0 votes 0 votes Let $A\varepsilon_{CFG} = \{ \langle{ G }\rangle \mid G\: \text{is a CFG that generates}\: \epsilon \}.$Show that $A\varepsilon_{CFG}$ is decidable. Theory of Computation michael-sipser theory-of-computation turing-machine context-free-grammar decidability proof + – admin asked Oct 15, 2019 • edited Oct 15, 2019 by Lakshman Bhaiya admin 193 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.