psi '(\mathscr {T})$ is an $\mathbf {LJTbu}$-derivation of .

The maps $\psi$ and $\psi '$ are defined in Figs. 10 and 11 respectively. The latter case in Fig. 10 defines the application of $\psi$ to an $\uparrow$-reduced $\mathbf {Nbu}$-derivation $\mathscr {D}$. Note that the derivation $\mathscr {D}$ is obtained by applying the rule $\mathrm {Id}$ to the open assumption of the $\uparrow$-reduced $\mathbf {Nbu}$-tree

Since is in $\mathscr {D}$ an axiom sequent, it holds that $A\in \varGamma$. To compute $\psi (\mathscr {D})$, we firstly apply $\psi '$ to $\mathscr {T}_0$, yielding by (P2) an $\mathbf {LJTbu}$-derivation of . Since $A\in \varGamma$, we can apply $\mathrm {Cont}$ to get . The map $\psi '$ destructures an $\uparrow$-reduced $\mathbf {Nbu}$-tree $\mathscr {T}$ with the open assumption $\sigma$ in the main branch, starting from $\sigma$ and moving downwards. The base cases (first three cases in Fig. 11) occur when $\sigma$ is a premise of the root rule of $\mathscr {T}$. Otherwise, $\sigma$ must be a premise of $\mathscr {R}\in \{\wedge E_k,\rightarrow E\}$ and we inductively apply $\psi '$ to the subtree of $\mathscr {T}$ having the conclusion of $\mathscr {R}$ as open assumption. One can immediately check that (P1) and (P2) hold.

Now, we define the maps $\theta$ and $\theta '$, corresponding to the inverse of $\psi$ and $\psi '$ respectively, so that the following properties hold:

• (T1) Let $\mathscr {E}$ be an $\mathbf {LJTbu}$-derivation of . Then, $\theta (\mathscr {E})$ is an $\mathbf {Nbu}$-derivation of .

• (T2) Let $\mathscr {E}$ be an $\mathbf {LJTbu}$-derivation of . Then, $\theta '(\mathscr {E})$ is an $\uparrow$-reduced $\mathbf {Nbu}$-tree $\mathscr {T}$ of the form

  

The maps $\theta$ and $\theta '$ are defined in Figs. 12 and 13 respectively. In the latter case of Fig. 12, by applying $\theta '$ to $\mathscr {E}_0$ we get an $\uparrow$-reduced $\mathbf {Nbu}$-tree $\mathscr {T}$ having the open assumption in the main branch and no other open assumption. Note that, since rule $\mathrm {Cont}$ has been applied at the root of $\mathscr {E}$, $A\in \varGamma$. To turn $\mathscr {T}$ into an $\mathbf {Nbu}$-derivation, we can apply $\mathrm {Id}$ to $\sigma$. The following facts can be easily checked:

• (1) $\theta (\psi (\mathscr {D}))\,=\,\mathscr {D}$, where $\mathscr {D}$ is an $\mathbf {Nbu}$-derivation;

• (2) $\theta '(\psi '(\mathscr {T}))\,=\,\mathscr {T}$, where $\mathscr {T}$ is an $\uparrow$-reduced $\mathbf {Nbu}$-derivation with an open assumption in the main branch;

• (3) $\psi (\theta (\mathscr {E})) \,=\,\mathscr {E}$, where $\mathscr {E}$ is an $\mathbf {LJTbu}$-derivation of ;

• (4) $\psi '(\theta '(\mathscr {E})) \,=\,\mathscr {E}$, where $\mathscr {E}$ is an $\mathbf {LJTbu}$-derivation of .

Accordingly, $\theta =\psi ^{-1}$ and $\theta '=\psi '^{-1}$. We conclude that $\psi$ is a 1-1 map between $\mathbf {Nbu}$ and $\mathbf {LJTbu}$.