Set Theory – Consistency Strength of Reflection Forms

lo.logicset-theory

Working in mono-sorted first order logic with equality $“="$ and membership $“\in"$:

  • Define: $set(x) \equiv_{df} \exists y \, (x \in y)$

  • Axiomatize:

  1. Extensionality: $( a \subseteq b \land b \subseteq a \to a=b)$

  2. Separation: $(set(a) \to \exists \ set \ x : \forall y \, (y \in x \leftrightarrow y \in a \land \phi ))$

  3. Reflection: $ (\varphi \to \exists \ set \ x : \text { trs}(x) \land \varphi^x)$

where formulas $\phi, \varphi$ do not use $“x"$; $\varphi$ do not use $“="$; $\varphi^x$ is obtained from $\varphi$ by merely bounding all of its quantifiers by $“\subseteq x"$; and $“\text { trs}" $ stands for is transitive.

This theory can prove: Set existence, Empty, Pairing, set Union, Power, Infinity, over the set world of it, also it proves Substitution. It can also prove class comprehension scheme. So it appears to prove all axioms of $\sf MK$–$\sf Reg.$–$\sf Choice$.

Is this theory consistent?

If so, is it equivalent to Bernay's reflection?

Best Answer

This theory is inconsistent.

We note that by 1 and 2 that if set(x) and y⊆x, then set(y).

(a) There is a v such that ∀x(set(x)-->x∈v).

Proof:Suppose not. Then ∀v∃s∃t(s∈t∧s∉v). By 3 there is transitive x such that

  set(x) and ∀v(v⊆x-->∃s∃t(s⊆x∧t⊆x∧s∈t∧s∉v). In particular 

  (x⊆x-->∃s∃t(s⊆x∧t⊆x∧s∈t∧s∉x).  But this is impossible.

Suppose that ∀x(set(x)-->x∈V). Then ∃w∀t(t∈V-->t∈w). By 3 there is transitive x

such that set(x) and ∃w(w⊆x∧∀t(t⊆x∧t∈V-->t∈w)). Since t⊆x implies set(t),

t⊆x implies t∈x. By 2, there is a c such that t∈c<-->(t∈x∧t∉t). Since set(c),

c∈c<-->c∉c.

Related Question