Dialogue games and deductive information: a dialogical account of the concept of virtual information

Because we are Kantian, because the space of possibilities within which our thinking moves continues to be defined by fundamentally Kantian theses (in ways we will be concerned to explicate), we cannot see that reason has now been realized as a power of knowing, in mathematics in the nineteenth century [...]. And because we do not see this, we cannot understand [...] contemporary mathematical practice, in particular the fact that its proofs can be at once deductive and ampliative, real extensions of our knowledge. (Macbeth, 2014, p. 6).

Abstract

There is a broad debate in contemporary philosophy of logic on the informativeness of proofs. In this context, informative proofs are demonstrations whose premises do not include the content of the conclusion. D’Agostino and Floridi (Synthese 167(2):271–315, 2009) claimed that proofs are informative if they use virtual information. In their terminology, this is the data carried by dischargeable hypotheses, assumptions entertained during proof and eliminated before concluding. Although these authors capture several cases of informative demonstrations, they do not explain the nature of virtual information. Why does the use of dischargeable hypotheses increase the informativeness of a demonstration? Following a Kantian inspiration, D’Agostino and Floridi claim that virtual information adds a synthetic feature to our proofs. However, this appeal to Kantianism is misleading: for Kant, mathematics is synthetic because it requires the construction of concepts in pure intuition. On the other hand, the entertainment of provisional hypotheses does not seem to involve the construction of figures in (transcendental) imagination. Exploring a dialogical characterization of logic, I suggest that virtual information results from the dynamics between a reasoner and her audience. When a reasoner provisionally assumes P, she enacts potential interlocutors who commit to P. Thus, “virtual information” denotes the content of some assumptions of the possible audience of a demonstration. This process of departing from one’s original premises to embrace the suppositions of other people is informative, but it is not synthetic: in this way, the reasoner does not entertain intuitions but only needs to reason through the conceptual content of other people’s premises.

Appendix A: A correspondence between proofs and dialogues

In this appendix, I prove that there is a correspondence between D’Agostino and Floridi’s hierarchy of tractable logics and the hierarchy of dialogical logics presented in Sect. 3.3. To do this, I do not provide a straightforward translation of proofs into dialogues. It’s very hard to obtain such translations: one of the few works exploring this path is (Rahman et al., 2009), but, there, the authors, from the start, diverge significantly from Lorenzen’s standard system. Instead of doing this, in this appendix, I show that D’Agostino and Floridi’s natural deduction system and the dialogical logics introduced here are complete for the same semantics.

In (2015), D’Agostino offered a semantic system for the minimal logic $L_0$ (hereafter, following D’Agostino, let us call it 3ND-semantics). 3ND-semantics is governed by the following matrices.

$\wedge$	1	1/2	0
1	1	1/2	0
1/2	1/2	$\{1/2,0\}$	0
0	0	0	0

$\vee$	1	1/2	0
1	1	1	1
1/2	1	$\{1, 1/2\}$	1/2
0	1	1/2	0

$\lnot$
1	0
1/2	1/2
0	1

$\rightarrow$	1	1/2	0
1	1	1/2	0
1/2	1	$\{1, 1/2\}$	1/2
0	1	1	1

3ND-semantics is non-deterministic: in the matrices displayed above, a place filled with $\{1, 1/2\}$, for example, means that it accepts anyone of these values.

D’Agostino showed that $ND(L_0)$ is complete for 3ND-semantics. So, in what follows, we only need to demonstrate that the dialogical logic for $L_0$ is also complete for this system of truth tables.

Theorem 1

P has a winning strategy in a game over $(P_1\wedge \ldots \wedge P_{n})\rightarrow Q$ in the dialogical logic for $L_0$ if, and only if, $(P_1\wedge \ldots \wedge P_{n})\rightarrow Q$ is logically valid in 3ND-semantics.

Proof

To prove this theorem, we pursue the following strategy. Assume P has no winning strategy in a game over $(P_1\wedge \ldots \wedge P_{n})\rightarrow Q$ in the dialogical logic for $L_0$. We will extend the set of formulae $\{P_1,\ldots ,P_n\}$ to a greater collection $\Delta ^*$ to which we have a model in 3ND-semantics. Subsequently, we will see that this model satisfies all formulae in $\Delta ^*$ but does not verify Q.

To do this, consider a well-ordering $\{\phi _m\}_{m\in \mathbb {N}}$ of the set of formulae of our propositional language. We define $\Delta ^*$ by recursion. First, assume that $\{P_1,\ldots , P_{n}\}=\Delta _0$. Secondly, for any already defined $\Delta _m$, let $\Delta _{m+1}$ be as follows.

