# Is acyclic ZF consistent?

I need to know if the following system is consistent, because I want to use it in presenting automorphisms over stratified versions of it.

The system I'd label as "Acyclic ZF", which is $$\small \sf ZF-Reg.+ acyclic \ AFA + Rank$$, is obtained by adding a two place predicate symbol $$\mathcal R$$ symbolizing is the rank of to the language of $$\small \sf ZF$$, then replace Regularity by the following two acyclic AFA axioms:

Acyclicity: $$\forall x_1,..,x_n: \neg (x_1 \in x_2 \in x_3 \in ... \in x_n \land x_1=x_n)$$

Acyclic construction: For every acyclic accessible pointed graph there exists a set whose membership graph is isomorphic to it, where the latter means the membership map on the transitive closure of that set.

Now we define the unary predicate ordinal, symbolized by $$\mathcal Ord$$, as transitive set of transitive sets. To be emphasized here is that an ordinal can be a von Neumann or may not be so! If the ordinal is well founded on $$\in$$, then it is a von Neumann ordinal, if not then its to be called as a non-standard ordinal, or even more outrageously an ill-founded ordinal. We make axioms to the effect that the ranking relation $$\mathcal R$$ constitute a partial function from ordinals to sets such that the indexed sets would correspond to iterative powers similar to the buildup of the cumulative hierarchy. Formally this is:

$$\forall a,b,c,d: \mathcal R(a,b) \land \mathcal R(c,d) \longrightarrow [a=c \Leftrightarrow b=d]$$

$$\forall x: \exists y (\mathcal R (x,y)) \iff \mathcal Ord(x)$$

$$\forall \alpha \forall x \forall y: \mathcal R(\alpha, x) \land \mathcal R (\alpha \cup \{\alpha \}, y) \longrightarrow y=\mathcal P(x)$$

$$\forall \alpha \forall x [(\not \exists \beta: \alpha=\beta \cup \{\beta\}) \land \mathcal R(\alpha,x) \longrightarrow \\ x= \bigcup \{y: \exists \beta \in \alpha ( \mathcal R(\beta,y)) \}]$$

$$\forall \alpha \forall x : \mathcal R(\alpha,x) \to \alpha \subseteq x \land \alpha \not \in x$$

$$\forall \alpha \forall x: \mathcal R (\alpha, x) \to \forall y \in x (y \subseteq x)$$

The last axiom is to restrict sets to those itrative stages, I'll consider it as a parallel to foundation, that is:

Para-foundation: $$\forall x \exists \alpha \exists v : \mathcal R(\alpha, v) \land x \in v$$

Now, is Acyclic ZF consistent?

Acyclic ZF is inconsistent. Let D be the directed graph whose vertices are the natural numbers where (a,b) is an edge if a=0 and a<b, or b>0 and a>b. By Acyclic construction, there is set {x0.x1,x2,...} such that xi∈𝑥j iff (i,j) is in D. Each xi is a transitive set and thus each xi is an ordinal.

(1)Suppose R(xi,yi) for all natural numbers i. Suppose j>0. Then every von Neumann ordinal is in yj.
Proof: Suppose it is not true and let b be the least von Neumann ordinal which is not in yi for all i>0. We note that for i>0, xi=x(i+1)U{x(i+1)}. By our assumption, for every i>0, b⊆y(i+1). Therefore b∈yi.

Therefore there is a set A of all von Neumann ordinals. But then A is a von Neumann ordinal and thus A∈A.

• D is a directed graph. What is the cycle? – Greg Kirmayer Feb 15 at 20:39
• Thank you for that answer. The graph you defined is cyclic so it won't apply. But anyhow you are right as far as regarding my intention behind developing this theory! It appears that in order to save this approach I must forbid $\mathcal R$ from appearing in separation and replacement. – Zuhair Al-Johar Feb 16 at 6:58

It is impossible to have a rank that is both:

1. Ill-founded, and
2. successive steps are obtained by the power set operation.

The reason is simple, it would define a "Specker-tree" which is not a tree. Recall that $$T$$ is a Specker tree if it is a collection of sets such that $$x$$ is the direct successor of $$y$$ if and only if $$|y|=|\mathcal P(x)|$$. Specker trees are well-founded, in the sense that they do not have infinite branches.

The reason, of course, is that $$x\mapsto\aleph(x)$$ is a function which is non-decreasing with power sets, so an infinite branch would correspond to a decreasing sequence of ordinals. And even if you decide that all hereditarily transitive sets should be called "ordinals", it does not change the fact that Hartogs numbers are well-founded ordinals.

• I think you want to say that $x\mapsto\aleph(x)$ is increasing with powersets, not just non-decreasing; otherwise I don't see how to get an infinite descending sequence of Hartogs numbers. – Noah Schweber Feb 16 at 1:49
• It's not increasing, because if $x$ is amorphous $\aleph(x)=\aleph(\mathcal P(x))$. The thing is that the tree is really going downards. – Asaf Karagila Feb 16 at 2:05
• The point is, if $\aleph(\mathcal{PPP}(x))=\kappa$, then $\aleph(x)<\kappa$. Now consider the fact that a Specker tree is in a sense logarithmic when it "grows up", an infinite branch is necessarily a decreasing sequence in Hartogs numbers. – Asaf Karagila Feb 16 at 2:15
• (You can probably get better gaps with Lindenbaum numbers, but it doesn't matter, does it?) – Asaf Karagila Feb 16 at 2:17
• Thanks for the answer. The definition of the Specker tree would be made inside a stage by using separation (or replacement), I think this definition would essentially use of the symbol $\mathcal R$ here. If we forbid using $\mathcal R$ in instances of separation and replacement would that objection still hold. – Zuhair Al-Johar Feb 16 at 9:22