Is ZF${}-{}$(Axiom of Infinity) consistent? Godel's theorem implies that Con(ZF) is not provable in ZF since it contains the axiom of infinity.
So is it consistent if the Axiom of infinity is removed?
 A: ZF with the negation of Infinity and Peano arithmetic are mutually interpretable. Richard Kaye and Tin Lok Wong's paper On interpretations of arithmetic and set theory has one of the classic expositions of how to interpret finite ZF in PA, and vice versa; Quine's Set Theory and its Logic also has a nice detailed exposition on getting Peano arithmetic without any infinite sets. But the consequence is that if finite ZF could prove its own consistency, it would imply that PA could prove its own consistency, which is not the case.
A: Your question is unclear.
It is true that $\sf ZF$ cannot prove its own consistency. But $\sf ZF$ can prove the consistency of $\sf ZF-Infinity$, simply by verifying that the set of hereditarily finite sets satisfies all the axioms of $\sf ZF$ except the axiom of infinity.
This set, often denoted by $HF$ or $V_\omega$ can be defined as follows, $V_0=\varnothing, V_{n+1}=\mathcal P(V_n)$, and $V_\omega=\bigcup V_n$. Note that all the elements of $V_\omega$ are finite, and their elements are finite and so on.
On the other hand, if you want to ask whether or not $\sf ZF-Infinity$ proves its own consistency, then the answer is no. The reason is that this theory satisfies the requirements of the incompleteness theorem, and therefore cannot prove its own consistency (unless it is inconsistent to begin with).