• 1. If $\phi _{m}$ is a disjunction $R\vee S$, then $\Delta _{m+1}$ is

  • $\Delta _m\cup \{\phi _m\}$, if $\lnot R, \lnot S \not \in \Delta _m$;

  • $\Delta _m\cup \{\phi _m\}$, if either $\lnot R\in \Delta _m$ or $\lnot S\in \Delta _m$ and P has no winning strategy in a game over $(\bigwedge \Delta _m\wedge \phi _m)\rightarrow Q$;

  • $\Delta _m$, otherwise.

• 2. If $\phi _{m}$ is a conditional $R\rightarrow S$, then $\Delta _{m+1}$ is

  • $\Delta _m\cup \{\phi _m\}$, if $R, \lnot S \not \in \Delta _m$;

  • $\Delta _m\cup \{\phi _m\}$, if either $R\in \Delta _m$ or $\lnot S\in \Delta _m$ and P has no winning strategy in a game over $(\bigwedge \Delta _m\wedge \phi _m)\rightarrow Q$;

  • $\Delta _m$, otherwise.

• 3. If $\phi _{m}$ is a denied conjunction $\lnot (R\wedge S)$, then $\Delta _{m+1}$ is

  • $\Delta _m\cup \{\phi _m\}$, if $R, S\not \in \Delta _m$;

  • $\Delta _m\cup \{\phi _m\}$, if either $R\in \Delta _m$ or $S\in \Delta _m$ and P has no winning strategy in a game over $(\bigwedge \Delta _m\wedge \phi _m)\rightarrow Q$;

  • $\Delta _m$, otherwise.

• 4. If $\phi _m$ is neither a disjunction nor an implication nor a denied conjunction, then $\Delta _{m+1}$ is

  • $\Delta _{m}\cup \{\phi _m\}$, if P has no winning strategy in a game over $(\bigwedge \Delta _m\wedge \phi _m)\rightarrow Q$;

  • $\Delta _m$, otherwise.

Finally, let $\Delta ^*=\underset{m< \omega }{\bigcup }\ \Delta _m$. $\square$

Some facts about $\Delta ^*$ will be important in the rest of this proof.

Proposition 1

The following holds for $\Delta ^*$:

• 5. The set $\Delta ^*$ may not be maximal;

• 6. $Q\not \in \Delta ^*$;

• 7. There is no formula $P'$ such that $P', \lnot P'\in \Delta ^*$;

• 8. $R\wedge S\in \Delta ^*$ if, and only if, $R, S\in \Delta ^*$;

• 9. If $R\vee S\in \Delta ^*$, then, if $\lnot R\in \Delta ^*$, then $S\in \Delta ^*$, and if $\lnot S\in \Delta ^*$, then $R\in \Delta ^*$;

• 10. If $R\rightarrow S\in \Delta ^*$, then, if $R\in \Delta ^*$, then $S\in \Delta ^*$, and if $\lnot S\in \Delta ^*$, then $\lnot R\in \Delta ^*$.

• 11. $\lnot \lnot R\in \Delta ^*$ if, and only if, $R\in \Delta ^*$;

• 12. If $\lnot (R\wedge S)\in \Delta ^*$, then, if $R\in \Delta ^*$, then $\lnot S\in \Delta ^*$, and if $S\in \Delta ^*$, then $\lnot R\in \Delta ^*$;

• 13. $\lnot (R\vee S)\in \Delta ^*$ if, and only if, $\lnot R, \lnot S\in \Delta ^*$;

• 14. $\lnot (R\rightarrow S)\in \Delta ^*$ if, and only if, $R,\lnot S\in \Delta ^*$

• 15. If either $R\in \Delta ^*$ or $S\in \Delta ^*$, then $R\vee S\in \Delta ^*$;

• 16. If either $\lnot R\in \Delta ^*$ or $S\in \Delta ^*$, then $R\rightarrow S\in \Delta ^*$;

• 17. If either $\lnot R\in \Delta ^*$ or $\lnot S\in \Delta ^*$, then $\lnot (R\wedge S)\in \Delta ^*$;

• 18. If $\lnot (R\wedge S)\not \in \Delta ^*$, then either R or S are in $\Delta ^*$;

• 19. If $R\vee S\not \in \Delta ^*$, then either $\lnot R$ or $\lnot S$ are in $\Delta ^*$;

• 20. If $R\rightarrow S\not \in \Delta ^*$, then either R or $\lnot S$ are in $\Delta ^*$.

