Ground-theoretic equivalence

Abstract

Say that two sentences are ground-theoretically equivalent iff they are interchangeable salva veritate in grounding contexts. Notoriously, ground-theoretic equivalence is a hyperintensional matter: even logically equivalent sentences may fail to be interchangeable in grounding contexts. Still, there seem to be some substantive, general principles of ground-theoretic equivalence. For example, it seems plausible that any sentences of the form $A \wedge B$ and $B \wedge A$ are ground-theoretically equivalent. What, then, are in general the conditions for two sentences to stand in the relation of ground-theoretic equivalence, and what are the logical features of that relation? This paper develops and defends an answer to these questions based on the mode-ified truthmaker theory of content presented in my recent paper ‘Towards a theory of ground-theoretic content’ (Krämer in Synthese 195(2):785–814, 2018).

Appendices

A Soundness and completeness

1.1 A.1 Definitions

Recall the definition of $\mathscr {L}_\approx$ as based on a propositional language with connectives $\wedge$, $\vee$, and $\lnot$, augmented by all expressions of the form $A \approx B$ where A,  B are sentences of the propositional language (and thus do not already include occurrences of $\approx$). We call expressions of the latter form equivalences, and reserve the label ‘sentence’ for the purely truth-functional sentences. The language comprising only the sentences of $\mathscr {L}_\approx$ will be called $\mathscr {L}_B$.

For ease of reference, we repeat some relevant definitions from Krämer (2018):

Definition 1

(Mode-Space) A mode-space is a pair $\langle M, V\rangle$ such that

• 1. M is a non-empty set

• 2. V is a non-empty, partial function taking non-empty countable sequences of non-empty subsets of M into members of M

• 3. the domain of V is closed under non-empty subsequences and countable concatenation of sequences

• 4. $V(\gamma _1^\frown \gamma _2^\frown \ldots ) = V(\delta _1^\frown \delta _2^\frown \ldots )$ whenever $V(\gamma _1) = V(\delta _1), V(\gamma _2) = V(\delta _2), \ldots$

Informally, M is the set of modes. Any non-empty set of modes is a proposition. V is the via-function, mapping some sequences of propositions $P_1,\,P_2,\,\ldots$ to the mode of verifying via $P_1,\,P_2,\,\ldots$. Modes which are never the value of V are called fundamental, and their set is denoted $M^F$. All other modes are called derivative and their set is denoted $M^D$. Of any derivative modes m and n we can form the fusion $m \sqcup n$ which is $V\langle P_1,\,P_2,\,\ldots ,\,Q_1,\,Q_2,\,\ldots \rangle$ when $V\langle P_1,\,P_2,\,\ldots \rangle = m$ and $V\langle Q_1,\,Q_2,\,\ldots \rangle = n$. If V is defined for $\langle P\rangle$, P is called raisable, and the set of raisable contents is denoted $\mathscr {R}$.

Conjunction, disjunction, and negation are now defined as follows. First, we define binary operations of fusion ($\sqcup$) and (disjunctive) addition ($+$) for propositions that are subsets of $M^D$:

Next, for raisable contents P, we define an operation of raising on unilateral contents, which given an input P yields a content $\uparrow \!\!P$ just like P except in that it may be verified via P, and in terms of raising, fusion, and addition, we define conjunction and disjunction on raisable unilateral contents:

Finally, we define conjunction, disjunction, and negation on bilateral contents in $\mathscr {R}\times \mathscr {R}$:

For the operations so defined to behave as desired, the background mode-space needs to satisfy two important conditions:

Definition 2

A mode-space $\langle M, V\rangle$ is called complete iff $P \sqcup Q \in \mathscr {R}$, $P + Q \in \mathscr {R}$, and $\uparrow \!\!P \in \mathscr {R}$ whenever $P, Q \in \mathscr {R}$.

Definition 3

A mode-space $\langle M, V\rangle$ is called constrained iff $V(\gamma ) = V(\delta )$ only if the same ground-set corresponds to $\gamma$ and $\delta$.

Note that in a constrained mode-space, every derivative mode m corresponds to a unique ground-set, which is denoted by |m|. We shall henceforth deal only with complete and constrained mode-spaces.

Definition 4

A unilateral proposition P is

• irreflexive iff: P does not occur in $\gamma$ whenever $V(\gamma ) \in P$,

• closed iff: P contains a mode with ground-set $\varGamma _1 \cup \varGamma _2 \cup \ldots$ whenever P contains modes with ground-sets $\varGamma _1, \,\varGamma _2, \,\ldots$,

• transitive iff: P includes a mode with ground-set $\varGamma ,\, \varDelta$ whenever P includes a mode with ground-set $\varDelta ,\, Q$ and Q includes a mode with ground-set $\varGamma$.

• normal iff: irreflexive, closed, and transitive.

We define the two classes of mode-spaces with respect to which we shall establish soundness and completeness results.

Definition 5

A mode-space $\langle M, V\rangle$ is

• intermediate iff: $V(\gamma ) = V(\delta )$ whenever the same multi-set underlies $\gamma$ and $\delta$

• extensional iff: $V(\gamma ) = V(\delta )$ whenever the same set underlies $\gamma$ and $\delta$

Note that the class of intermediate mode-spaces is exactly the class of mode-spaces compatible with an order-insensitive but repetition-sensitive conception of modes, whereas the class of extensional mode-spaces correspondingly reflects the repetition-insensitive conception of modes.

We are now in a position to give a mode-space semantics for $\mathscr {L}_\approx$ by defining the notions of a model, truth in a model, and validity in a class of models.

Definition 6

If $\langle M, V\rangle$ is a mode-space, then $\mathscr {M}= \langle M, V, [\cdot ]\rangle$ is a model based on $\langle M, V\rangle$ just in case $[\cdot ]$ is a function mapping every sentence in $\mathscr {L}_\approx$ to a member of $\mathscr {R}\times \mathscr {R}$ so that for all sentences $A, B \in \mathscr {L}_\approx$:

• $[\lnot A] = \lnot [A]$

• $[A \wedge B] = [A] \wedge [B]$

• $[A \vee B] = [A] \vee [B]$

$\mathscr {M}$ is a model just in case $\mathscr {M}$ is a model based on some mode-space.

Definition 7

(Truth and Validity) For any equivalence $A \approx B \in \mathscr {L}_\approx$:

• $A \approx B$ is true in a model $\mathscr {M}$ ($\mathscr {M}\models A \approx B$) iff $[A] \approx [B]$

• $A \approx B$ is valid in a class of models $\mathscr {C}$ ($\models _\mathscr {C}A \approx B$) iff true in every model in $\mathscr {C}$.

