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Let $S(i,x_1,\ldots, x_n)$ be a primitive recursive predicate. \begin{equation} f(i_1,i_2,x_1,\ldots, x_n) = \begin{cases} 1 &\text{ when for all i, }\; i_1 \le i \le i_2,\; S(i,x_1, \ldots, x_n)=1\\ 0 &\text{ otherwise} \end{cases} \end{equation}

Show that $f(i_1,i_2,x_1,\ldots, x_n)$ is also primitive recursive

I use Davis Computability and Complexity book. I get I need to write as described in this page. Show $x^y$ is a primitive recursive function

But how to do for predicates? Even a similar example which has predicates would be very helpful.

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David Hamide
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Do not get confused by the word "predicate". It is actually quite common to show that a function is primitive recursive by using another function for which we already know that it is primitive recursive in its definition.

For example, in the proof that multiplication is primitive recursive one usually defines multiplication using addition, which is not a basic primitive recursive function.

If "predicate" confuses you check out its definition, e.g., wikipedia. It is just a function $f: \mathbb{N} \rightarrow \{0,1\}$. So you can use a primitive recursive predicate just as a primitive recursive function of that kind.

All you need to show is, as mentioned in the comments, that the bounded quantification is primitive recursive.

Dino Rossegger
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