Proof of Proposition 1

• 5. For instance, if $\Delta _0$ was $\{P\rightarrow Q, \lnot P\rightarrow Q\}$, then this set could be increased neither with P nor $\lnot P$.

• 6. Suppose Q is in some $\Delta _m\subseteq \Delta ^*$. Then, P has the following winning strategy in a game over $(\bigwedge \Delta _{m-1}\wedge Q)\rightarrow Q$: she must force O to discharge Q and, subsequently, she must assert the same formula. This strategy induces the following dialogue:

0	D: P - ! - $(\bigwedge \Delta _{m-1}\wedge Q)\rightarrow Q$
1	A: O - ! - $(\bigwedge \Delta _{m-1}\wedge Q)$	0
2	A: P - ? - $\wedge _2$	1
3	D: O - ! - Q	2
4	D: P - ! - Q	1

Therefore, $Q\not \in \Delta ^*$.

• 7. Including contradictory pairs in any $\Delta _m$ creates a winning strategy for P in the dialogical game. We prove this fact by induction on the complexity of formulae. Consider the case of a sentence of the form $R\vee S$ (the cases of $R\wedge S$ and $R\rightarrow S$ work in a similar way). Assume that, for some $\Delta _m\subseteq \Delta ^*$, $\Delta _m=\Gamma \cup \{R\vee S,\lnot (R\vee S)\}$. In a game over $\bigwedge \Delta _m\rightarrow Q$, P can force O to discharge $\lnot R$, $\lnot S$, and either R or S. By inductive hypothesis, P has a winning strategy in a match in which O defends this collection of formulae.

• 8. The use of a conjunction in a dialogue involves discharging its conjunctives. Consequently, the introduction of the conjunctives in $\Delta ^*$ is redundant. Similar arguments justify items 9 – 14.

• 9. Without loss of generality, assume that $R\in \Delta ^*$. Then, $\lnot R\not \in \Delta ^*$, by item 7. So, the introduction of $R\vee S$ in $\Delta ^*$ is, in a sense, redundant. Without $\lnot R$, one cannot use the disjunction to discharge S. Moreover, the other disjunctive, R, is already in $\Delta ^*$. Hence, including $R\vee S$ does not create a winning strategy for P. A similar argument justifies item 17.

• 10. We need to consider cases here:

  • Case 1: $\lnot R, S\in \Delta ^*$. So, by item 2, $R,\lnot S\not \in \Delta ^*$. Hence, by definition of $\Delta ^*$, $R\rightarrow S$ is in this set.

  • Case 2: $\lnot R, \lnot S\in \Delta ^*$. With the inclusion of $R\rightarrow S$ in $\Delta ^*$, P may force O to defend $\lnot R$, which is a redundant move since $\lnot R$ is already in $\Delta ^*$. Therefore, including the conditional in $\Delta ^*$ does not create a winning strategy for P.

  • Case 3: $R, S\in \Delta ^*$. With the inclusion of $R\rightarrow S$ in $\Delta ^*$, P may force O to defend S, which is a redundant move since S is already in $\Delta ^*$. Therefore, including the conditional in $\Delta ^*$ does not create a winning strategy for P.

• 11. P can attack $\lnot (R\wedge S)$ only in the presence of either R or S. Similar reasoning justifies items 19 and 20.

Let us come back to the proof of Theorem 1. Now, we need to define a model in 3ND-semantics for $\Delta ^*$. So, consider the following truth-assignment v (in 3ND-semantics) for the atomic formulae of our propositional language. For any atomic formula p,

• $v(p)=1$ if, and only if, $p\in \Delta ^*$;

• $v(p)=0$ if, and only if, $\lnot p \in \Delta ^*$;

• $v(p)=1/2$ if, and only if, $p,\lnot p \not \in \Delta ^*$.

Due to the non-deterministic feature of 3ND-semantics, we need to add some semantic clauses for v(P) when P is either a conjunction, a disjunction, or an implication.

• If P is $R\wedge S$ and $v(R)=v(S)=1/2$, then $v(P)=0$;

• If P is either $R\vee S$ or $R\rightarrow S$ and $v(R)=v(S)=1/2$, then $v(P)=1$.

