Let $U\subset\mathbb{R}^{d}$ be an open set. The singular support of a distribution $u\in\mathcal{D}^{\prime}(U)$ is defined to be the set of points, which have a neighbourhood in which $u$ can be identified with a smooth function. In a paper Fourier integral operators. I of Hörmander, it is claimed that

$$\DeclareMathOperator\singsupp{sing supp}\singsupp(u)=\bigcap_{\varphi\in C^{\infty}_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}.$$

However, I don't see why it is the case. Does anyone know how to see that this is the case?

My attempt:

The direction $\Leftarrow$ is quite clear I think: Let $x\notin\singsupp(u)$. Then there is an open neighbourhood $N$ of $x$ on which $u\vert_{N}\in C^{\infty}$. Let $\chi\in C_{c}^{\infty}(U)$ be supported in $N$ such that $\chi(x)\neq 0$. Then $\chi u\vert_{N}\in C^{\infty}(U)$ and hence $x\notin\bigcap_{\varphi\in C^{\infty}_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}$.

What is left is the other direction. I essentially have to show that if there exists a test function $\varphi\in C^{\infty}_{c}(U)$ such that $\varphi u\in C^{\infty}(U)$ and $\varphi(x)\neq 0$, then $x\notin\singsupp(u)$.