1.2 A.2 Preparatory results

For ease of reference, we repeat the central theorems of Krämer (2018) that we shall need. Throughout, we tacitly restrict attention to complete and constrained mode-spaces.

Lemma 1

The introduction and elimination principles for bilateral propositions hold (theorems 5 and 6):

($<\!\!\wedge$):

$\varvec{\varGamma }< \mathbf {P}\wedge \mathbf {Q}$ iff $\varvec{\varGamma }\le \{\mathbf {P}, \mathbf {Q}\}$

($<\!\!\vee$):

$\varvec{\varGamma }< \mathbf {P}\vee \mathbf {Q}$ iff $\varvec{\varGamma }\le \mathbf {P}$ or $\varvec{\varGamma }\le \mathbf {Q}$ or $\varvec{\varGamma }\le \{\mathbf {P}, \mathbf {Q}\}$

($<\!\!\lnot \lnot$):

$\varvec{\varGamma }< \lnot \lnot \mathbf {P}$ iff $\varvec{\varGamma }\le \mathbf {P}$

($<\!\!\lnot \wedge$):

$\varvec{\varGamma }< \lnot (\mathbf {P}\wedge \mathbf {Q})$ iff $\varvec{\varGamma }\le \lnot \mathbf {P}$ or $\varvec{\varGamma }\le \lnot \mathbf {Q}$ or $\varvec{\varGamma }\le \{\lnot \mathbf {P}, \lnot \mathbf {Q}\}$

($<\!\!\lnot \vee$):

$\varvec{\varGamma }< \lnot (\mathbf {P}\vee \mathbf {Q})$ iff $\varvec{\varGamma }\le \{\lnot \mathbf {P}, \lnot \mathbf {Q}\}$

Lemma 2

The following structural principles hold for normal unilateral propositions (theorem 7):

• 1. $P \not \prec P$

• 2. If $\varGamma _1 < P$, $\varGamma _2 <P$, ..., then $\varGamma _1,\, \varGamma _2,\, \ldots < P$.

• 3. If $\varGamma \le P$ and $Q \prec P$ for all $Q \in \varGamma$, then $\varGamma < P$.

• 4. If $\varGamma ,\, P \le P$, then $\varGamma \le P$.

• 5. If $\varGamma _1 \le P,~\varGamma _2 \le P,~\ldots$, then $\varGamma _1 \cup \varGamma _2 \cup \ldots \le P$.

• 6. If $\varGamma < P$ and $\varDelta ,\, P < Q$, then $\varGamma ,\, \varDelta < Q$.

• 7. If $\varGamma _1 \le P_1$, $\varGamma _2 \le P_2,$ ..., and $P_1,\, P_2,\, \ldots \le Q$ then $\varGamma _1,\, \varGamma _2,\, \ldots \le Q$

• 8. If $P \preceq Q$ and $Q \prec R$ then $P \prec R$

• 9. If $P \prec Q$ and $Q \preceq R$ then $P \prec R$

• 10. If $P \preceq Q$ and $Q \preceq R$ then $P \preceq R$

Lemma 3

Normality of unilateral propositions is preserved under $\wedge$, $\vee$, and $\uparrow \!\!~$ (theorem 8).

We state without proof the following straightforward facts about ground-theoretic equivalence:

Lemma 4

$\approx$ has the following properties:

• 1. $\approx$ is an equivalence relation—for all $\mathbf {P},\,\mathbf {Q},\,\mathbf {R}$ we have (i) $\mathbf {P}\approx \mathbf {P}$, (ii) if $\mathbf {P}\approx \mathbf {Q}$ then $\mathbf {Q}\approx \mathbf {P}$, and (iii) if $\mathbf {P}\approx \mathbf {Q}$ and $\mathbf {Q}\approx \mathbf {R}$ then $\mathbf {Q}\approx \mathbf {R}$

• 2. $\approx$ is preserved under conjunction, disjunction, and double negation—for all $\mathbf {P},\,\mathbf {Q},\,\mathbf {R}$, if $\mathbf {P}\approx \mathbf {Q}$, then (i) $\mathbf {P}\wedge \mathbf {R}\approx \mathbf {Q}\wedge \mathbf {R}$, (ii) $\mathbf {P}\vee \mathbf {R}\approx \mathbf {Q}\vee \mathbf {R}$, and (iii) $\lnot \lnot \mathbf {P}\approx \lnot \lnot \mathbf {Q}$

• 3. $\approx$ satisfies the DeMorgan rules—for all $\mathbf {P},\,\mathbf {Q}$: (i) $\lnot (\mathbf {P}\wedge \mathbf {Q}) \approx \lnot \mathbf {P}\vee \lnot \mathbf {Q}$ and (ii) $\lnot (\mathbf {P}\vee \mathbf {Q}) \approx \lnot \mathbf {P}\wedge \lnot \mathbf {Q}$

We now establish some easy sufficient identity conditions for unilateral contents:

Lemma 5

Let P,  Q be raisable unilateral propositions.

• (i) If the mode-space is intermediate: $P \wedge Q = Q \wedge P$ and $P \vee Q = Q \vee P$.

• (ii) If the mode-space is extensional: $P \wedge P = P \vee P$.

• (iii) If the mode-space is extensional and P closed: $P \vee P = \uparrow \!\!P$

Proof

For (i): We show that the operation of fusion on the modes is commutative. Let m and n be derivative modes, and suppose $m = V(\gamma )$ and $n = V(\delta )$. By the definition of fusion, $m \sqcup n = V(\gamma ^\frown \delta )$ and $n \sqcup m = V(\delta ^\frown \gamma )$. Since the mode-space is assumed to be intermediate, V maps two sequences to the same mode if they correspond to the same multi-set, and since $\gamma ^\frown \delta$ and $\delta ^\frown \gamma$ correspond to the same multi-set, it follows that $m \sqcup n = n \sqcup m$. From this, the result follows immediately by definition of $\wedge$ and $\vee$.

For (ii): By application of the definitons, $P \wedge P = \uparrow \!\!P \sqcup \uparrow \!\!P \subseteq \uparrow \!\!P \cup \uparrow \!\!P \cup (\uparrow \!\!P \sqcup \uparrow \!\!P) = \uparrow \!\!P + \uparrow \!\!P = P \vee P$. It remains to show that $\uparrow \!\!P \subseteq \uparrow \!\!P \sqcup \uparrow \!\!P$. This follows from the idempotence of $\sqcup$ as defined on derivative modes in extensional mode-spaces. Let m be a derivative mode and assume $m = V(\gamma )$. Then $m \sqcup m = V(\gamma ^\frown \gamma )$ and since the same set underlies $\gamma$ and $\gamma ^\frown \gamma$, by extensionality of the mode-space, $m = V(\gamma ) = V(\gamma ^\frown \gamma ) = m \sqcup m$.

