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Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trscl}(X) \times \operatorname {trscl}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$  (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?

Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trscl}(X) \times \operatorname {trscl}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$;

$\operatorname {trscl}(X)=\{y \mid y=X \lor \forall T \,(trs(T) \land X \subseteq T \to y \in T) \}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$  (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?

A related question is to adapt this to accommodate existence of Ur-elements, and ask the same question about consistency of $\sf Strong \ NFU$. In this case we require the ismorphisms in the axiom of Strong Extensionality to fix Ur-elements.

Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trs}(X) \times \operatorname {trs}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$ (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?

Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trscl}(X) \times \operatorname {trscl}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$ (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?

Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trs}(X) \times \operatorname {trs}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$ (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?

Let's work in Quine's $\sf ML$, we can define classes after nonstratified formulas. Now for every set $X$ we can define the membership graph $\operatorname {MG}(X)$ on its transitive closure, that is the restriction of the graph of membership relation (which is a class) to the transitive closure class of $X$, so $$\operatorname {MG}(X)= E \cap (\operatorname {trs}(X) \times \operatorname {trs}(X))$$; Where $E = \{(x,y) \mid x \in y, x \in V, y \in V\}$

Now what I want to do is to instead of just stipulating Extensionality over $\sf SF$ (i.e.; $\sf NF-Extensionality$), we stipulate:

$$ \operatorname {MG}(X) \approx \operatorname {MG}(Y) \to X=Y.$$

where $(,)$ is defined after Wiener pairs; and $\approx$ stand for graph ismorphism.

I'd label this axiom as Strong Extensionality, and "Strong Extensionality + Stratified Comprehension" as $\sf Strong \ NF$.

Clearly Strong Extensionality proves Extensionality. But if we assume that $\sf NF$ is consistent, then would this be enough to prove the consistency of $\sf Strong \ NF$?