Is operational really isomorphic to a free monad?

Question

Proofs

In this blog post, Tekmo makes the point that we can prove that ExitSuccess exits because (I presume) it\'s like the Const functor for

Answer 1

The answer is yes, but only if you use a different translation of TeletypeF:

data TeletypeI a where
    PutStrLn :: String -> TeletypeI ()
    GetLine :: TeletypeI String
    ExitSuccess :: TeletypeI Void

The argument of TeletypeI is what the operation will/must provide to the rest of the program. It is the type of the argument of the continuation k in

eval (ExitSuccess :>>= k) = ...

Since there are no values of type Void, we can be sure that k will never be called. (As always we'll have to ignore undefined here.)

An equivalent type is:

data TeletypeI a where
    PutStrLn :: String -> TeletypeI ()
    GetLine :: TeletypeI String
    ExitSuccess :: TeletypeI a

Now we would have to provide a value to k that matches any type, and we can't do that either. This can be more practical since singleton ExitSuccess now has the flexible type Program TeletypeI a.

Similarly, exitSuccess could be fixed by giving it the type IO Void, or IO a.