0 votes 0 votes Let $L_1$ be the language $L_1 =$ {$a^nb^mc^k : n = m$ or $m ≤ k$} and $L_2$ the language $L_2 =$ {$a^nb^mc^k : n + 2m = k$}. Show that $L_1 ∪ L_2$ is a context-free language. Theory of Computation peter-linz peter-linz-edition4 theory-of-computation context-free-grammar context-free-language + – Naveen Kumar 3 asked Apr 14, 2019 Naveen Kumar 3 289 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes S→ AB | CDE | F A→aAb | λ B→ Bc | λ C→ aC | λ D→bDc | λ E→ Ec | λ F→aFc | G G→bGcc | λ JAINchiNMay answered Oct 11, 2020 JAINchiNMay comment Share Follow See all 0 reply Please log in or register to add a comment.