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Can this naivenaïve-like set theory using acyclic membership be consistent?

The following theory contains a comprehension axiom that is a naive likenaïve-like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$)$\operatorname{FOL}(=, \in^*)$.

Formal workup:

Language: the first order language of set theory.

ExtesnionalityExtensionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$$\forall x \exists t: t=\operatorname{TC}(x)$

$\DeclareMathOperator\TC{TC}\DeclareMathOperator\trs{trs}$Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$$t=\TC(x) \iff \trs(t) \land x \subseteq t \land \forall k (\trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$$\trs$" stands for "is transitive", that is: closure under relation $\in$.

Induction: if $\phi$ is a formula, then:

$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in \TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$$y \in^* x \iff y \in x \land \neg \, x \in \TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$$x$, whose predicates are among $=, \in^*$$=$, $\in^*$ symbols.

Questions:Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

Can this naive-like set theory using acyclic membership be consistent?

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

Can this naïve-like set theory using acyclic membership be consistent?

The following theory contains a comprehension axiom that is a naïve-like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language $\operatorname{FOL}(=, \in^*)$.

Formal workup:

Language: the first order language of set theory.

Extensionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=\operatorname{TC}(x)$

$\DeclareMathOperator\TC{TC}\DeclareMathOperator\trs{trs}$Define: $t=\TC(x) \iff \trs(t) \land x \subseteq t \land \forall k (\trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$\trs$" stands for "is transitive", that is: closure under relation $\in$.

Induction: if $\phi$ is a formula, then:

$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in \TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in \TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $x$, whose predicates are among $=$, $\in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

added 3 characters in body
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Zuhair Al-Johar
  • 11.3k
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The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x \ (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x \ (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

added 2 characters in body
Source Link
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$$$\forall y \in x \ (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

The following theory contains a comprehension axiom that is a naive like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language FOL(=, $\in^*$).

Formal workup:

Language: the first order language of set theory.

Extesnionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=TC(x)$

Define: $t=TC(x) \iff trs(t) \land x \subseteq t \land \forall k (trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$trs$" stands for "is transitive", that is: closure under relation $\in$

Induction: if $\phi$ is formula, then:

$$\forall y \in x \ (\phi) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using $``x"$, whose predicates are among $=, \in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

added 21 characters in body
Source Link
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47
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Source Link
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47
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