Paths to Triviality

Abstract

This paper presents a range of new triviality proofs pertaining to naïve truth theory formulated in paraconsistent relevant logics. It is shown that excluded middle together with various permutation principles such as A → (B → C)⊩B → (A → C) trivialize naïve truth theory. The paper also provides some new triviality proofs which utilize the axioms ((A → B)∧(B → C)) → (A → C) and (A → ¬A) → ¬A, the fusion connective and the Ackermann constant. An overview over various ways to formulate Leibniz’s law in non-classical logics and two new triviality proofs for naïve set theory are also provided.

Appendices

Appendix A: Naïve Set Theory and Substructural Logics

The goal of this appendix is to show that cutting structural contraction while retaining ∘ and t, is not sufficient to avoid trivializing naïve set theory. The structural rule of contraction is the rule that $A \vdash B$ follows from $A, A \vdash B$. Structural contraction holds in Hilbertian proof systems since such systems take the antecentent of $\vdash$ to be sets. So in order to restrict this rule, we need a new notion of antecedent structure.

In order to make the transition as smooth as possible I have kept the notion of a proof intact. However, it is now sequents and not formulas which are the objects of proofs. $\vdash$ will in the following be the sequent symbol. The proof system will (by and large) be that of Restall’s An Introduction to Substructural Logics ([38]).

Definition 13

(Structure)

• 0 is a structure (but not a formula)

• If A is a formula, then A is a structure

• If X and Y are structures, then so is (X;Y)

• If X is a structure, and A is a formula, then $X \Vdash A$ is a sequent.

Substructure is defined in the obvious way.

• X(Y) indicates that Y is a substructure of X.

• X(Y/Z) is the structure got by replacing every substructure Y in X with Z.

The system $\mathfrak {S}$ consists of the following rules:

$$\begin{array}{ll} \mathrm{Operational \,\,rules} &\left\{\begin{array}{ccc} \frac{~} {A \Vdash A }(ID) \\\\ \frac{X; A\Vdash B}{X \Vdash A \rightarrow B } (\rightarrow\!I) &\frac{X\Vdash A \rightarrow B\qquad Y\Vdash A}{X; Y \Vdash B}(\rightarrow\!E) \\\\ \frac{X\Vdash A \qquad Y\Vdash B} {X; Y \Vdash A \circ B}(\circ I) &\frac{X\Vdash A \circ B \qquad Y(A; B)\Vdash C} {Y(A; B/X) \Vdash C}(\circ E) \\\\ \frac{~}{0 \Vdash \mathbf{t}}(\mathbf{t} I) &\frac{X\Vdash \mathbf{t}\qquad Y(0)\Vdash A}{Y(0/X) \Vdash A}(\mathbf{t} E) \\\\ \frac{X\Vdash \bot}{X \Vdash A}(\bot E) \\\\ \frac{X\Vdash A(x/y)}{X \Vdash \forall x A}(\forall I) &\frac{X; A(x/a)\Vdash B}{X; \forall x A\Vdash B} (\forall E)\\ (y ~\text{not free in}\,\,X \Vdash \forall x A)&\quad(a ~\text{is any term free for}~ x~ \text{in}~ A) \end{array} \right. \end{array} $$

$$\begin{array}{cl} \mathrm{Structural~ rules} &\left\{\begin{array}{ccc} \frac{X\Vdash A \qquad Y(A)\Vdash B} {Y(A/X)\Vdash B }(cut) \\\\ \frac{0; X\Vdash A }{X \Vdash A}(Left\, Pop) &\frac{X\Vdash A} {0; X \Vdash A} (Left\, Push) \end{array} \right. \\\\ \textnormal{Set-theoretic~ rules} &\left\{ \begin{array}{ccc} \frac{X; A(x/a)\Vdash B }{X; a \in \{x|A\} \Vdash B}(\in\!L)\qquad \frac{X\Vdash A(x/a)} {X \Vdash a \in \{x | A\}}(\in\!R) \\\\ \frac{x \in a\Vdash x \in b \qquad x \in b\Vdash x \in a} {0 \Vdash a = b }(=\in) \end{array} \right. \\\\ \mathrm{Identity~ rules} &\left\{\begin{array}{ccc} \frac{~} {0 \Vdash a = a}(=\!ID)\qquad \frac{0\Vdash a = b \qquad 0\Vdash A(a)}{0 \Vdash A(b)}(subLL) \end{array} \right. \end{array} $$

Definition 14

(Proof) A proof of a sequent $X \Vdash A$ from a set of sequents Γ in the system $\mathfrak {S}$ is defined to be a finite list α ₁,α ₂,…,α _n such that α _n is $X \Vdash A$ and every α _m≤n is either a member of Γ, or there is a set ${\varDelta } \subseteq \{\alpha _{i}\,|\, i < m\}$ such that α _m follows from Δ by one of the rules of $\mathfrak {S}$. The existential claim that there is such a proof will be written

$${\Gamma} \Vvdash_{\mathfrak{S}} (X \Vdash A). $$

I will now show that the sequent $0 \Vdash \bot$ is derivable in the system $\mathfrak {S}$. The proof will mimic the proof of Lemma 7.

