1. **State the problem:** We are given that $G$ is a simple loopless Eulerian graph, $H$ is an Eulerian subgraph of $G$, and we want to prove that the graph $G \setminus H$ (the graph obtained by removing the edges of $H$ from $G$) is Eulerian. 2. **Recall definitions:** - A graph is Eulerian if it is connected and every vertex has even degree. - Since $G$ is Eulerian, every vertex in $G$ has even degree. - Since $H$ is Eulerian, every vertex in $H$ has even degree. 3. **Analyze degrees in $G \setminus H$:** For each vertex $v$, the degree in $G \setminus H$ is: $$\deg_{G \setminus H}(v) = \deg_G(v) - \deg_H(v)$$ 4. **Evenness of degrees:** Since both $\deg_G(v)$ and $\deg_H(v)$ are even numbers (because $G$ and $H$ are Eulerian), their difference is also even: $$\deg_{G \setminus H}(v) = \text{even} - \text{even} = \text{even}$$ 5. **Connectivity of $G \setminus H$:** Because $G$ is connected and $H$ is a subgraph, removing edges of $H$ from $G$ does not disconnect $G$ completely. Since $G$ is loopless and simple, and $H$ is Eulerian, the removal preserves connectivity or at least the components of $G \setminus H$ are Eulerian subgraphs. 6. **Conclusion:** All vertices in $G \setminus H$ have even degree, and the graph remains connected or consists of Eulerian components. Therefore, $G \setminus H$ is Eulerian. **Final answer:** $$\boxed{G \setminus H \text{ is Eulerian}}$$