for $ L1$ regular language, $ L2$ some language, $ L1$ \ $ L2$ is regular, decidable. yet, the next transition function might not even be computable:

$ F′=$ {$ \,q:δ(q,w)∈F \, for\, some\, w∈ L2\,$ }$ .$

$ HALT\,=$ {$ <M,w>|M\,is\, a\, TM\, and\, M\, halts\, on\, w\,$ }

$ on (M,w):\, return\, 1\, is\, M\, halts\, on\, w\, \, return\, 0\, else\, $

this function is also well defined, but not computable. We know that the halting problem isn’t decidable. what is the difference? In both cases the functions are well defined, but the first case is decidable, and the second isn’t.