Show that function is recursive 
Recursive functions: $\mathbb{N}_0^k \to \mathbb{N}_0$
Class of recursive functions:
The minimal subclass of the class of all the functions $f:
\mathbb{N}_0^k \to \mathbb{N}, k=1,2, \dots$ that contains the
  
  
*
  
*$f(x)=c$ (constants)
  
*$S(x)=x+1$ (successor)
  
*$P_n^m(x_1, \dots, x_m)=x_n \ (m \geq n)$ (projections)
  
  
  and is closed under the following
  
  
*
  
*If $h,g_1, \dots, g_n: \mathbb{N}_0^m \to \mathbb{N}_0$ are recursive then $f(\overline{x})=h(g_1(\overline{x}), \dots,
g_n(\overline{x})): \mathbb{N}_0^m \to \mathbb{N}$ is recursive .
  (composition)
  
*If $g: \mathbb{N}_0^{k-1} \to \mathbb{N}_0$ and $h: \mathbb{N}_0^{k+1} \to \mathbb{N}_0$ are recursive then $f:
\mathbb{N}_0^k \to \mathbb{N}_0$ is also recursive
  $$f(0,\overline{y})=g(\overline{y}) \\ f(x+1, \overline{y})=h(x, f(x,
\overline{y}), \overline{y})$$   (recursion)
  
*If $g$ is recursive and $\forall \overline{x} \  \exists z \ g(z, \overline{x})=0\\ f(\overline{x})=\mu z \ (g(z, \overline{x})=0)= \min
\{ z \in \mathbb{N}_0 \mid g(z, \overline{x})=0\}$
If $A \subset \mathbb{N}_0^k$ then $A$ is a recursive set iff the
  characteristic function of $A$ is recursive.

I want to show that the following functions and sets are recursive:


*

*$f(x,y)=x+y$

*$f(x,y)=xy$

*$\DeclareMathOperator{sign}{sign}\sign(x)=\begin{cases}
1 &, x=0 \\ 
0 & , x>0
\end{cases}.$

*$f(x)=\begin{cases}
1 &, g(x)=0 \\ 
0 &, h(x)=0 
\end{cases}$, under the term that $g,h: \mathbb{N}_0 \to \mathbb{N}_0$ are recursive and $\forall x(g(x)h(x)=0, g(x)+h(x)>0)$

*$f(y)=\prod_{x=0}^y g(x)$

*$B = \{ y \mid \exists x \leq y \  x \in A \}, C = \{ y \mid \forall x \leq y \  x \in A\}$ if $A$ recursive.

*$x \dot{-} y=\left\{\begin{matrix}
x-y &, x \geq y \\ 
0 &, \text{otherwise} 
\end{matrix}\right.$

*$f(y)=\sum_{x=0}^y g(x)$ if $g$ is recursive


Since I haven't seen any example, could you maybe explain to me with an example how we show that a specific function is recursive using the above definition?

Re second set:
We note 
$$
\chi_C(y) = \prod_{x=0}^y \chi_A(x)
$$
which is recursive because $A$ recursive implies its characteristic function $\chi_A$ is recursive. The used product results in a recursive function, as shown for the fifth task above. Thus $C$ is recursive.
Re first set:
The idea here is to use logical negation to express the needed OR expression by an AND expression:
$$
\chi_B(y) = \sign\left( \prod_{x=0}^y \sign\left(\chi_A(x) \right) \right)
$$
