Emptiness of the intersection of two CFGs is undecidable

From the PCP instance build two context-free grammars over the alphabet consisting of the original symbols together with index symbols $1,\dots ,n$: $S_T\to u_i{S}_{T}i\mid u_{i}i,S_B\to v_i{S}_{B}i\mid v_{i}i(1\le i\le n)\mathrm{.}$ Thus $G_T$ generates exactly the words $u_{i_1}\cdots u_{i_k}{i}_k\cdots i_1$, and $G_B$ generates exactly the words $v_{i_1}\cdots v_{i_k}{i}_k\cdots i_1$, for nonempty index sequences $i_1,\dots ,i_k$.

If $i_1,\dots ,i_k$ is a PCP match, then $u_{i_1}\cdots u_{i_k}{i}_k\cdots i_1=v_{i_1}\cdots v_{i_k}{i}_k\cdots i_1,$ so that common word lies in $L(G_T)\cap L(G_B)$. Conversely, if a word lies in $L(G_T)\cap L(G_B)$, its reversed index suffix uniquely determines the same index sequence on both sides, and removing that suffix leaves $u_{i_1}\cdots u_{i_k}=v_{i_1}\cdots v_{i_k}\mathrm{.}$ Hence the original PCP instance has a match if and only if $L(G_T)\cap L(G_B)\mathrm{\ne }\mathrm{\varnothing }$.

The map from the PCP instance to the pair $(G_T,G_B)$ is total and computable by step 1.1. If intersection-emptiness for pairs of CFGs were decidable, step 2.1 and [L3] would make PCP decidable, contradicting [L2].

Therefore emptiness of the intersection of two CFGs is undecidable.