For (iii): $\uparrow \!\!P \subseteq P \vee P$ is immediate by definition of $\vee$. It remains to show that $\uparrow \!\!P \sqcup \uparrow \!\!P \subseteq \uparrow \!\!P$. So let $m \in \uparrow \!\!P \sqcup \uparrow \!\!P$. Then for some $m_1,\,m_2$: $m = m_1 \sqcup m_2$ and $m_1 \in \uparrow \!\!P$ and $m_2 \in \uparrow \!\!P$. By assumption, P is closed. It is straightforward to show that $\uparrow \!\!P$ is then also closed, and so $\uparrow \!\!P$ contains some mode with ground-set $|m_1| \cup |m_2| = |m_1 \sqcup m_2| = |m|$. Since in an extensional mode-space, no two modes have the same ground-set, it follows that $m \in \uparrow \!\!P$ and thus $P \vee P \subseteq \uparrow \!\!P$. $\square$

We may also establish some substantive necessary conditions for certain kinds of unilateral propositions to be identical:

Lemma 6

Let P,  Q,  R,  S be normal raisable unilateral propositions.

• (i) $\uparrow \!\!P = \uparrow \!\!Q$ implies $P = Q$

• (ii) $P \wedge Q = \uparrow \!\!R$ implies $P = Q = R$

• (iii) $P \vee Q = \uparrow \!\!R$ implies ($P = R$ and $Q \le R$) or ($Q = R$ and $P \le R$)

• (iv) $P \wedge Q = R \wedge S$ implies $\{P,\,Q\} = \{R,\,S\}$

• (v) $P \wedge Q = R \vee S$ implies $P = Q$ and (($R = P$ and $S \le R$) or ($S = P$ and $R \le S$))

• (vi) $P \vee Q = R \vee S$ implies that either

  • a. $\{P,\,Q\} = \{R,\,S\}$, or

  • b. $P = R$ and $Q \le P$ and $S \le R$, or

  • c. $P = S$ and $Q \le P$ and $R \le S$, or

  • d. $Q = R$ and $P \le Q$ and $S \le R$, or

  • e. $Q = S$ and $P \le Q$ and $R \le S$

Proof

By use of the equivalences noted at the beginning of this section and the transitivity and antisymmetry of $\le$ and $\preceq$.

For (i): Since $P < \uparrow \!\!P$, it follows from the antecedent that $P < \uparrow \!\!Q$ and hence $P \le Q$. Likewise since $Q < \uparrow \!\!Q$, it follows that $Q < \uparrow \!\!P$ and hence $Q \le P$. By antisymmetry of $\le$, $P = Q$.

For (ii): Since $P, Q < P \wedge Q$, it follows from the antecedent that $P, Q < \uparrow \!\!R$ and hence $P, Q \le R$. So $P \preceq R$ and $Q \preceq R$. Moreover, $R < \uparrow \!\!R$ so $R < P \wedge Q$, so $R \le \{P, Q\}$. It follows that $R \le P$ and $R \le Q$, so $R \preceq P$ and $R \preceq Q$, and hence by antisymmetry of $\preceq$ that $R = P$ and $R = Q$.

For (iii): In similar fashion as before, it follows from the antecedent that $P \le R$ and $Q \le R$, as well as that either (a) $R \le P$ or (b) $R \le Q$. If (a), then by antisymmetry of $\le$ we have $P = R$, and if (b), we have $Q = R$.

For (iv): From the antecedent it is straightforward to show (1) that either (1a) $P \preceq R$ and $Q \preceq S$ or (1b) $P \preceq S$ and $Q \preceq R$ and (2) that either (2a) $R \preceq P$ and $S \preceq Q$ or (2b) $R \preceq Q$ and $S \preceq P$. If (1a) and (2a), then by antisymmetry of $\preceq$, $P = R$ and $Q = S$ follows. If (1b) and (2b), then it follows that $P = S$ and $Q = R$. If (1b) and (2a), we have $P \preceq S \preceq Q \preceq R \preceq P$ and so $P = Q = R = S$. Likewise if (1a) and (2b), we have $P \preceq R \preceq Q \preceq S \preceq P$ and so again $P = Q = R = S$. So in all four cases, $\{P, Q\} = \{R, S\}$.

For (v): From the antecedent it follows that $R \le P$ and $R \le Q$ and $S \le P$ and $S \le Q$. Moreover, either (a) $P, Q \le R$ or (b) $P, Q \le S$ or (c) $P, Q \le \{R, S\}$. If (a), then $P \preceq R$ and so $P = R$. Similarly $Q \preceq R$, and so $Q = R$. Hence $S \le P = Q = R$, establishing the consequent. If (b), then $P \preceq S$, so $P = S$, and $Q \preceq S$, so $Q = S$. Hence $R \le P = Q = S$, again establishing the consequent. Finally, if (c), then it easy to show that either $P \preceq R$ and $Q \preceq S$ or $P \preceq S$ and $Q \preceq R$. In the first case, $P = R \le Q = S \le P$, so $P = Q = R = S$. In the second case, $P = S \le Q = R \le P$, so again $P = Q = R = S$. Either way, the consequent is again established.

For (vi): From the antecedent it follows (1) that (1a) $P \le R$ or (1b) $P \le S$, and (2) that (2a) $Q \le R$ or (2b) $Q \le S$, and (3) that (3a) $R \le P$ or (3b) $R \le Q$, and (4) that (4a) $S \le P$ or (4b) $S \le Q$. It can now be shown that under each of the 16 possible combinations, one of the conditions a.–e. obtain. We shall restrict ourselves to an illustrative four cases; the remaining ones follow the same pattern. Suppose first that (1a), (2a), (3a), and (4a) obtain. Then by (1a) and (3a), $P = R$. By (2a), $Q \le R = P$. By (4a), $S \le P = R$. So case b. above obtains. Suppose now that instead of (4a), (4b) obtains. We still have $P = R$ and $Q \le P$ as before. By (4b), $S \le Q$, and since $Q \le P$, we obtain $S \le P = R$, as required for case b. For a different sort of case, suppose (1a), (2b), (3a), and (4b) obtain. Then still $P = R$, and by (2b), $Q \le S$, as well as by (4b) $S \le Q$, so $S = Q$. It follows that case a. above obtains. Finally, suppose that instead of (3a) and (4b), we have (3b) and (4a). Then $P \le R$, $Q \le S$, $R \le Q$, and $S \le P$. That is, $P \le R \le Q \le S \le P$, and thus $P = R = Q = S$. Again it follows that case a. obtains. $\square$

Finally, in an extensional mode-space, binary weak full ground (among unilateral contents) can be characterized in terms of disjunction and identity:

Lemma 7

For P,  Q normal raisable unilateral propositions in an extensional mode-space: $P \le Q$ iff $P \vee Q = Q \vee Q$.