Definition 15

$$\begin{array}{rll} \mathfrak{p}_{A} &=_{df} \{x | A\}\\ a \triangleright b &=_{df} \forall x (a \in x \rightarrow b \in x)\\ \mathfrak{c} &=_{df} \{y | (\mathfrak{p}_{y \in y} \triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t}\}\\ \mathfrak{C} &=_{df} \mathfrak{c} \in \mathfrak{c} \end{array} $$

The following two lemmas are easily proven and are therefore left for the reader.

Lemma 8

For sentences A and B, $\{A \Vdash B, B \Vdash A\} \Vvdash _{\mathfrak {S}} (0 \Vdash \mathfrak {p}_{A} \triangleright \mathfrak {p}_{B})$

Lemma 9

$\{0 \Vdash A(a)\} \Vvdash _{\mathfrak {S}} (a \triangleright b; \mathbf {t} \Vdash A(b))$

Theorem 16

$\emptyset \Vvdash _{\mathfrak {S}} (0 \Vdash \bot )$

Proof

$$\begin{array}{rll} (1) &\, \mathfrak{C} \Vdash (\mathfrak{p}_{\mathfrak{C}}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t} & (\in\!L)\\ (2) &\, (\mathfrak{p}_{\mathfrak{C}}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t} \Vdash \mathfrak{C} & (\in\!R)\\ (3) &\, 0 \Vdash \{x | (\mathfrak{p}_{\mathfrak{C}}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t} \} \triangleright \mathfrak{p}_{\mathfrak{C}} & \text{1, 2, Lemma 8}\\ (4) &\, \mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot}; \mathbf{t}\Vdash \{x | (\mathfrak{p}_{\bot}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t}\} \triangleright \mathfrak{p}_{\bot}\;& \text{3, Lemma 9}\\ (5) &\, 0 \Vdash \mathfrak{p}_{\bot} \in \{x | (\mathfrak{p}_{\bot}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t}\}& \text{fiddling}\\ (6) &\, \{x | (\mathfrak{p}_{\bot}\triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t} \} \triangleright \mathfrak{p}_{\bot}; \mathbf{t}\Vdash \mathfrak{p}_{\bot} \in \mathfrak{p}_{\bot} \;& \text{5, Lemma 9}\\ (7) &\, (\mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot}; \mathbf{t}); \mathbf{t}\Vdash \mathfrak{p}_{\bot} \in \mathfrak{p}_{\bot}& \text{4, 6, (cut)}\\ (8) &\, (\mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot}; \mathbf{t}); \mathbf{t}\Vdash \bot& \text{7, fiddling}\\ (9) &\, (\mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot}\circ \mathbf{t})\circ \mathbf{t}\Vdash \bot& \text{8, \((\circ E)\) + fiddling}\\ (10) &\, \mathfrak{C} \Vdash \bot& \text{1, 9, (cut)}\\ (11) &\, \bot \Vdash \mathfrak{C}& \text{ID + \((\bot\!E)\)}\\ (12) &\, 0 \Vdash \mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot} &\text{10, 11, Lemma 8}\\ (13) &\, 0 \Vdash (\mathfrak{p}_{\mathfrak{C}} \triangleright \mathfrak{p}_{\bot} \circ \mathbf{t}) \circ \mathbf{t} &\text{12, fiddling}\\ (14) &\, 0 \Vdash \mathfrak{C}& \text{2, 13, (cut)}\\ (15) &\, 0 \Vdash \bot& \text{10, 14, (cut)} \end{array} $$

One could, just as in the structural setting, dispense with t and ∘ provided one adds a substructural counterpart of R8, namely the rule

$$\begin{array}{c} \frac{X; 0\Vdash A} {X \Vdash A}(Right\, Pop) \end{array} $$

I leave the proof to the reader. However, doing away with t and ∘ seems quite a drastic measure for the substructuralist. After all, these connectives in a sense represent the basic building blocks of a structure, namely 0 and ; respectively. Eliminating them would be comparable to removing ∧ in the structural setting.

