Finding a “free theorem”
How do I derive the free theorem for the type:
data F a = C1 Nat | C2 Bool Nat a
where Nat is simply data Nat = Z | S Nat?
In principle, this can be answered by the Haskell 'free-theorems' package, but it's too elderly to compile under any GHC version I can reasonably install.
haskell free-theorem
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
add a comment |
How do I derive the free theorem for the type:
data F a = C1 Nat | C2 Bool Nat a
where Nat is simply data Nat = Z | S Nat?
In principle, this can be answered by the Haskell 'free-theorems' package, but it's too elderly to compile under any GHC version I can reasonably install.
haskell free-theorem
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
2
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.fmap f = fmap fwherefmap :: (a->b)-> F a -> F bis the functor instance.
– chi
Nov 17 '18 at 14:39
add a comment |
How do I derive the free theorem for the type:
data F a = C1 Nat | C2 Bool Nat a
where Nat is simply data Nat = Z | S Nat?
In principle, this can be answered by the Haskell 'free-theorems' package, but it's too elderly to compile under any GHC version I can reasonably install.
haskell free-theorem
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
How do I derive the free theorem for the type:
data F a = C1 Nat | C2 Bool Nat a
where Nat is simply data Nat = Z | S Nat?
In principle, this can be answered by the Haskell 'free-theorems' package, but it's too elderly to compile under any GHC version I can reasonably install.
haskell free-theorem
haskell free-theorem
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
asked Nov 17 '18 at 13:58
NietzscheanAINietzscheanAI
431415
431415
2
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.fmap f = fmap fwherefmap :: (a->b)-> F a -> F bis the functor instance.
– chi
Nov 17 '18 at 14:39
add a comment |
2
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.fmap f = fmap fwherefmap :: (a->b)-> F a -> F bis the functor instance.
– chi
Nov 17 '18 at 14:39
2
2
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.
fmap f = fmap f where fmap :: (a->b)-> F a -> F b is the functor instance.– chi
Nov 17 '18 at 14:39
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.
fmap f = fmap f where fmap :: (a->b)-> F a -> F b is the functor instance.– chi
Nov 17 '18 at 14:39
add a comment |
1 Answer
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There is an online generator for free theorems at, and when it was down a little while ago I created an alternative UI that runs completely in the browser (using reflex-dom).
But the deeper problem is that free theorems, in the sense of these packages, are properties of polymorphic functions, so in order to answer your question, you have to give a function (like map) whose free theorem you are interested in.
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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votes
There is an online generator for free theorems at, and when it was down a little while ago I created an alternative UI that runs completely in the browser (using reflex-dom).
But the deeper problem is that free theorems, in the sense of these packages, are properties of polymorphic functions, so in order to answer your question, you have to give a function (like map) whose free theorem you are interested in.
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
add a comment |
There is an online generator for free theorems at, and when it was down a little while ago I created an alternative UI that runs completely in the browser (using reflex-dom).
But the deeper problem is that free theorems, in the sense of these packages, are properties of polymorphic functions, so in order to answer your question, you have to give a function (like map) whose free theorem you are interested in.
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
add a comment |
There is an online generator for free theorems at, and when it was down a little while ago I created an alternative UI that runs completely in the browser (using reflex-dom).
But the deeper problem is that free theorems, in the sense of these packages, are properties of polymorphic functions, so in order to answer your question, you have to give a function (like map) whose free theorem you are interested in.
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
There is an online generator for free theorems at, and when it was down a little while ago I created an alternative UI that runs completely in the browser (using reflex-dom).
But the deeper problem is that free theorems, in the sense of these packages, are properties of polymorphic functions, so in order to answer your question, you have to give a function (like map) whose free theorem you are interested in.
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
answered Nov 17 '18 at 14:40
Joachim BreitnerJoachim Breitner
20.4k563110
20.4k563110
add a comment |
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2
Free theorems are usually associated to polymorphic function types. Otherwise, IIRC, you get a trivial theorem, e.g.
fmap f = fmap fwherefmap :: (a->b)-> F a -> F bis the functor instance.– chi
Nov 17 '18 at 14:39