Proof

For the right-to-left-direction, assume $P \vee Q = Q \vee Q$. Then $P < P \vee Q$ so $P < Q \vee Q$, so $P \le Q$. For the left-to-right direction, assume $P \le Q$. Then either $P = Q$ or $P < Q$. If $P = Q$, then evidently $P \vee Q = Q \vee Q$. So suppose $P < Q$. Given extensionality, to show that $P \vee Q = Q \vee Q$ it suffices to show for arbitrary $\varGamma$ that $\varGamma < P \vee Q$ iff $\varGamma < Q \vee Q$. But if $\varGamma < Q \vee Q$, then $\varGamma \le Q$, so $\varGamma < P \vee Q$. If $\varGamma < P \vee Q$, then either (a) $\varGamma \le P$, or (b) $\varGamma \le Q$, or (c) $\varGamma \le \{P, Q\}$, so $\varGamma _P \le P$ and $\varGamma _Q \le Q$ for some $\varGamma _P,~\varGamma _Q$ with $\varGamma = \varGamma _P \cup \varGamma _Q$. If (a), then since $P < Q$ and $Q < Q \vee Q$, it is easy to show that $\varGamma < Q \vee Q$. if (b), then since $Q < Q \vee Q$, it is equally straightforward that $\varGamma < Q \vee Q$. If (c), then by the previous reasoning, $\varGamma _P \le Q$ and $\varGamma _Q \le Q$ and so $\varGamma _P \cup \varGamma _Q = \varGamma < Q \vee Q$. $\square$

1.3 A.3 Adequacy of the intermediate system

For $\varphi$ an equivalence in $\mathscr {L}_\approx$, we write $\vdash _\mathfrak {I}\varphi$ to say that $\varphi$ is derivable within $\mathfrak {I}$, and we write $\models _\mathfrak {I}\varphi$ to say that $\varphi$ is valid in the class of models based on complete, constrained, intermediate mode-spaces. We show first that $\mathfrak {I}$ is sound with respect to that class of models.

Theorem 8

(Soundness of the Intermediate System) $\models _\mathfrak {I}\varphi$ whenever $\vdash _\mathfrak {I}\varphi$.

Proof

The soundness of (Comm.$\vee$) and (Comm.$\wedge$) is immediate from Lemma 5(i). The soundness of (Reflexivity), (Symmetry), (Transitivity), the preservation rules and the DeMorgan rules is immediate from Lemma 4. $\square$

To prove completeness, we construct a canonical intermediate mode-space and model, and show that every equivalence that is true in this model is derivable within $\mathfrak {I}$. The canonical mode-space is defined as follows.

Definition 8

The canonical intermediate mode-space for $\mathscr {L}_\approx$ is the pair $\langle M_I, V_I\rangle$, where

• $M_0 := \{A \in \mathscr {L}_\approx : A \text { is a literal}\}$

• $C_n := \wp (M_n)\!\setminus \!\{\emptyset \}$

• $M_{n+1} := \{m: m \in M_0$ or m is a non-empty multi-set of members of $C_n\}$

• $C := \bigcup _{n\in N} C_n$

• $M_I := \{m: m \in M_0$ or m is a non-empty multi-set of members of $C\}$

• $V_I(\gamma ) = \varGamma$ if $\gamma$ is a non-empty countable sequence of members of C and $\varGamma$ is the underlying multi-set.

• $V_I(\gamma )$ is undefined otherwise.

We now establish that $\langle M_I, V_I\rangle$ is indeed a complete, constrained, intermediate mode-space. We show first that it is a mode-space.

Lemma 9

$\langle M_I, V_I\rangle$ is a mode-space.

Proof

First, $M_I$ is non-empty, since $M_I$ includes all the literals in $\mathscr {L}_\approx$.

Second, $V_I$ is a non-empty partial function that maps non-empty sequences of subsets of $M_I$ to members of $M_I$. For since there are non-empty countable sequences of members of C and corresponding underlying multi-sets, $V_I$ is non-empty. Since the members of C are subsets of $M_I$, the arguments of $V_I$ are sequences of subsets of $M_I$. Since any multi-set underlying a non-empty countable sequence of members of C is a non-empty multi-set of members of C, the values of $V_I$ are subsets are members of $M_I$.

Third, the domain of $V_I$ is closed under non-empty subsequences and countable concatenations, for non-empty subsequences and countable concatenations of non-empty countable sequences of members of C are themselves such sequences.

Finally, $V_I(\gamma _1^\frown \gamma _2^\frown \ldots ) = V_I(\delta _1^\frown \delta _2^\frown \ldots )$ whenever $V_I(\gamma _1) = V_I(\delta _1),~V_I(\gamma _2) = V_I(\delta _2), \ldots$ For the multi-set underlying $\gamma _1^\frown \gamma _2^\frown \ldots$ is determined by which items occur how many times in $\gamma _1$, and in $\gamma _2$, and $\ldots$, which is to say that it is determined by the multi-sets underlying $\gamma _1, \gamma _2, \ldots$. Since the multi-sets corresponding underlying $\gamma _1, \gamma _2, \ldots$ are the same as those underlying $\delta _1, \delta _2, \ldots$ whenever $V_I(\gamma _1) = V_I(\delta _1), V_I(\gamma _2) = V_I(\delta _2), \ldots$, the result is then immediate. $\square$

Next, we show that the recursive construction of the canonical mode-space is cumulative in the following sense:

Lemma 10

In the construction of $\langle M_I, V_I\rangle$, for all n, $M_n \subseteq M_{n+1}$ and $C_n \subseteq C_{n+1}$.

Proof

Suppose $m \in M_0$. Then by definition, $m \in M_1$. Suppose $P \in C_0$. Then $\emptyset \subset P \subseteq M_0 \subseteq M_1$ and hence $P \in C_1$. Now assume that the claim holds up to n. Suppose $m \in M_{n+1}$. Then either $m \in M_0$, in which case $m \in M_{n+2}$ follows by definition, or m is a non-empty multi-set of members of $C_n$. By IH, $C_n \subseteq C_{n+1}$, hence m is a non-empty multi-set of members of $C_{n+1}$. By definition of $M_{n+2}$, it follows that $m \in M_{n+2}$. Suppose finally that $P \in C_{n+1}$. Then just as before, $\emptyset \subset P \subseteq M_{n+1} \subseteq M_{n+2}$, hence $P \in C_{n+2}$. $\square$

Lemma 11

The mode-space $\langle M_I, V_I\rangle$ is complete, constrained, and intermediate.