Finally, we will show that the following holds: for any formula P of our language,

$$\begin{aligned}{} & {} v(P)=1 \text { if, and only if, } P\in \Delta ^*. \\{} & {} v(P)=0 \text { if, and only if, } \lnot P\in \Delta ^*. \\{} & {} v(P)=1/2 \text { if, and only if, } P, \lnot P\not \in \Delta ^* \end{aligned}$$

We prove this fact by induction on the complexity of formulae.

If P is atomic, the property holds by definition of v.

Assume that P is a conjunction $R\wedge S$.

(Proof of necessity)

• Suppose $R\wedge S\in \Delta ^*$. By Prop. 1.8, $R, S\in \Delta ^*$. By the inductive hypothesis, $v(R)=v(S)=1$. So, $v(R\wedge S)=1$.

• Suppose $\lnot (R\wedge S)\in \Delta ^*$. We need to consider cases:

  • Case 1: $R, S\not \in \Delta ^*$.

    • Subcase 1: either $\lnot R$ or $\lnot S$ are in $\Delta ^*$. So, by the inductive hypothesis, either v(R) or v(S) is 0. Then, $v(R\wedge S)=0$.

    • Subcase 2: $\lnot R,\lnot S\not \in \Delta ^*$. By the inductive hypothesis, $v(R)=v(S)=1/2$. By definition of v, $v(R\wedge S)=0$.

  • Case 2: either R or S are in $\Delta ^*$. By Prop. 1.12, we have either $\lnot R$ or $\lnot S\in \Delta ^*$. By the inductive hypothesis, v(R) or v(S) is 0. So, $v(R\wedge S)=0$.

• Suppose $R\wedge S, \lnot (R\wedge S)\not \in \Delta ^*$. By Prop. 1.17, $\lnot R,\lnot S\not \in \Delta ^*$. Moreover, by Prop. 1.8, either R or S is not in $\Delta ^*$. However, due to Prop. 1.18, at least one of them is in $\Delta ^*$. Without loss of generality, assume $R\not \in \Delta ^*$ and $S\in \Delta ^*$. By inductive hypothesis, $v(R)=1/2$ and $v(S)=1$. So, $v(R\wedge S)=1/2$.

(Proof of sufficiency)

• Suppose $v(R\wedge S)=1$. Then, $v(R)=v(S)=1$. By the inductive hypothesis, $R, S\in \Delta ^*$. By Prop. 1.8, $R\wedge S\in \Delta ^*$.

• Suppose $v(R\wedge S)=0$. We need to consider cases:

  • Case 1: Either $v(R)=0$ or $v(S)=0$. Without loss of generality, consider $v(R)=0$. By the inductive hypothesis, $\lnot R\in \Delta ^*$. By Prop. 1.17, $\lnot (R\wedge S)\in \Delta ^*$.

  • Case 2: $v(R)=v(S)=1/2$. By the inductive hypothesis, $R, S, \lnot R, \lnot S\not \in \Delta ^*$. So, by the construction of $\Delta ^*$, $\lnot (R\wedge S)\in \Delta ^*$.

• Suppose $v(R\wedge S)=1/2$. Then, either $v(R)=1/2$ and $v(S)=1$ or $v(R)=1$ and $v(S)=1/2$. Without loss of generality, assume that $v(R)=1/2$ and $v(S)=1$. By the inductive hypothesis, $R, \lnot R\not \in \Delta ^*$ and $S\in \Delta ^*$. By Prop. 1.8, $R\wedge S\not \in \Delta ^*$. By Prop. 1.12, $\lnot (R\wedge S)\not \in \Delta ^*$.

Assume that P is a disjunction $R\vee S$.

(Proof of necessity)

• Suppose $R\vee S\in \Delta ^*$. We need to consider cases:

  • Case 1: $\lnot R,\lnot S\not \in \Delta ^*$.

    • Subcase 1: either R or S are in $\Delta ^*$. By the inductive hypothesis, v(R) or v(S) is 1. So, $v(R\vee S)=1$.

    • Subcase 2: $R, S\not \in \Delta ^*$. By inductive hypothesis, $v(R)=v(S)=1/2$. By definition of v, $v(R\vee S)=1$.

  • Case 2: either $\lnot R$ or $\lnot S$ are in $\Delta ^*$. By Prop. 1.9, we have either S or $R\in \Delta ^*$. By the inductive hypothesis, v(S) or v(R) is 1. So, $v(R\vee S)=1$.

• Suppose $\lnot (R\vee S)\in \Delta ^*$. By Prop. 1.13, $\lnot R, \lnot S\in \Delta ^*$. By inductive hypothesis, $v(R)=v(S)=0$. Consequently, $v(R\vee S)=0$.

