If $\phi$ is any formula of set theory with just one free variable $x$, the abstraction term $A\_{\phi}=\lbrace x | \phi(x) \rbrace$ is either a set or a proper class. Assume that ZFC is consistent, or any large cardinal axiom you like. Then my question is, are there two formulas 
$\phi$ and $\psi$ such that ZFC+($A\_{\phi}$ is a set) is consistent, ZFC+($A\_{\psi}$ is a set) is consistent also, but ZFC+($A\_{\phi}$ and $A\_{\psi}$ are both sets) is not?


UPDATE 09/15/2011 : to avoid "cheating" as in François Dorais' answer, we may introduce the following additional constraint : if $T$ is any theory extending $ZFC$, say that the abstraction term $A\_{\phi}=\lbrace x | \phi(x) \rbrace$ is small in $T$ if $T$ proves that $A\_{\neg \phi}$ is not a set ; for example, if $\phi(x)$ is "x is an accessible ordinal" or "all cardinals below the ordinal $x$ are not measurable" or "all cardinals below the ordinal $x$ are not Mahlo" then $A\_{\phi}$ will be small, but this will not be the case if $\phi$ is an undecidable statement independent of $x$ as in Francois Dorais' answer.


The question then becomes, are there two formulas $\phi$ and $\psi$ such that 
$A\_{\phi}$ and $A\_{\psi}$ are both small in $ZFC$, ZFC+($A\_{\phi}$ is a set) is consistent, ZFC+($A\_{\psi}$ is a set) is consistent also, but ZFC+($A\_{\phi}$ and $A\_{\psi}$ are both sets) is not?



mutually incompatible abstraction terms?
If by "consistent" you mean "consistent relative to the consistency of ZFC," then there is a simple example. Let $\phi$ be the [Rosser sentence](http://en.wikipedia.org/wiki/Rosser%27s_trick#The_Rosser_sentence) for ZFC, and let $\psi$ be its negation. Then, in any model of ZFC, $A\_\phi$ is either $V$ (if $\phi$ is true) or $\varnothing$ (if $\phi$ is false); similarly for $A\_\psi$. However, we cannot have both $A\_\phi = \varnothing$ and $A\_\psi = \varnothing$ since one of the two sentences must be true.


