Show this function is partially computable Let B(x) be a computable predicate. Show that 
\begin{equation}
G_B(r)=
\begin{cases}
1 \;\;\;\;\;\text{ if there are at least r numbers n such that B(k) = 1 }  \\
\uparrow \;\;\;\;\; \text{ otherwise}
\end{cases}
\end{equation}
I want to write a program with just these four instructions to compute $G_B$(r)

\begin{array} \\
\;\;\;\;\;\;\;\;\;Y \gets 0 \\
\;\;\;\;\;\;\;\;\;\text{IF } X \neq 0 \text{ GOTO } A \\
[E] \;\;\;\text{ GOTO } E \\
[A]\;\;\;\; Y \gets Y+1 
\end{array}

The book gives examples for writing programs to compute $G(x_1,x_2)$ but does not give any when predicates are involved. Could someone give just an example for even a completely different function with predicates or just point me in the correct direction.
Since $B$ is a computable predicate, there is one procedure that computes it. We can call this procedure in any point of our program.

K &lt;- k
i &lt;- 0
LOOP
    RESULT &lt;- B(i)
    IF RESULT != 0 GOTO TRUE
        GOTO ELSE
        TRUE
        K&lt;-K-1
    ELSE
    IF K != 0 GOTO KEEP_GOING
        GOTO BREAK
    KEEP_GOING
    i &lt;- i+1
GOTO LOOP

BREAK
RETURN 1


You will need to substitute the instruction K &lt;- K-1 by an appropriate procedure.

To understand what is going on, we are looping through every natural number and checking whether it satisfies the predicate $B$, keeping a counter that diminishes every time we find one such number. So when the counter K reaches zero we know that the function should return 1, and we proceed to do exactly that.