I mentioned in Section 10 that Mares and Paoli differentiated between the external and internal consequence relation of their logic. The following defines these relations for the system $\mathfrak {S}$.

Definition 16

External vs. internal consequence

• (Internal consequence) $X \vdash ^{I}_{\mathfrak {S}} A$ is defined for any structure X and formula A to hold just in case $\emptyset \Vvdash _{\mathfrak {S}} (X \Vdash A)$.

• (External consequence) ${\Theta } \vdash ^{E}_{\mathfrak {S}} A$ is defined for any set of formulas Θ and formula A to hold just in case $\{0 \Vdash \theta | \theta \in {\Theta }\} \Vvdash _{\mathfrak {S}} (0 \Vdash A)$.

Thus A is an internal consequence of X just in case the sequent $X \Vdash A$ is derivable without assumptions. That A is a logical truth (according to $\mathfrak {S}$) is recorded as $0 \Vdash A$. Furthermore, A is an external consequence of the Θ’s if it is provable that A is a logical truth upon assuming the Θ’s to be logical truths.

Mares and Paoli wanted the rule

$$\begin{array}{llll} (Ext_{\in}) & \frac{\varGamma, x \in a \vdash x \in b, \varDelta \quad \varGamma, x \in b \vdash x \in a, \varDelta} {\varGamma \vdash a = b, \varDelta} \end{array} $$

to hold provided $\vdash$ was interpreted as the internal consequence relation of their logic. In their system Γ and Δ are multiset. For empty Γ and Δ this then amounts to validating the inference that if the sequents $x \in a \Vdash x \in b$ and $x \in b \Vdash x \in a$ are derivable without using assumptions, then so is $0 \Vdash a = b$ which is precisely what the rule (=∈) above does. To avoid irrelevant formulas such as $a = b \rightarrow (A \rightarrow A)$, Mares and Paoli restricted the use of the rule

$$\begin{array}{llll} \frac{\phi(a) \vdash \varDelta}{\Gamma, a = b, \phi(b) \vdash \varDelta} & \text{(=L\(_{l}\))} \end{array} $$

to context where $\vdash$ is the external consequence relation. It would therefore license the inference of $0 \Vdash B$ from the assumptions $0 \Vdash a = b$ and $0 \Vdash A(b)$, provided one has inferred $0 \Vdash B$ from $0 \Vdash A(a)$. This is however easily seen to be equivalent to what the rule (s u b L L) licenses.

Substructural approaches to the paradoxes are sometimes deemed more radical than the structural. Restricting structural contraction is arguably a radical approach if the internal consequence relation is used to interpret what logical entailment amounts to. With regards to which logic does or does not trivialize the naïve theories of truth, properties and sets, the two approaches are however equivalent—all the logics set forth in Section 2 can be formulated as substructural logics in such a way that if $\mathfrak {L}$ is the set of “substructural rules” for L, then for any set of formulas Θ,

$${\Theta} \vdash_{\mathbf{L}} A \Leftrightarrow {\Theta}\vdash^{E}_{\mathfrak{L}} A. $$

²⁸ From this it is easy to see that by adding rules for the naïve theory $\mathcal {M}$ to $\mathfrak {L}$ instead of adding its axioms, one obtains that

$$\mathcal{M} \vdash_{\mathbf{L}} \bot \Leftrightarrow \emptyset \Vvdash_{\mathfrak{L}} (0 \Vdash \bot).$$

For instance, let $\mathfrak {T}$, in addition to (D), $(\rightarrow \!I)$, $(\rightarrow \!E)$, (⊥E) and (c u t), consist of the rules of structural permutation and weak reductio

$$\begin{array}{llll} \mathrm{[R10](Permutation\,\, rule)} \frac{X; Y \Vdash C}{Y; X \Vdash C}&\quad \frac{A \Vdash \neg A}{0 \Vdash \neg A}\mathrm{(Weak\,\, reductio)[Ax13]} \end{array} $$

and the rules for a simple negation satisfying double negation elimination, that is ¬ satisfies the rules

$$\begin{array}{llll} [R5](\neg I/\neg E) \frac{A \Vdash \neg B \qquad X \Vdash B} {X \Vdash \neg A} &\qquad \frac{X \Vdash \neg \neg A}{X \Vdash A}(\neg\neg E)[Ax4] \end{array} $$

$\mathfrak {T}$ is the substructural version of the implication-negation fragment of the logic B B X[R10], and it is easy to show that the sequent $0 \Vdash \bot$ is derivable from the sequents $\neg (C \rightarrow \bot ) \rightarrow \bot \Vdash C$ and $C \Vdash \neg (C \rightarrow \bot ) \rightarrow \bot$ in it. Thus also substructural B B X[R10] trivializes any naïve theory (cf. Theorem 3).