Proof

Complete: We show first that $P \sqcup Q$ and $P + Q$ are raisable whenever P and Q are derivative, raisable contents. Since P and Q are derivative contents, $P \in C_n$ for some $n > 0$ and $Q \in C_m$ for some $m > 0$. Now let $j = max(m, n)$. By Lemma 10, P and Q are both in $C_j$ and therefore non-empty subsets of $M_j$. But then it follows that $P \sqcup Q \in \mathscr {R}$. For suppose $m_1 \in P$ and $m_2 \in Q$, so $m_1, m_2 \in M_j$. Then since every $M_n$ with $n > 0$ is closed under fusion, $m_1 \sqcup m_2 \in M_j$. It follows that $P \sqcup Q \subseteq M_j$, and thus that $\{P \sqcup Q\} \subseteq C_j$. Hence V is defined for $\langle P \sqcup Q\rangle$, and so $P \sqcup Q$ is raisable. Moreover, together with the fact that $C_j$ is closed under union, it also follows from this result that $P + Q$ is raisable. Finally, we show that $\uparrow \!\!P$ is raisable if P is. Firstly, $\{V\langle P\rangle \} \in C_{n+1}$, and hence $\{V\langle P\rangle \}$ is raisable. If $P \cap M^D$ is empty, then $\uparrow \!\!P = \{V\langle P\rangle \}$, so $\uparrow \!\!P$ is raisable. If $P \cap M^D$ is non-empty, then $P \cap M^D \in C_n$, and so $P \cap M^D$ is raisable. Then by the above result, so is $\{V\langle P\rangle \} + (P \cap M^D) = \uparrow \!\!P$.

Constrained: Since $V(\gamma )$, when defined, is the multi-set underlying $\gamma$, and since $\gamma$ and $\delta$ determine the same multi-set only if they determine the same set, it is immediate that $V(\gamma ) = V(\delta )$ only if $\gamma$ and $\delta$ determine the same set.

Intermediate: Since $V(\gamma )$ is the multi-set underlying $\gamma$, and $V(\delta )$ is the multi-set underlying $\delta$, if the same multi-set underlies $\gamma$ and $\delta$, then $V(\gamma ) = V(\delta )$. $\square$

The canonical intermediate model of $\mathscr {L}_\approx$ is now defined as follows:

Definition 9

The canonical intermediate model $\mathscr {M}_I$ of $\mathscr {L}_\approx$ is $\langle M_I, V_I, [\cdot ]_I\rangle$, where for sentences $A, B \in \mathscr {L}_\approx$

• $[A]_I = \langle \{A\},\{\lnot A\}\rangle$ if is atomic

• $[\lnot A]_I = \lnot [A]_I$

• $[A \wedge B]_I = [A]_I \wedge [B]_I$

• $[A \vee B]_I = [A]_I \vee [B]_I$

Equivalences that are true in the canonical intermediate model will also be called canonical.

Note that all contents assigned by $\mathscr {M}_I$ to some sentence are normal. For it is easy to see that all contents assigned to atomic sentences are normal, and so by the result that normality is preserved under truth-functional operations, it follows that all assigned contents are.

We now establish a correlation between the syntactic complexity of the formulas of the propositional language $\mathscr {L}_B$ and the level at which their contents are constructed in the recursive definition of the mode-space.

Definition 10

For all $P \in C$, let rank(P) be the lowest n with $P \in C_n$.

Lemma 12

For all $P \in \mathscr {R}$ and $Q, R \in \mathscr {R}\cap \mathscr {C}^D$:

• (i) $rank(Q \sqcup R) = max \{rank(Q), rank(R)\}$

• (ii) $rank(Q + R) = max \{rank(Q), rank(R)\}$

• (iii) $rank(\uparrow \!\!P) = rank(P) + 1$

Proof

For (i): Since each $M_n$ is closed under fusion of modes, if $Q, R \in C_n$, then $Q \sqcup R \in C_n$, so the rank of $Q \sqcup R$ cannot be higher than the maximum rank of Q and R. Since each $M_n$ is closed under non-empty submulti-sets, the rank of $Q \sqcup R$ also cannot be lower than the maximum rank of Q and R.

For (ii): In addition to the observations under (i), note that each $C_n$ is closed under unions and non-empty subsets to see that the rank of $Q + R$ can be neither higher nor lower than the maximum rank of Q and R.

For (iii): Assume $rank(P) = n$. Firstly, by construction, $\{V\langle P\rangle \} = \{\lfloor P\rfloor \} \in C_{n+1}$.⁵⁸ Moreover, $\{\lfloor P\rfloor \} \notin C_n$. For suppose otherwise. Then $\lfloor P\rfloor \in M_n$. Now if $n = 0$, it follows that $\lfloor P\rfloor$ is a literal in $\mathscr {L}_\approx$, which it is not. If $n > 0$, it follows that $P \in C_{n-1}$, contrary to the supposition that $rank(P) = n$. So $rank(\{V\langle P\rangle \}) = n+1$. Now clearly, if $P \cap M^D$ is non-empty, $rank(P \cap M^D) \le rank(P) = n$, and thus by part (ii), since $\uparrow \!\!P = \{V\langle P\rangle \} + (P \cap M^D)$, $rank(\uparrow \!\!P) = n+1$. If $P \cap M^D$ is empty, $\uparrow \!\!P = \{V\langle P\rangle \}$, so again $rank(\uparrow \!\!P) = n+1$. $\square$

Definition 11

We define in a simultaneous induction the positive degree pdeg(A) and the negative degree ndeg(A) of a formula $A \in \mathscr {L}_B$.

• $pdeg(A) = ndeg(A) = 0$ if A is atomic

• $pdeg(\lnot A) = ndeg(A)$

• $ndeg(\lnot A) = pdeg(A) + 1$

• $pdeg(A \wedge B) = pdeg(A \vee B) = max \{pdeg(A), pdeg(B)\} + 1$

• $ndeg(A \wedge B) = ndeg(A \vee B) = max \{ndeg(A), ndeg(B)\} + 1$

The positive degree of a formula $A \in \mathscr {L}_B$ will sometimes also simply be called A’s degree, and denoted deg(A). The degree of an equivalence $A \approx B$ is $max \{deg(A), deg(B)\}$.⁵⁹

Lemma 13

In $\mathscr {M}_I$, for all $A \in \mathscr {L}_B$, $pdeg(A) = rank([A]_I^+)$ and $ndeg(A) = rank([A]_I^-)$.

