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. Otherwise, if it has rules of the form $S \rightarrow 1^aS1^b$ such that $a &gt; b$, and rules of the form $S \rightarrow 1^cS1^d$ such that $c &lt; d$, then accept... But is there a simpler way to solve this problem?
Maybe you're confusing two different problems.
The algorithm you are describing shows that the problem of testing whether a CFG generates some string from $1^∗$ is decidable (e.g., see here at page 21). Anyway, the problem of testing whether a CFG generates all the strings of the language $1^∗$ is truly decidable. Consider all production rules to be simple (I mean, no &quot;$|$&quot; on the right), and eliminate all productions including a teminal character different from 1 on the right. Now you have a CFG over a terminal alphabet of a single letter, and then it generates a regular language (see here). All that remains is to test if two regular languages are equal. Moreover, there are effective procedures to convert a CFG over a one letter (terminal) alphabet into a regular grammar (e.g., see here).

I was forgetting the last part of your question: the fact that it is undecidable whether $R\subseteq G$, where $R$ is a regular language, and $G$ is a CFG, does not implies that for some specific regular languages you can not decide the inclusion. For example, if $R$ is finite (possibly empty), then testing for $R\subseteq G$ is decidable (see the first link above).