Appendix B: Unrestricted Pair-abstraction and the Fixed-point Theorem

Unrestricted abstraction is sometimes generalized so as to allow for impredicative definitions as it were—by generalizing the notion of an abstract to {x;y|A}, the schema of generalized unrestricted abstraction may be stated as the universal closure of

$$\forall x (x \in \{x; y | A\} \leftrightarrow A(y/\{x; y | A\}) $$

where {x;y|A} is free for y in A. This schema guarantees the existence of fixed-point terms modulo $\overset {e}{=}$; for every formula A there is a term t _A such that $t_{A} \overset {e}{=} A(t_{A})$. That every formula has such a fixed-point term is however easily derived from $\mathcal {P}$ alone provided the logic is sufficiently strong:

Theorem 17

(Fixed-point theorem) If $\forall x\forall y(\langle x, y\rangle \in \{\langle x, y\rangle | A\} \leftrightarrow A)$ holds for some definition of 〈x,y〉 and {〈x,y〉|A}, then $\mathcal {P}$ suffices for the the existence of fixed-points modulo $\overset {e}{=}$ ; for every formula A there is a term t _A such that $t{\!_{A}} \overset {e}{=} \{x| A(y/t{\!_{A}})\}$.²⁹

Proof

Let

$$\begin{array}{ll} r{\!_{A}} =_{df}& \{\langle u, v\rangle | A(x/u,y/\{w| \langle w, v\rangle \in v\})\}\\ t{\!_{A}} =_{df}& \{w| \langle w, r{\!_{A}}\rangle \in r{\!_{A}}\}. \end{array} $$

$$\begin{array}{rlll} {(1)} &\,x \in t{\!_{A}} &\leftrightarrow \langle x, r{\!_{A}}\rangle \in r{\!_{A}} & \;\mathcal{P} \text{+ def.~of} ~t{\!_{A}}\\ {(2)} &\, &\leftrightarrow A(x/x, y/\{w| \langle w, r{\!_{A}}\rangle \in r{\!_{A}}\})& \;\text{1, assumption}\\ {(3)} &\, &\leftrightarrow A(y/t{\!_{A}})& \;2, \text{def.~of}\,\, t{\!_{A}}\\ {(4)} &\, &\leftrightarrow x \in \{x| A(y/t{\!_{A}})\}& \;3, \mathcal{P}\\ {(5)} &&\, t{\!_{A}} \overset{e}{=} \{x| A(y/t{\!_{A}})\} & \;\text{1--4, RQ} \end{array}$$

□

The purpose of this appendix is to show that the logic ∀B ^t∘ is sufficiently strong to provide a definition of both 〈x,y〉 and {〈x,y〉|A} so that unrestricted pair-abstraction,

$$\forall x\forall y(\langle x, y\rangle \in \{\langle x, y\rangle| A\} \leftrightarrow A), $$

is derivable. Since ∀B ^t∘ is a rather weak logic the definitions and proofs will however be quite baroque. After presenting the proofs I will give some quick comments on the prospects of finding other definitions suitable for logics which might treat naïve set theory as a non-trivial theory.

Definition 17

$$\begin{array}{rlll} a \overset{i}{=} b =_{df}& \forall x(a \in x \leftrightarrow b \in x)\\ \{a\} =_{df}& \{x| x \overset{i}{=} a\}\\ \{a, b\} =_{df}& \{x| x \overset{i}{=} a \vee x \overset{i}{=} b\}\\ \quad \langle a, b\rangle =_{df}& \{\{a\}, \{a, b\}\}\\ \partial(A)=_{df}& \left[(A \circ \mathbf{t}) \circ \mathbf{t}\right]\circ \big(\big(\big(\big(\left[(A \circ \mathbf{t}) \circ \mathbf{t}\right] \circ \left[(A \circ \mathbf{t}) \circ \mathbf{t}\right]\big) \circ \mathbf{t}\big) \circ \left[(A \circ \mathbf{t}) \circ \mathbf{t}\right]\big) \circ \mathbf{t}\big)\\ \{\langle x, y\rangle | A\} =_{df}& \{z | \exists x\exists y(\partial(\langle x, y \rangle \overset{i}{=} z) \circ A)\} \end{array} $$

Lemma 10

$$\begin{array}{ll} A\vdash_{\mathbf{BB}^{\mathbf{t}\circ}} \partial(A) \end{array} $$

