3 3 votes $L= { <G>$ is a CFG & not ambiguous $}$ RE or not? We can have 2 TMs where $T_{yes}\subset T_{no}$ just by adding additional ambiguous productions to an existing non ambigous CFG. So it seems to me it is not RE, but given answer is RE. Where I am wrong? Theory of Computation theory-of-computation decidability + – Aghori 474 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.