Proof

(For readability, I drop the subscript I.) By induction on the complexity of A. Suppose first that A is atomic. Then $pdeg(A) = ndeg(A) = 0$, and $[A]^+ = \{A\} \in C_0$, so $rank([A]^+) = 0$, and $[A]^- = \{\lnot A\} \in C_0$, so $rank([A]^-) = 0$. Suppose now the thesis holds for A and B (IH). Then it also holds for $A \wedge B$, $A \vee B$, and $\lnot A$. I give the proof for $pdeg(A \wedge B)$, the other cases are similar.

$\square$

For our purposes, the most important bit is the immediate corollary that only equivalences between formulas of equal degree are canonical:

Corollary 14

For all sentences $A, B \in \mathscr {L}_\approx$, if $\mathscr {M}_I \models A \approx B$, then $deg(A) = deg(B)$.

We are now in a position to prove completeness.

Theorem 15

(Completeness of the intermediate system) $\vdash _\mathfrak {I}\varphi$ whenever $\models _\mathfrak {I}\varphi$.

Proof

As indicated earlier, we prove this by showing that every canonical equivalence is derivable. So assume that $\varphi$ is canonical. Suppose first that $deg(\varphi ) = 0$. Then $\varphi$ has one of these forms, with A and B atomic:

• (i) $A \approx B$

• (ii) $A \approx \lnot B$

• (iii) $\lnot A \approx B$

• (iv) $\lnot A \approx \lnot B$

Cases (ii) and (iii) cannot obtain, for $[A]^+ = \{A\} \ne \{\lnot B\} = [B]^- = [\lnot B]^+$, and likewise $[\lnot A]^+ = [A]^- = \{\lnot A\} \ne \{B\} = [B]^+$. Equivalences of forms (i) and (iv) are canonical only if $A = B$, so they take the forms $A \approx A$ and $\lnot A \approx \lnot A$, respectively. But equivalences of these forms are derivable by (Reflexivity).

So suppose that equivalences of degree $\le n$ are derivable if canonical, and suppose $\varphi$ is of degree $n+1$. Then $\varphi$ is an equivalence between two formulas of degree $n+1$. Each of them can be either a conjunction, or a disjunction, or a negated conjunction or disjunction, or a double negation. However, given (Symmetry) we need not separately consider, say, the case of $A \wedge B \approx C \vee D$ and that of $A \vee B \approx C \wedge D$. Moreover, given the DeMorgan equivalences, we also need not consider equivalences with a negated conjunction or a negated disjunction, since these cases may be reduced using the DeMorgan rules to cases of disjunctions or conjunctions of negations. So the cases we need to consider are these:

• (i) $A \wedge B \approx C \wedge D$

• (ii) $A \vee B \approx C \vee D$

• (iii) $A \wedge B \approx C \vee D$

• (iv) $A \wedge B \approx \lnot \lnot C$

• (v) $A \vee B \approx \lnot \lnot C$

Case (i): By Lemma 6(iv), if $A \wedge B \approx C \wedge D$ is canonical, then so are either (a) $A \approx C$ and $B \approx D$ or (b) $A \approx D$ and $B \approx C$. These equivalences are at most degree n, and so by IH, they are derivable. If (a), then by (Preservation $\wedge$), $A \wedge B \approx C \wedge B$ and $C \wedge B \approx C \wedge D$ are derivable. By (Transitivity), $A \wedge B \approx C \wedge D$ is derivable. If (b), then by (Preservation $\wedge$) we obtain $A \wedge B \approx D \wedge B$ and $B \wedge D \approx C \wedge D$. By (Commutativity $\wedge$) and (Transitivity), we may again derive $A \wedge B \approx C \wedge D$.

Case (ii): By Lemma 6(vi), there are five ways for $A \vee B \approx C \vee D$ to be canonical. The first is that as in case (i), either (a) $A \approx C$ and $B \approx D$ or (b) $A \approx D$ and $B \approx C$ are canonical. By IH, these will be derivable, and similarly as before, using (Preservation $\vee$) and (Commutativity $\vee$) in place of the corresponding rules for conjunctions, we may derive $A \vee B \approx C \vee D$. The other four cases exhibit a common structure, so I shall confine myself to treating one of them, which is that $A \approx C$, $B \le A$, and $D \le C$ are canonical. Assuming that this is not also an instance of the first case, it follows that $B \approx D$ is not canonical. We can now show that both $[B]^+ \le [D]^+$ and $[D]^+ \le [B]^+$, which entails that $B \approx D$ is canonical, contrary to assumption. For since $[A]^+ = [C]^+$, for any $m \in [C]^+$, $m \sqcup \lfloor [B]^+\rfloor \in [A \vee B]^+ = [C \vee D]^+$. By construction of the canonical model, no mode in $[C]^+$ contains any content more than finitely many times, so $m \sqcup \lfloor [B]^+\rfloor$ is always distinct from m. Since $[C]^+$ moreover includes only finitely many modes, $m \sqcup \lfloor [B]^+\rfloor \notin [C]^+$. It follows that $\lfloor [B]^+\rfloor \in [D]^+$ and therefore $[B]^+ \le [D]^+$. Similarly, for any $m \in [A]^+$, $m \sqcup \lfloor [D]^+\rfloor \in [C \vee D]^+ = [A \vee B]^+$. By analogous reasoning as before, $\lfloor [D]^+\rfloor \in [B]^+$ and hence $[D]^+ \le [B]^+$.

The remaining cases (iii)–(v) cannot obtain. For cases (iii) and (iv), it suffices to note that both $[\lnot \lnot C]^+$ and $[C \vee D]^+$ always include the mode corresponding to the multi-set including only $[C]^+$, and exactly once, whereas every mode in a conjunction corresponds to a multi-set which either contains at least two elements, or contains one element at least twice.

For case (v), by Lemma 6(iii), if $A \vee B \approx \lnot \lnot C$ is canonical, so is either $A \approx C$ or $B \approx C$. So suppose $[A]^+ = [C]^+$; the other case is analogous. Then whenever $m \in [C]^+$, $[A \vee B]^+$ also includes $m \sqcup \lfloor [B]^+\rfloor$. As before, no mode in $[C]^+$ contains any content more than finitely many times, so $m \sqcup \lfloor [B]^+\rfloor$ is always distinct from m, and since $[C]^+$ includes only finitely many modes, $[A \vee B]^+$ and $[C]^+$ are distinct. $\square$

1.4 A.4 Adequacy of the extensional system

For $\varphi$ an equivalence in $\mathscr {L}_\approx$, we write $\vdash _\mathfrak {E}\varphi$ to say that $\varphi$ is derivable within $\mathfrak {E}$, and we write $\models _\mathfrak {E}\varphi$ to say that $\varphi$ is valid in the class of models based on complete, constrained, extensional mode-spaces. We show first that $\mathfrak {E}$ is sound with respect to that class of models.