Proof

This holds essentially because $A, B \vdash _{\mathbf {BB}^{\mathbf {t}\circ }} A \circ B$ holds. That this is so is easily seen by noting that $A \rightarrow (B \rightarrow A \circ B)$ is derivable using Ax1 and R13. $\ \Box$

Lemma 11

Assuming a and b to be free for x in A, then

$$\begin{array}{rll} (1) & \mathcal{P}\vdash_{\forall\mathbf{BB}} a \overset{i}{=} b \rightarrow (A(x/a) \rightarrow A(x/b))\\ \vspace*{-2mm}\\ \text{(2)} & \frac{\mathcal{P}\vdash_{\forall\mathbf{BB}^{\mathbf{t}}} A(x/a)} {\mathcal{P}\vdash_{\forall\mathbf{BB}^{\mathbf{t}}} a \overset{i}{=} b \rightarrow (\mathbf{t} \rightarrow A(x/b))} \end{array} $$

Proof

Obvious. □

Lemma 12

$$\begin{array}{ll} \mathcal{P} \vdash_{\forall\mathbf{BB}^{\mathbf{t}\circ}} \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \textbf{t}\right] \rightarrow a \overset{i}{=} c \end{array} $$

Proof

$$\begin{array}{rll} {(1)} &\, \{a\} \in \langle a, b \rangle& \text{def.\,\,of\,\,ord.\,\,pair} + \mathcal{P}\\ {(2)} &\, \langle a, b \rangle \overset{i}{=} \langle c, d \rangle \rightarrow (\mathbf{t} \rightarrow \{a\} \in \langle c, d \rangle) & \text{1, Lemma 11(2)}\\ {(3)} &\, \{a\} \in \langle c, d \rangle \rightarrow (\{a\} \overset{i}{=} \{c\} \vee \{a\} \overset{i}{=} \{c, d\})& \text{def.\,\,of\,\,ord.\,\,pair} + \mathcal{P}\\ {(4)} &\, a \in \{a\}& \text{def.\,\,of\,\,singleton} + \mathcal{P}\\ {(5)} &\, \{a\} \overset{i}{=} \{c\} \rightarrow (\mathbf{t} \rightarrow a \in \{c\})& \text{4, Lemma 11(2)}\\ {(6)} &\, a \in \{c\} \rightarrow a \overset{i}{=} c& \text{def.\,\,of\,\,singleton} + \mathcal{P}\\ {(7)} &\, \{a\} \overset{i}{=} \{c\} \rightarrow (\mathbf{t} \rightarrow a \overset{i}{=} c)& \text{5, 6, rightER}\\ {(8)} &\, \{a\} \overset{i}{=} \{c, d\} \rightarrow (\mathbf{t} \rightarrow a \overset{i}{=} c)& \text{similar to (7)}\\ {(9)} &\, (\{a\} \overset{i}{=} \{c\} \vee \{a\} \overset{i}{=} \{c, d\}) \rightarrow (\mathbf{t} \rightarrow a \overset{i}{=} c)& \text{7, 8, R7}\\ {(10)} &\, \langle a, b \rangle \overset{i}{=} \langle c, d \rangle \rightarrow (\mathbf{t} \rightarrow (\mathbf{t} \rightarrow a \overset{i}{=} c))& \text{2, 3, 9, rightER \& transitivity}\\ {(11)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow a \overset{i}{=} c& \text{10, R13} \end{array} $$

□

Lemma 13

$$\begin{array}{ll} {(1)} & \mathcal{P} \vdash_{\forall\mathbf{BB}^{\mathbf{t}\circ}} \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right] \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d\\ {(2)} & \mathcal{P} \vdash_{\forall\mathbf{BB}^{\mathbf{t}\circ}} \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right] \rightarrow a \overset{i}{=} d \vee b \overset{i}{=} d \end{array} $$

Proof

The proof of (2) is similar to that of (1).

