This is almost certainly a silly question, but:

I am currently reading Moschovakis' article "Kleene's amazing second recursion theorem" (http://www.math.ucla.edu/~ynm/papers/1602-002-1.pdf) and there is a footnote in it which confuses me.

In footnote 10, on page 195, Moschovakis writes:

. . . [T]he operation $$\Phi(p)=\begin{cases} 1, & \text{if $p(0)\downarrow$ or $p(1)\downarrow$,}\\ \perp, & \text{otherwise}\\ \end{cases}$$ is effective but not computable by a deterministic Turing machine.

Here "operations" are functions on indices of programs which take the same value on indices representing the same partial computable function (bottom of pg. 194): so in more conventional terminology, $\Phi(e)=1$ if $\varphi_e(0)\downarrow$ or $\varphi_e(1)\downarrow$, and is undefined otherwise.

The problem I'm having is that this seems completely false. Deterministic Turing machines compute exactly the same functions as nondeterministic Turing machines.

But of course Moschovakis knows this, so my question is:

What is Moschovakis actually saying?

I am sure the answer is quite simple, but I don't see it.