# CFG that generates $1^*$ is decidable

I read somewhere that the problem which asks whether or not a $$CFG$$ $$G$$ generates $$1^*$$ is decidable. The proof goes like this:

$$1^* \cap G$$ is context free since it is the intersection of a regular language and a $$CFG$$, therefore we can test if $$1^* \cap G$$ is empty since it is decidable to check if a $$CFG$$ is empty. If $$1^* \cap G$$ is empty, reject, otherwise, accept

I have doubts however with this proof since it only shows that some string in $$1^*$$ is generated by $$G$$, but not whether $$G$$ generates all strings in $$1^*$$.

Moreover, if this proof is correct, we can use the same proof outline under the alphabet $$\Sigma=\{0,1\}$$ to show that $$G$$ generates $$\Sigma^*$$, or that $$\Sigma^* \subseteq G$$. However, it is known that it is undecidable whether $$R \subseteq G$$, where $$R$$ is a regular language, and $$G$$ is a $$CFG$$ (by setting $$R=\Sigma^*$$, and $$\Sigma=\{0,1\}$$.

But to show that a $$CFG$$ generates $$1^*$$ is decidable, the only way I can think of is to use something similar to the proof that it is decidable for a $$PCP$$ instance to generate some string in $$1^*$$ (i.e., $$w=v$$, where $$w,v \in 1^*$$), i.e. we can check if the $$CFG$$ has rule $$S \rightarrow S1$$, then accept. O.w. if it has rules of the form $$S \rightarrow 1^aS1^b$$ such that $$a > b$$, and rules of the form $$S \rightarrow 1^cS1^d$$ such that $$c < d$$, then accept… But is there a simpler way to solve this problem ?