$$\begin{array}{rll} {(1)} &\,\{a, b\} \in \langle a, b \rangle & \text{def.\,\,of\,\,ord. pair} + \mathcal{P}\\ {(2)} &\, \langle a, b \rangle \overset{i}{=} \langle c, d \rangle \rightarrow (\mathbf{t} \rightarrow \{a, b\} \in \langle c, d\rangle)& \text{1, Lemma 11(2)}\\ {(3)} &\, \{a, b\} \in \langle c, d\rangle \rightarrow \{a, b\} \overset{i}{=} \{c\} \vee \{a, b\} \overset{i}{=} \{c, d\} & \text{def.\,\,of\,\,ord. pair} + \mathcal{P}\\ {(4)} &\, b \in \{a, b\} & \text{def.\,\,of\,\,ord. pair} + \mathcal{P}\\ {(5)} &\, \{a, b\} \overset{i}{=} \{c\} \rightarrow (\mathbf{t} \rightarrow b \in \{c\})& \text{4, Lemma 11(2)}\\ {(6)} &\, b \in \{c\} \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d& \text{def.\,\,of\,\,singleton}, \mathcal{P} \& Ax2\\ {(7)} &\, \{a, b\} \overset{i}{=} \{c\} \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d)& \text{5, 6, rightER}\\ {(8)} &\, \{a, b\} \overset{i}{=} \{c, d\} \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d)& \text{similar to (7)}\\ {(9)} &\, \{a, b\} \overset{i}{=} \{c\} \vee \{a, b\} \overset{i}{=} \{c, d\} \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d)& \text{4, 5, R7}\\ {(10)} &\, \langle a, b \rangle \overset{i}{=} \langle c, d \rangle \rightarrow (\mathbf{t} \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d))& \text{2, 3, 6, rightER\,\,\& transitivity}\\ (11) &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d& \text{10, R13} \end{array} $$

$\ \Box$

Lemma 14

$$\begin{array}{ll} \mathcal{P} \vdash_{\forall\mathbf{B}^{\mathbf{t}}} a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))) \end{array} $$

Proof

$$\begin{array}{rll} {(1)} & b \overset{i}{=} b& \text{Theorem}\\ {(2)} & b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} b)& \text{1, Lemma 11(2)}\\ {(3)} & b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))) & \text{2, Lemma 11(2)}\\ {(4)} & a \overset{i}{=} c \rightarrow (b \overset{i}{=} d \rightarrow b \overset{i}{=} d)& \text{Lemma 11(1)}\\ {(5)} & a \overset{i}{=} c \rightarrow (b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))))& \text{3, 4, rightER}\\ {(6)} & b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} c)& \text{1, Lemma 11(2)}\\ {(7)} & c \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))& \text{6, Lemma 11(2)}\\ {(8)} & a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \rightarrow c \overset{i}{=} d)& \text{Lemma 11(1)}\\ {(9)} & a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))))& \text{7, 8, rightER}\\ {(10)} & a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))))& \text{5, 9, R6\,\,\& Ax7}\\ {(11)} & a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))))& \text{Similar to (10)}\\ {(12)} & a \overset{i}{=} c \rightarrow (a \overset{i}{=} d \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d))))& \text{10, 11, Ax6\,\,\& Ax7} \end{array} $$

□

Lemma 15

$$\begin{array}{ll} A \rightarrow (B \rightarrow (C \rightarrow (D \rightarrow E))), F \rightarrow D \vdash_{\mathbf{BB}} A \rightarrow (B \rightarrow (C \rightarrow (F \rightarrow E))) \end{array} $$

Proof

Use R3, R4 and rightER. □

Lemma 16

$$\begin{array}{lll} \mathcal{P} \vdash_{\forall\mathbf{B}^{\mathbf{t}\circ}} &&\left( \left( \left( \left( \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \textbf{t}) \circ \textbf{t}\right] \circ \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \textbf{t}) \circ \textbf{t}\right]\right) \circ \textbf{t}\right) \circ\right. \right. \\ &&\left.\left. \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right]\right) \circ \mathbf{t}\right) \rightarrow b \overset{i}{=} d \end{array} $$

Proof

$$\begin{array}{rll} {(1)} &\, a \overset{i}{=} c \rightarrow ((a \overset{i}{=} d \vee b \overset{i}{=} d) \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))& \text{Lemma 14}\\ {(2)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow a \overset{i}{=} d \vee b \overset{i}{=} d& \text{Lemma 13}\\ {(3)} &\, a \overset{i}{=} c \rightarrow (\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow (\mathbf{t} \rightarrow (b \overset{i}{=} c \vee b \overset{i}{=} d \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))& \text{1, 2, leftER}\\ {(4)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow b \overset{i}{=} c \vee b \overset{i}{=} d& \text{Lemma 13}\\ {(5)} &\, a \overset{i}{=} c \rightarrow (\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow \\ &\, \qquad (\mathbf{t} \rightarrow (\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))& \text{3, 4, Lemma 15}\\ {(7)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow a \overset{i}{=} c& \text{Lemma 12}\\ {(8)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow (\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow \\ &\, \qquad (\mathbf{t} \rightarrow (\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow (\mathbf{t} \rightarrow b \overset{i}{=} d)))& \text{6, 7, trans.}\\ {(9)} &\, \Bigg(\bigg(\Big(\big(\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \circ \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\big) \circ \mathbf{t}\Big) \circ \\ &\, \qquad \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\bigg) \circ \mathbf{t}\Bigg) \rightarrow b \overset{i}{=} d& \text{8, R13} \end{array} $$

