Expressing partial decidability using existential quantification def.
A predicate M(x,y) is partially decidable if the function f given by " f(x,y) = 1(if M(x,y) holds), f(x,y) = undefined(otherwise) " is computable.

Thm.
If M(x,y) is partially decidable, then so is the predicate ∃yM(x,y).

proof.
 Take a decidable predicate R(x,y,z) such that M(x,y) iff ∃zR(x,y,z).
Then...



I can not imagine the R(x,y,z)...
Please explain the way of thinking.

( Page115, Computability by Nigel Cutland)
The predicate $R(x,y,z)$ states that "$M$ accepts the input $x,y$ within $z$ steps". It is computable since we can simulate a given machine on a given input for a bounded number of steps.