Division by two in set theory

Let $A,B$ be two sets such that $2A \cong 2B$ (here $2A := A \coprod A$). Then $A \cong B$. This can be proven without the axiom of choice, which means that one can explicitly construct a bijection $A \to B$ out of a bijection $2A \to 2B$. This is non-trivial and interesting, see the wonderful paper by Conway, Doyle, also for generalizations. The construction is infinitary, and therefore the following question comes into my mind.

Question. Is the assertion also true in ZF - {Axiom of Infinity}?

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Hm, I don't understand. How do you define "infinite" at all without refering to $\mathbb{N}$? So what is the formalization of your first sentence in ZF? And how does your remark explicitly produce a proof of the assertion which does not use $\mathbb{N}$? – Martin Brandenburg Oct 17 '12 at 8:07