□

Lemma 17

$$\begin{array}{c} \mathcal{P} \vdash_{\forall\mathbf{B}^{\mathbf{t}\circ}} \partial\left( \langle a, b \rangle \overset{i}{=} \langle c, d \rangle\right) \circ A(a, b) \rightarrow A(c, d), i.e.\\ \mathcal{P} \vdash_{\forall\mathbf{B}^{\mathbf{t}\circ}} \left( \left[\left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right]\circ \right.\right.\\ \left.\left( \left( \left( \left( \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right] \circ \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right]\right) \circ \mathbf{t}\right) \circ \left[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\right]\right) \circ \mathbf{t}\right)\!\right] \circ \\ \left.A(a, b)\right) \rightarrow A(c, d) \end{array} $$

Proof

$$\begin{array}{rll} {(1)} &\, b \overset{i}{=} d \rightarrow (A(a, b) \rightarrow A(a, d))& \text{Lemma 11(1)}\\ {(2)} &\, a \overset{i}{=} c \rightarrow ((A(a, b) \rightarrow A(a, d)) \rightarrow (A(a, b) \rightarrow A(c, d)))& \text{Lemma 11(1)}\\ {(3)} &\, a \overset{i}{=} c \rightarrow (b \overset{i}{=} d \rightarrow (A(a, b) \rightarrow A(c, d)))& \text{1, 2, leftER}\\ {(4)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow a \overset{i}{=} c& \text{Lemma 12}\\ {(5)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow \Big[b \overset{i}{=} d \rightarrow (A(a, b) \rightarrow A(c, d))\Big]& \text{3, 4, trans.}\\ {(6)} &\, \Bigg(\bigg(\Big(\big(\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \circ \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\big) \circ \mathbf{t}\Big) \circ \\ &\, \qquad \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\bigg) \circ \mathbf{t}\Bigg) \rightarrow b \overset{i}{=} d& \text{Lemma 16}\\ {(7)} &\, \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \rightarrow \\ &\,\quad \Big[\Bigg(\bigg(\Big(\big(\big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big] \circ \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\big) \circ \mathbf{t}\Big) \circ \\ &\, \qquad \big[(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \circ \mathbf{t}) \circ \mathbf{t}\big]\bigg) \circ \mathbf{t}\Bigg) \rightarrow (A(a, b) \rightarrow A(c, d))\Big]& \text{5, 6, leftER}\\ {(8)} &\, \partial\big(\langle a, b \rangle \overset{i}{=} \langle c, d \rangle \rangle\big) \circ A(a, b) \rightarrow A(c, d)& \text{7, R13} \end{array} $$

□

Theorem 18

$\mathcal {P} \vdash _{\forall \mathbf {B}^{\mathbf {t}\circ }}\langle a, b\rangle \in \{\langle x, y\rangle | A\} \leftrightarrow A(a, b)$

Proof

$$\begin{array}{rll} {(1)} &\, \partial\big(\langle x, y\rangle \overset{i}{=} \langle a, b\rangle\big) \circ A(x, y) \rightarrow A(a, b) & \;\text{Lemma 17}\\ {(2)} &\, \exists x\exists y(\partial\big(\langle x, y\rangle \overset{i}{=} \langle a, b\rangle\big) \circ A(x, y)) \rightarrow A(a, b) & \;\text{1, RQ \& Q8}\\ {(3)} &\,\langle a, b\rangle \in \{\langle x, y\rangle | A\} \rightarrow A(a, b) & \;2, \mathcal{P} \text{+ def.\, of}\,\,\{\langle x, y\rangle | A\}\\ {(4)} &\, \partial\big(\langle a, b\rangle \overset{i}{=} \langle a, b\rangle\big) \rightarrow (A(a, b) \rightarrow \partial\big(\langle a, b\rangle \overset{i}{=} \langle a, b\rangle\big) \circ A(a, b))& \;\text{Ax1\& R13}\\ {(5)} &\, \partial\big(\langle a, b\rangle \overset{i}{=} \langle a, b\rangle\big)& \;\text{Lemma 10}\\ {(6)} &\, A(a, b) \rightarrow \partial\big(\langle a, b\rangle \overset{i}{=} \langle a, b\rangle\big) \circ A(a, b)& \;\text{4, 5, R2}\\ {(7)} &\, A(a, b) \rightarrow \exists x\exists y(\partial\big(\langle x, y\rangle \overset{i}{=} \langle a, b\rangle\big) \circ A(x, y))& \;\text{6, Q4}\\ {(8)} &\, A(a, b) \rightarrow \langle a, b\rangle \in \{\langle x, y\rangle | A\}& \;7, \mathcal{P} \text{+ def.\,of}~\,\{\langle x, y\rangle | A\}\\ {(9)} &\, \langle a, b\rangle \in \{\langle x, y\rangle | A\} \leftrightarrow A(a, b)& \;\text{3, 8, R1} \end{array}$$

$\ \Box$

Corollary 3

∀B ^t∘ suffices for the fixed-point theorem.