• Suppose $R\vee S, \lnot (R\vee S)\not \in \Delta ^*$. Then, by Prop. 1.15, $R, S\not \in \Delta ^*$. Furthermore, by Prop. 1.13, either $\lnot R$ or $\lnot S$ are not in $\Delta ^*$. On the other hand, due to Prop. 1.19, at least one of these formulae is in $\Delta ^*$. Without loss of generality, assume $\lnot R\not \in \Delta ^*$ and $\lnot S\in \Delta ^*$. By inductive hypothesis, $v(R)=1/2$ and $v(S)=0$. So, $v(R\vee S)=1/2$.

(Proof of sufficiency)

• Suppose $v(R\vee S)=1$. We need to consider cases:

  • Case 1: either $v(R)=1$ or $v(S)=1$. Without loss of generality, let $v(R)=1$. By the inductive hypothesis, $R\in \Delta ^*$. By Prop. 1.15, $R\vee S\in \Delta ^*$.

  • Case 2: $v(R)=v(S)=1/2$. By the inductive hypothesis, $R, S, \lnot R, \lnot S\not \in \Delta ^*$. By the construction of $\Delta ^*$, $R\vee S$ is in this set.

• Suppose $v(R\vee S)=0$. Then, $v(R)=v(S)=0$. By the inductive hypothesis, $\lnot R,\lnot S\in \Delta ^*$. Then, by Prop. 1. 13, $\lnot (R\vee S)\in \Delta ^*$.

• Suppose $v(R\vee S)=1/2$. Then, either $v(R)=1/2$ and $v(S)=0$ or $v(R)=0$ and $v(S)=1/2$. Without loss of generality, let us assume that $v(R)=1/2$ and $v(S)=0$. By the inductive hypothesis, $R,\lnot R\not \in \Delta ^*$ and $\lnot S\in \Delta ^*$. By Prop. 1. 13, $\lnot (R\vee S)\not \in \Delta ^*$. By Prop. 1. 9, $R\vee S\not \in \Delta ^*$.

Assume that P is an implication $R\rightarrow S$.

(Proof of necessity).

• Suppose $R\rightarrow S\in \Delta ^*$. We need to consider cases:

  • Case 1: $R, \lnot S\not \in \Delta ^*$.

    • Subcase 1: either $\lnot R$ or S are in $\Delta ^*$. By the inductive hypothesis, either $v(R)=0$ or $v(S)=1$. So, $v(R\rightarrow S)=1$.

    • Subcase 2: $\lnot R, S\not \in \Delta ^*$. By the inductive hypothesis, $v(R)=v(S)=1/2$. So, by the definition of v, $v(R\rightarrow S)=1$.

  • Case 2: either R or $\lnot S$ are in $\Delta ^*$. Then, by Prop. 1.10, either S or $\lnot R$ is in $\Delta ^*$. By the inductive hypothesis, either $v(S)=1$ or $v(R)=0$. So, $v(R\rightarrow S)=1$.

• Suppose $\lnot (R\rightarrow S)\in \Delta ^*$. By Prop. 1.14, $R, \lnot S\in \Delta ^*$. By the inductive hypothesis, $v(R)=1$ and $v(S)=0$. Consequently, $v(R\rightarrow S)=0$.

• Suppose $R\rightarrow S, \lnot (R\rightarrow S)\not \in \Delta ^*$. By Prop. 1.20, either R or $\lnot S$ are in $\Delta ^*$. Moreover, by Prop. 1. 14, either R or $\lnot S$ is not in $\Delta ^*$. Without loss of generality, assume $R\in \Delta ^*$ and $\lnot S\not \in \Delta ^*$. In this case, the inclusion of $R\rightarrow S\in \Delta ^*$ creates a winning strategy for P by enabling him to discharge S. Hence, $S\not \in \Delta ^*$. By the inductive hypothesis, $v(R)=1$ and $v(S)=1/2$. Consequently, $v(R\rightarrow S)=1/2$.

(Proof of sufficiency)

• Suppose $v(R\rightarrow S)=1$. We need to consider cases:

  • Case 1: either $v(R)=0$ or $v(S)=1$. Then, by the inductive hypothesis, either $\lnot R$ or S are in $\Delta ^*$. By Prop. 1.16, $R\rightarrow S\in \Delta ^*$.

  • Case 2: $v(R)=v(S)=1/2$. By the inductive hypothesis, $R, S, \lnot R, \lnot S\not \in \Delta ^*$. By the construction of $\Delta ^*$, $R\rightarrow S \in \Delta ^*$.

