Exercise 10:

{gCFG({a,b})L(g)={a,b}}{G1,G2L(G1)L(G2)}\{g\in\mathtt{CFG}(\{a,b\})\mid\mathcal{L}(g)=\{a,b\}^*\}\quad\leq\quad\{\langle G_1,G_2\rangle\mid\mathcal{L}(G_1)\subseteq\mathcal{L}(G_2)\}
Reduce the universality problem on CFGs over {a,b}\{a,b\} to the inclusion problem between two CFGs, in order to prove that such problem is not semi-decidable (not recursively enumerable).
Authors: Guillem Godoy / Documentation:
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