Finding a “free theorem”
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2
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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 at 13:58
NietzscheanAI
424415
add a comment |
up vote
2
down vote
favorite
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 at 13:58
NietzscheanAI
424415
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 at 14:39
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 at 13:58
NietzscheanAI
424415
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 at 13:58
NietzscheanAI
424415
asked Nov 17 at 13:58
NietzscheanAI
424415
asked Nov 17 at 13:58
NietzscheanAI
424415
asked Nov 17 at 13:58
NietzscheanAI
424415
asked Nov 17 at 13:58
NietzscheanAI
424415
424415
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 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 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 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 at 14:39
add a comment |
1 Answer
1
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4
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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 at 14:40
Joachim Breitner
20.3k562109
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
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 at 14:40
Joachim Breitner
20.3k562109
add a comment |
up vote
4
down vote
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 at 14:40
Joachim Breitner
20.3k562109
add a comment |
up vote
4
down vote
up vote
4
down vote
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 at 14:40
Joachim Breitner
20.3k562109
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 at 14:40
Joachim Breitner
20.3k562109
answered Nov 17 at 14:40
Joachim Breitner
20.3k562109
answered Nov 17 at 14:40
Joachim Breitner
20.3k562109
answered Nov 17 at 14:40
Joachim Breitner
20.3k562109
20.3k562109
add a comment |
add a comment |
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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 at 14:39