$\operatorname{func}(f)\to((x,y)\in f\to f(x)=y) $ and $\operatorname{func}(f)\to((x \in \operatorname{dom}(f) \wedge f(x)=y)\to (x,y)\in f)$

Question

In $\mathrm{NBG}$. Let be $f,x$ classes, I define: $$f(x):=\cap\{z|(x,z)\in f\}$$ Now I proof:

1) $\operatorname{func}(f)\to \forall x,y:((x,y)\in f \to f(x)=y)$

2) $\operatorname{func}(f)\to \forall x \in \operatorname{dom}(f),y: (f(x)=y \to (x,y) \in f)$

($\operatorname{func}(f)$ for $\forall x,z,y:((x,y)\in f\wedge (x,z)\in f) \to z=y)$

proof 1) $(x,y)\in f\to \{z|(x,z)\in f\} \neq \emptyset \to \cap\{z|(x,z)\in f\}$ is Set, but $\{z|(x,z)\in f\}=\{y\}$ (because $r \in \{z|(x,z)\in f\} \to (x,r)\in f \to r=y \to r \in \{y\}$ and $r \in \{y\} \to r=y \to (x,r) \in f \to r \in \{z|(x,z)\in f\}$ (QED)) then $f(x)=\cap\{z|(x,z)\in f\}=\cap\{y\}=y$ (QED)

proof 2) $x \in \operatorname{dom}(f) \to \exists w:((x,w)\in f )\to f(x)=w \to w=y \to (x,y) \in f$ (QED).

Is it correct?