2 added the least fixed point axiom; added 9 characters in body

I asked this question on math.stackexchange.com and got some great comments from Andres Caicedo, but no answers. His comments pointed me toward class forcing, but that's way out of my field. I'm basically an infant in advanced set theory and model theory - I'm a computer scientist that loves logic and set theory - but it seems to me that there should be an easier way.

The original problem starts this way: I would like to define a unary, recursive function $F$ on a proper class. But the recursion is necessarily not well-founded, so I can't.

But I can get by with something less fully specified than a function.

Instead, I've introduced a new primitive binary relation $R$, and stated a family of defining axioms, which . All but one would correspond with rules defining $F$ F$, if I could define$F$. Every axiom F$, and these can be written in the form

$$\forall x_1, x_2, \dots, x_n. Q(x_1,x_2,\dots,x_n) \implies R(x_1,x_2)$$

where $Q$ is an axiom-specific condition that usually refers to $R$. I'm not sure how to formulate the last axiom at the class level. It's that $R$ is the intersection of all relations that satisfy the previous axioms.

Here's my original problem: I'm unsure whether this extension of ZFC is conservative. Does it prove the existence of any new sets?

Andres suggested class forcing. The strategy would be: if $R$ is defined without parameters and is a partial order (I can easily make it so), I could possibly prove it is tame; it therefore preserves ZFC; QED. But proving tameness is beyond my abilities right now, and I have a hard time justifying the time I would spend learning how. I haven't a clue whether $R$ is tame, anyway.

The axioms have a simple, regular form that it seems I should be able to exploit to more easily prove the extension is conservative. In particular, they all but the last one have these properties:

1. None of them conclude or assert that a set exists.
2. Every $Q(x_1,x_2,\dots,x_n)$ is a $\Delta_0$ sentence.
3. Therefore, every axiom is a $\Pi_1$ sentence.

Using these properties, can I prove the extension is conservative?

1

# How to prove an extension of ZFC is conservative

I asked this question on math.stackexchange.com and got some great comments from Andres Caicedo, but no answers. His comments pointed me toward class forcing, but that's way out of my field. I'm basically an infant in advanced set theory and model theory - I'm a computer scientist that loves logic and set theory - but it seems to me that there should be an easier way.

The original problem starts this way: I would like to define a unary, recursive function $F$ on a proper class. But the recursion is necessarily not well-founded, so I can't.

But I can get by with something less fully specified than a function. I've introduced a new primitive binary relation $R$, and stated a family of defining axioms, which would correspond with rules defining $F$ if I could define $F$. Every axiom can be written in the form

$$\forall x_1, x_2, \dots, x_n. Q(x_1,x_2,\dots,x_n) \implies R(x_1,x_2)$$

where $Q$ is an axiom-specific condition that usually refers to $R$.

Here's my original problem: I'm unsure whether this extension of ZFC is conservative. Does it prove the existence of any new sets?

Andres suggested class forcing. The strategy would be: if $R$ is defined without parameters and is a partial order (I can easily make it so), I could possibly prove it is tame; it therefore preserves ZFC; QED. But proving tameness is beyond my abilities right now, and I have a hard time justifying the time I would spend learning how. I haven't a clue whether $R$ is tame, anyway.

The axioms have a simple, regular form that it seems I should be able to exploit to more easily prove the extension is conservative. In particular, they have these properties:

1. None of them conclude or assert that a set exists.
2. Every $Q(x_1,x_2,\dots,x_n)$ is a $\Delta_0$ sentence.
3. Therefore, every axiom is a $\Pi_1$ sentence.

Using these properties, can I prove the extension is conservative?