Theorem 16

(Soundness of the extensional system)

For every equivalence $\varphi \in \mathscr {L}_\approx$, if $\vdash _\mathfrak {E}\varphi$, then $\models _\mathfrak {E}\varphi$.

Proof

Given the previous soundness result in theorem 8 and the fact that every extensional mode-space is also intermediate, it suffices to establish soundness for the additional rules in $\mathfrak {E}$, i.e. (Collapse $\wedge /\vee$), (Collapse $\vee /\lnot \lnot$), (Introduction $\le \!\!\wedge$), and (Introduction $\le \!\!\vee$). The soundness of (Collapse $\wedge /\vee$), (Collapse $\vee /\lnot \lnot$) is immediate from lemma 5(ii)-(iii). The soundness of (Introduction $\le \!\!\wedge$) and (Introduction $\le \!\!\vee$) is straightforward given Lemma 7 and the principles ($<\!\!\wedge$) and ($<\!\!\vee$) in lemma 1. $\square$

The completeness proof proceeds in close analogy to that for the semi-extensional system. We first define the canonical extensional mode-space, simply replacing any reference to multi-sets in the definition of the canonical semi-extensional mode-space by reference to the corresponding set.

Definition 12

The canonical extensional mode-space for $\mathscr {L}_\approx$ is the pair $\langle M_E, V_E\rangle$, where

• $M_0 := \{A \in \mathscr {L}_\approx : A \text { is a literal}\}$

• $C_n := \wp (M_n)\!\setminus \!\{\emptyset \}$

• $M_{n+1} := \{m: m \in M_0$ or m is a non-empty set of members of $C_n\}$

• $C := \bigcup _{n\in N} C_n$

• $M_E := \{m: m \in M_0$ or m is a non-empty set of members of $C\}$

• $V_E(\gamma ) = \varGamma$ if $\gamma$ is a non-empty countable sequence of members of C and $\varGamma$ is the underlying set.

• $V_E(\gamma )$ is undefined otherwise.

By straightforward adjustments to the earlier proof, it may be shown that $\langle M_E, V_E\rangle$ is a mode-space of the desired kind.

Lemma 17

$\langle M_E, V_E\rangle$ is a complete, constrained, and extensional mode-space.

The canonical extensional model of $\mathscr {L}_\approx$ is defined in the obvious way:

Definition 13

The canonical extensional model $\mathscr {M}_E$ of $\mathscr {L}_\approx$ is $\langle M_E, V_E, [\cdot ]_E\rangle$, where for sentences $A, B \in \mathscr {L}_\approx$

• $[A]_E = \langle \{A\},\{\lnot A\}\rangle$ if is atomic

• $[\lnot A]_E = \lnot [A]_E$

• $[A \wedge B]_E = [A]_E \wedge [B]_E$

• $[A \vee B]_E = [A]_E \vee [B]_E$

The lemmata concerning the correspondence of the degree of syntactic complexity of an $\mathscr {L}_\approx$-sentence to the rank in the hierarchy of propositions in the construction of the mode-space unproblematically carries over to the extensional setting, so we again obtain the desired corollary:

Corollary 18

For all sentences $A, B \in \mathscr {L}_\approx$, if $\mathscr {M}_E \models A \approx B$, then $deg(A) = deg(B)$.

In preparation of the completeness proof, it helps to first prove the following lemma.

Lemma 19

If sentences $A,\,B \in \mathscr {L}_B$ are degree $\le n$, and if every equivalence up to and including degree n is derivable if canonical, then $B \le A$ is also derivable if canonical.

Proof

Assume the antecedent. By definition of $\le$, $B \le A$ is canonical just in case either $B \approx A$ is canonical or $[B]^+ < [A]^+$. By assumption, if $B \approx A$ is canonical, then it is derivable. But then by (Preservation $\vee$), so is $A \vee B \approx A \vee A$, which is $B \le A$. So suppose that $[B]^+ < [A]^+$. Then $deg(A) > 0$, and so A takes one of these forms

• (a) $D \vee E$

• (b) $D \wedge E$

• (c) $\lnot \lnot D$

• (d) $\lnot (D \vee E)$

• (e) $\lnot (D \wedge E)$

where D and E are degree $< n$.

If (a), and thus $[B]^+ < [D \vee E]^+ = [D]^+ \vee [E]^+$, it follows that either $[B]^+ \le [D]^+$ or $[B]^+ \le [E]^+$ and hence that either (i) $B \le D$ is canonical or (ii) $B \le E$ is canonical. Since D and E are degree $<n$, $D \vee D$ and $E \vee E$ are degree $\le n$, so by corollary 18, the equivalences $B \le D$ and $B \le E$ are degree $\le n$. So by assumption, if (i), then $B \le D$ is derivable, and if (ii), then $B \le E$ is derivable. Suppose (i). Then by (Introduction $\le \!\!\vee$), $B \le D \vee E$ is derivable, which is $B \le A$. Suppose (ii). Then by (Introduction $\le \!\!\vee$), $B \le E \vee D$ is derivable. Using (Commutativity $\vee$), (Transitivity), and (Preservation $\vee$), we may derive from this $B \le D \vee E$, that is $B \le A$.

If (b), and thus $\{[B]^+\} \in [D \wedge E]^+ = [D]^+ \wedge [E]^+$, it follows that $[B]^+ \le [D]^+$ and $[B]^+ \le [E]^+$ and hence that both $B \le D$ and $B \le E$ are canonical. As before, these equivalences are degree $\le n$ and thus derivable. By (Introduction $\le \!\!\wedge$), so is $B \le D \wedge E = A$.

The remaining cases can be reduced to the previous ones using the DeMorgan identities and Lemma 5. For illustration, suppose that case (c) obtains. Then $[B]^+ < [\lnot \lnot D]^+$. But $[\lnot \lnot D]^+ = [D \vee D]^+ = [D]^+ \vee [D]^+$. By the reasoning in case (a), $B \le D \vee D$ is derivable. Using the derivable equivalence of $D \vee D$ to $\lnot \lnot D$, we may derive $B \le \lnot \lnot D$, i.e. $B \le A$. $\square$

Theorem 20

(Completeness of the Extensional System)

For every equivalence $\varphi \in \mathscr {L}_\approx$, if $\models _\mathfrak {E}\varphi$, then $\vdash _\mathfrak {E}\varphi$.