Proof

This follows from Theorem 17 and Theorem 18. □

∀B ^t∘ trivializes naïve set theory (Theorem 14). The question then is if there are logics weak enough not to trivialize naïve set theory, yet strong enough so as to make unrestricted pair-abstraction derivable. As the definition of {〈x,y〉|A} above should make clear, there are countless non-equivalent ways of defining sets of ordered pairs provided the logic is weak enough. What should also be clear is that defining {〈x,y〉|A} as {z|∀x∀y(∂ ^′(〈x,y〉=z)→A)} for some variant $\partial ^{\prime }$ of ∂, would not improve the situation: $\mathcal {P}$ in $\overset {=}{\forall }\mathbf {BB}$ suffices for deriving 〈a,b〉∈{〈x,y〉|A}→(∂ ^′(〈a,b〉=〈a,b〉)→A(a,b)). To get $\langle a, b\rangle \in \{\langle x, y\rangle | A\} \rightarrow A(a, b)$ from this one would then most certainly require the weak permutation rule R8 which trivializes naïve set theory (Theorem 15). The only option left as I see it would be to use {〈x,y〉|A}= _{d f} {z|∃x∃y(∂ ^′(〈x,y〉=z)∧A)}. However, this is not an option for the naïve set theoriest either: in order to have that A(a,b)⊩〈a,b〉∈{〈x,y〉|A}, one would need a function $\partial ^{\prime }$ such that $A \vdash \partial ^{\prime }(A)$. Furthermore, in order to prove $\langle a, b\rangle \in \{\langle x, y\rangle | A\} \rightarrow A(a, b)$, one would need $\partial ^{\prime }(\langle a, b\rangle = \langle c, d\rangle ) \land A(a, b) \rightarrow A(c, d)$. These two assumptions suffice for a triviality proof:

$$\begin{array}{rll} {(1)} &\, C \leftrightarrow \partial^{\prime}(\langle \mathfrak{p}_{C}, \mathfrak{p}_{C}\rangle = \langle \mathfrak{p}_{\bot}, \mathfrak{p}_{\bot}\rangle)& \;\mathcal{N}\\ {(2)} &\, C \rightarrow \partial^{\prime}(\langle \mathfrak{p}_{C}, \mathfrak{p}_{C}\rangle = \langle \mathfrak{p}_{\bot}, \mathfrak{p}_{\bot}\rangle) \land \mathfrak{p}_{C} \in \mathfrak{p}_{C}& \;\text{1, fiddling}\\ {(3)} &\, \partial^{\prime}(\langle \mathfrak{p}_{C}, \mathfrak{p}_{C}\rangle = \langle \mathfrak{p}_{\bot}, \mathfrak{p}_{\bot}\rangle) \land \mathfrak{p}_{C} \in \mathfrak{p}_{C} \rightarrow \mathfrak{p}_{\bot} \in \mathfrak{p}_{\bot}& \;\text{assumed theorem}\\ {(4)} &\, C \rightarrow \bot& \;\text{2, 3, fiddling}\\ {(5)} &\, \mathfrak{p}_{C} = \mathfrak{p}_{\bot}& \;\text{4, extensionality + fiddling}\\ {(6)} &\, \langle \mathfrak{p}_{C}, \mathfrak{p}_{C}\rangle = \langle \mathfrak{p}_{\bot}, \mathfrak{p}_{\bot}\rangle& \;5, {LL}_{2\vdash\vdash}\\ {(7)} &\, \partial^{\prime}(\langle \mathfrak{p}_{C}, \mathfrak{p}_{C}\rangle = \langle \mathfrak{p}_{\bot}, \mathfrak{p}_{\bot}\rangle)& \;\text{6, assumed rule}\\ {(8)} &\, \bot& \;\text{1, 4, 7, R2} \end{array}$$

I therefore conclude that the prospects of finding a logic and a way of defining {〈x,y〉|A} so that the logic does not trivialize naïve set theory, yet is strong enough to make unrestricted pair-abstraction a theorem thereof, are dim, at best.