Let $G$ and $G_{0}$ be context-free grammars. Necessary and sufficient conditions on $G_{0}$ are obtained for the decidability of $L(G_{0}) \subseteq L(G)$. It is also shown that it is undecidable for which $G_{0},L(G) \subseteq L($G_{0})$ is decidable. Furthermore, given that $L(G) \subseteq L($G_{0})$ is decidable for a fixed $G_{0}$, there is no effective procedure to determine the algorithm which decides $L(G) \subseteq L($G_{0})$. If $L(G_{0})$ is a regular set, $L(G)=L(G_{0})$ is decidable if and only if $L(G_{0})$ is bounded. However, there exist non-regular, unbounded $L(G_{0})$ for which $L(G) = L(G_{0})$ is decidable.