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Oct 10 at 9:56 vote accept Sam Sanders
Oct 10 at 3:36 answer added Dmytro Taranovsky timeline score: 4
Oct 9 at 17:38 history became hot network question
Oct 9 at 17:36 comment added Hanul Jeon But also note that, this uniformization theorem does not guarantee the existence of a choice function (as you can see, a naive way to construct a choice function from $\hat{\phi}$ requires $\Pi^1_1$-comprehension.)
Oct 9 at 17:34 comment added Hanul Jeon The point is that it is a uniformization not for real numbers, but for natural numbers. You may find its proof in Simpson's $\Sigma^1_1$ and $\Pi^1_1$ transfinite induction (Logic Colloquium '80, Stud. Logic Found. Math., 108), Lemma 2.3.
Oct 9 at 17:30 comment added Sam Sanders @HanulJeon I thought that kind of uniformisation is only provable in $\Pi_1^1$-comprehension?
Oct 9 at 17:28 comment added Hanul Jeon Relatedly, the numerical $\Pi^1_1$-uniformization theorem is a theorem of $\mathsf{ATR}_0$ (which states that if $\phi(x)$ is a $\Pi^1_1$-formula then we can find another $\Pi^1_1$-formula $\hat{\phi}(x)$ such that $\mathsf{ATR}_0$ proves $\forall^0 x [\hat{\phi}(x)\to\phi(x)]$ and $(\exists^0 x\phi(x))\to (\exists!^0 x \hat{\phi}(x))$.
Oct 9 at 14:57 answer added aws timeline score: 4
Oct 9 at 14:12 comment added Dmytro Taranovsky The principle is provable in $Π^1_1-\text{CA}_0$, and its restriction to $m∈\{0,1\}$ is equivalent to $Σ^1_1$ separation and thus ATR$_0$. The upper bound version proves at least ACA$_0$.
Oct 9 at 11:33 comment added Sam Sanders @aws Results in intuitionist logic would be fine too.
Oct 9 at 11:22 comment added aws I assume you're thinking of classical logic? In intuitionistic logic there are some conservativity results in this area.
Oct 9 at 9:37 history asked Sam Sanders CC BY-SA 4.0