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$\{\langle g_1,g_2\rangle\in\mathtt{CFG}(\{a,b\})^2\mid\mathcal{L}(g_1)\cap\mathcal{L}(g_2)\neq\emptyset\}\quad\leq\quad\{\langle G_1,G_2,G_3\rangle\in\mathtt{CFG}(\{a,b\})^3\mid\mathcal{L}(G_1)\cap\mathcal{L}(G_2)\neq\emptyset\;\wedge\;\mathcal{L}(G_2)\cap\mathcal{L}(G_3)\neq\emptyset\;\wedge\;\mathcal{L}(G_1)\cap\mathcal{L}(G_3)=\emptyset\}$
Reduce the non-empty intersection problem on CFGs over $\{a,b\}$ to the set of
tuples of three CFGs $G_1,G_2,G_3$ whose alphabet is $\{a,b\}$ and such that
$G_1$ and $G_2$ generate at least one common word, $G_2$ and $G_3$ generate at
least one common word, but $G_1$ and $G_3$ do not generate a common word, in
order to prove that such set is undecidable (not recursive).
Authors: Guillem Godoy
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