In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is divided into levels. The comprehension schema for level $0$ sets does not allow any formulas with second-order quantifiers. For any natural number $n$, the comprehension schema for level $n+1$ sets allows quantification over sets of level $n$ and below. This ensures that no set is defined using quantification over itself. The resulting system, however, proved too weak to do much mathematics. (Although it turns out it can do more than Russell assumed; see here.) So Russell adopted his controversial axiom of reducibility, actually an axiom schema which for each natural number $n$, states that for any set $X$ of level $n$, there exists a set $Y$ of level $0$ such that $X$ and $Y$ contain the same elements.

Now, it is commonly asserted that the axiom of reducibility is equivalent to simply eliminating the ramified hierarchy and just working in standard second-order logic. But I don't see why. At first glance, it seems to me that saying that every set, period, is coextensional with a level $0$ set, is a stronger statement than saying that every set of any given level is coextensional with a level $0$ set. Aren't there sets that aren't coextensional with sets of any level?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: Here's another way to phrase my question. Consider the following two possible axiom schemata:

1. For any formula $\phi(x)$ in the language of second-order arithmetic, there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$
2. For any formula $\phi(x)$ with only graded quantifiers (i.e. quantifiers of the form "for all sets of level..." or "there exists a set of level..."), there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$.

Is the first axiom schema stronger than the second, or are they equivalent?

In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is divided into levels. The comprehension schema for level $0$ sets does not allow any formulas with second-order quantifiers. For any natural number $n$, the comprehension schema for level $n+1$ sets allows quantification over sets of level $n$ and below. This ensures that no set is defined using quantification over itself. The resulting system, however, proved too weak to do much mathematics. (Although it turns out it can do more than Russell assumed; see here.) So Russell adopted his controversial axiom of reducibility, actually an axiom schema which for each natural number $n$, states that for any set $X$ of level $n$, there exists a set $Y$ of level $0$ such that $X$ and $Y$ contain the same elements.

Now, it is commonly asserted that the axiom of reducibility is equivalent to simply eliminating the ramified hierarchy and just working in standard second-order logic. But I don't see why. At first glance, it seems to me that saying that every set, period, is coextensional with a level $0$ set, is a stronger statement than saying that every set of any given level is coextensional with a level $0$ set. Aren't there sets that aren't coextensional with sets of any level?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: Here's another way to phrase my question. Consider the following two possible axiom schemata:

1. For any formula $\phi(x)$ in the language of second-order arithmetic, there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$
2. For any formula $\phi(x)$ with only graded quantifiers (i.e. quantifiers of the form "for all sets of level..." or "there exists a set of level..."), there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$.

Is the first axiom schema stronger than the second, or are they equivalent?

In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is divided into levels. The comprehension schema for level 0 sets does not allow any formulas with second-order quantifiers. For any natural number n, the comprehension schema for level n+1 sets allows quantification over sets of level n and below. This ensures that no set is defined using quantification over itself. The resulting system, however, proved too weak to do much mathematics. (Although it turns out it can do more than Russell assumed; see here.) So Russell adopted his controversial axiom of reducibility, actually an axiom schema which for each natural number n, states that for any set X of level n, there exists a set Y of level 0 such that X and Y contain the same elements.

Now, it is commonly asserted that the axiom of reducibility is equivalent to simply eliminating the ramified hierarchy and just working in standard second-order logic. But I don't see why. At first glance, it seems to me that saying that every set, period, is coextensional with a level 0 set, is a stronger statement than saying that every set of any given level is coextensional with a level 0 set. Aren't there sets that aren't coextensional with sets of any level?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: Here's another way to phrase my question. Consider the following two possible axiom schemata:

1. For any formula $\phi(x)$ in the language of second-order arithmetic, there exists a set of level 0 whose elements are the ones that satisfy $\phi(x)$
2. For any formula $\phi(x)$ with only graded quantifiers (i.e. quantifiers of the form "for all sets of level..." or "there exists a set of level..."), there exists a set of level 0 whose elements are the ones that satisfy $\phi(x)$.

Is the first axiom schema stronger than the second, or are they equivalent?

In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is divided into levels. The comprehension schema for level $0$ sets does not allow any formulas with second-order quantifiers. For any natural number $n$, the comprehension schema for level $n+1$ sets allows quantification over sets of level $n$ and below. This ensures that no set is defined using quantification over itself. The resulting system, however, proved too weak to do much mathematics. (Although it turns out it can do more than Russell assumed; see here.) So Russell adopted his controversial axiom of reducibility, actually an axiom schema which for each natural number $n$, states that for any set $X$ of level $n$, there exists a set $Y$ of level $0$ such that $X$ and $Y$ contain the same elements.

Now, it is commonly asserted that the axiom of reducibility is equivalent to simply eliminating the ramified hierarchy and just working in standard second-order logic. But I don't see why. At first glance, it seems to me that saying that every set, period, is coextensional with a level $0$ set, is a stronger statement than saying that every set of any given level is coextensional with a level $0$ set. Aren't there sets that aren't coextensional with sets of any level?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: Here's another way to phrase my question. Consider the following two possible axiom schemata:

1. For any formula $\phi(x)$ in the language of second-order arithmetic, there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$
2. For any formula $\phi(x)$ with only graded quantifiers (i.e. quantifiers of the form "for all sets of level..." or "there exists a set of level..."), there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$.

Is the first axiom schema stronger than the second, or are they equivalent?