Proof

We show by induction on the degree of equivalences that every canonical equivalence is derivable. Suppose $\varphi$ is a canonical equivalence. The case of $deg(\varphi ) = 0$ is exactly as in the semi-extensional case.

Now assume all canonical equivalences of degree $\le n$ are derivable and suppose $\varphi$ is of degree $n+1$. So $\varphi$ is an equivalence between two formulas of degree $n+1$. Each of them can be either a conjunction, or a disjunction, or a negated conjunction or disjunction, or a double negation. The last three cases can be reduced to the first two in the same way we did in the proof of Lemma 19. So we only have three kinds of equivalences of degree $n+1$ to consider, namely instances of the following forms, where $A,\,B,\,C$, and D are each of some degree $\le n$:

• (i) $A \wedge B \approx C \wedge D$

• (ii) $A \vee B \approx C \vee D$

• (iii) $A \wedge B \approx C \vee D$

Case (i): By Lemma 6(iv), if $A \wedge B \approx C \wedge D$ is canonical, then so are either (a) both $A \approx C$ and $B \approx D$, or (b) both $A \approx D$ and $B \approx C$. So suppose (a). The equivalences $A \approx C$ and $B \approx D$ are both at most degree n, so by IH, $A \approx C$ and $B \approx D$ are derivable. Using (Preservation $\wedge$), $A \wedge B \approx C \wedge B$ and $C \wedge B \approx C \wedge D$ are derivable. Using (Transitivity), $A \wedge B \approx C \wedge D$ is derivable. Now suppose (b) $A \approx D$ and $B \approx C$ are canonical. Then these are at most degree n and thus derivable. Using (Preservation $\wedge$), $A \wedge B \approx D \wedge B$ and $B \wedge D \approx C \wedge D$ are derivable. Using (Commutativity $\wedge$) and (Transitivity), $A \wedge B \approx C \wedge D$ is derivable.

Case (ii): By Lemma 6(vi), if $A \vee B \approx C \vee D$ is canonical, there are five ways this can come about. One is that $A \wedge B \approx C \wedge D$ is canonical, in which case as before, either $A \approx C$ and $B \approx D$ are canonical, or $A \approx D$ and $B \approx C$ are canonical. These will then be derivable, and much as in case (i) but using (Commutativity $\vee$) instead of (Commutativity $\wedge$), $A \vee B \approx C \vee D$ is derivable from them. A second way in which $A \vee B \approx C \vee D$ can be canonical is by $A \approx C$, $B \le A$, and $D \le A$ being canonical; the remaining cases are analogous and will be omitted. Then by IH and lemma 19, $A \approx C$, $B \le A$ and $D \le A$ are all derivable. From these, using mainly (Commutativity $\vee$) and (Preservation $\vee$), we may then derive $A \vee B \approx C \vee D$.

Case (iii): By Lemma 6(v), if $A \wedge B \approx C \vee D$ is canonical, then so is $A \approx B$, which, by IH, is derivable. But then also $[A \wedge B]^+ = [A]^+ \wedge [B]^+ = [A]^+ \wedge [A]^+ = [A]^+ \vee [A]^+ = [A \vee A]^+$, so $A \vee A \approx C \vee D$ is also canonical, and by case (ii) derivable. From these, using mainly (Collapse $\wedge /\vee$) and (Preservation $\wedge$), we may derive $A \wedge B \approx C \vee D$. $\square$

B Comparison of deductive systems

Theorem 21

For every equivalence $\varphi \in \mathscr {L}_\approx$, if $\vdash _\mathfrak {R}\varphi$ then $\vdash _\mathfrak {I}\varphi$.

Proof

Call a theorem $\varphi$ of $\mathfrak {R}$ unproblematic if the theorem produced by applying the rule (Pres.$\lnot$) to $\varphi$ can also be derived within $\mathfrak {I}$. We show by an induction on the length of derivations that all theorems of $\mathfrak {R}$ are unproblematic. From this it follows straightforwardly that all theorems of $\mathfrak {R}$ are theorems of $\mathfrak {I}$. Consider first the case of a derivation D of length 1. There are three case:

• 1. D consists in an application of (Reflexivity). Then the result of applying (Preservation $\lnot$) can also be achieved simply by an application of (Reflexivity).

• 2. D consists in an application of (Commutativity $\vee$), so the theorem established by D is $A \vee B \approx B \vee A$. Application of (Preservation $\lnot$) yields $\lnot (A \vee B) \approx \lnot (B \vee A)$. This can be derived within $\mathfrak {I}$ from $A \vee B \approx B \vee A$ by application of the DeMorgan rules and (Commutativity $\wedge$): $(\lnot (A \vee B) \approx \lnot A \wedge \lnot B \approx \lnot B \wedge \lnot A \approx \lnot (B \vee A))$

• 3. D consists in an application of (Commutativity $\wedge$). Analogous to the previous case.

Suppose then that derivations up to length n produce only unproblematic theorems, and suppose D has length $n+1$. The cases in which the final step in D consists in the application of one of the premise-less rules just discussed are exactly as before. The remaining cases are five, according as the final step in D is an application of

• 1. (Symmetry) Then application of (Preservation $\lnot$) produces $\lnot B \approx \lnot A$. By IH, $\lnot A \approx \lnot B$ can be derived within $\mathfrak {I}$, and thus by (Symmetry), so can $\lnot B \approx \lnot A$.

• 2. (Transitivity) Then application of (Preservation $\lnot$) produces $\lnot A \approx \lnot C$. By IH, $\lnot A \approx \lnot B$ and $\lnot B \approx \lnot C$ can be derived within $\mathfrak {I}$, and thus by (Transitivity), so can $\lnot A \approx \lnot C$.

• 3. (Preservation $\vee$) Then application of (Preservation $\lnot$) produces $\lnot (A \vee C) \approx \lnot (B \vee C)$. By IH, $\lnot A \approx \lnot B$ can be derived within $\mathfrak {I}$. By (Preservation $\wedge$), $\lnot A \wedge \lnot C \approx \lnot B \wedge \lnot C$ can then be derived, and by DeMorgan, so can $\lnot (A \vee C) \approx \lnot (B \vee C)$.

• 4. (Preservation $\wedge$) Analogous to the previous case.

• 5. (Preservation $\lnot$) Then application of (Preservation $\lnot$) produces $\lnot \lnot A \approx \lnot \lnot B$, which can be derived within $\mathfrak {I}$ from $A \approx B$ by (Preservation $\lnot \lnot$).$\square$