• Suppose $v(R\rightarrow S)=0$. Then, $v(R)=1$ and $v(S)=0$. By the inductive hypothesis, $R,\lnot S\in \Delta ^*$. By Prop. 1.14, $\lnot (R\rightarrow S)\in \Delta ^*$.

• Suppose $v(R\rightarrow S)=1/2$. Then, either $v(R)=1/2$ and $v(S)=0$ or $v(R)=1$ and $v(S)=1/2$. Without loss of generality, assume $v(R)=1/2$ and $v(S)=0$. By the inductive hypothesis, $R,\lnot R\not \in \Delta ^*$ and $\lnot S\in \Delta ^*$. By Prop. 1. 14, $\lnot (R\rightarrow S)\not \in \Delta ^*$. By Prop. 1.10, $R\rightarrow S\not \in \Delta ^*$.

Assume that P is a negation $\lnot R$.

(Proof of necessity)

• Suppose $\lnot R\in \Delta ^*$. Then, by the inductive hypothesis, $v(R)=0$. So, $v(\lnot R)=1$.

• Suppose $\lnot \lnot R\in \Delta ^*$. By Prop. 1.11, $R\in \Delta ^*$. By the inductive hypothesis, $v(R)=v(\lnot \lnot R)=1$.

• Suppose $\lnot R, \lnot \lnot R\not \in \Delta ^*$. Then, $R\not \in \Delta ^*$. By the inductive hypothesis, $v(R)=v(\lnot R)=1/2$.

(Proof of sufficiency)

• Suppose $v(\lnot R)=1$. Then, $v(R)=0$. By the inductive hypothesis, $\lnot R\in \Delta ^*$.

• Suppose $v(\lnot R)=0$. Then, $v(R)=1$. By the inductive hypothesis, $R\in \Delta ^*$. Then, By Prop. 1. 11, $\lnot \lnot R\in \Delta ^*$.

• Suppose $v(\lnot R)=1/2$. Then, $v(R)=1/2$. By the inductive hypothesis, $R,\lnot R\not \in \Delta ^*$.

So, v attributes 1 for every formula in $\Delta ^*$. Since $\Delta _0\subseteq \Delta ^*$, then v defines a model in 3ND-semantics that verifies the sentences $P_1,\ldots , P_n$ but does not satisfy Q. $\square$

Corollary 1

For any set of sentences $\{P_1,\ldots , P_n, Q\}$, the deductive sequence $P_1,\ldots ,P_n\Rightarrow Q$ holds in $ND(L_0)$ if, and only if, there is a winning strategy for P in the $L_0$-game over $(P_1\wedge \ldots \wedge P_n) \rightarrow Q$.

Finally, let us prove the correspondence between D’Agostino and Floridi’s hierarchy and the dialogical logics presented here.

Theorem 2

For any set of sentences $\{P_1,\ldots , P_m, Q\}$, for every $n\ge 0$, the deduction $P_1,\ldots ,P_m\Rightarrow _n Q$ holds if, and only if, there is a winning strategy for P in the $L_n$-game over $(P_1\wedge \ldots \wedge P_m) \rightarrow Q$.

Proof

Assume that $P_1,\ldots ,P_m\Rightarrow _{n+1} Q$ holds. Then, there is a formula $R\in$ Sub($\{P_1,\ldots , P_m, Q\}$) such that

• $P_1,\ldots ,P_m, R\Rightarrow _{n} Q$;

• $P_1,\ldots ,P_m, \lnot R\Rightarrow _{n} Q$.

By the inductive hypothesis, there are winning strategies for P in the games $G_n((P_1\wedge \ldots \wedge P_m\wedge R)\rightarrow Q)$ and $G_n((P_1\wedge \ldots \wedge P_m\wedge \lnot R)\rightarrow Q)$. It is easy to verify that, based on these strategies, we can define winning strategies for P in the games $G_n(R\rightarrow [(P_1\wedge \ldots \wedge P_m)\rightarrow Q])$ and $G_n(\lnot R \rightarrow [(P_1\wedge \ldots \wedge P_m)\rightarrow Q])$. So, P has a winning strategy in the game $G_n^+((P_1\wedge \ldots \wedge P_m)\rightarrow Q, R)$. Therefore, P has a winning strategy in $G_{n+1}((P_1\wedge \ldots \wedge P_m)\rightarrow Q)$. $\square$