Using polymorphic functions in definitions
Following my question here, I have several functions with different types of arguments which I defined the Inductive type formula on them. Is there anyway to use Inductive formula in compute_formula. I am doing this to make proving easier by decreasing the number of constructors that I have to handle in proofs. Thank you.
Fixpoint add (n:type1) (m:type2): type3 :=
match n with
(*body for add*)
end.
Fixpoint mul (n:type1) (m:type4): type5 :=
match n with
(*body for mul*)
end.
Inductive formula : Type :=
| Formula {A B}: type1-> A -> (type1->A->B) -> formula.
(* How should I write this *)
Definition compute_formula {A B} (f: formula) (extraArg:A) : B :=
match f with
|Formula {A B} part1 part2 part3=>
if (A isof type2 && B isof type3) then add part1 part2+extraArg
if (A isof type4 && B isof type5) then mul part1 part2+extraArg
end.
coq
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
add a comment |
Following my question here, I have several functions with different types of arguments which I defined the Inductive type formula on them. Is there anyway to use Inductive formula in compute_formula. I am doing this to make proving easier by decreasing the number of constructors that I have to handle in proofs. Thank you.
Fixpoint add (n:type1) (m:type2): type3 :=
match n with
(*body for add*)
end.
Fixpoint mul (n:type1) (m:type4): type5 :=
match n with
(*body for mul*)
end.
Inductive formula : Type :=
| Formula {A B}: type1-> A -> (type1->A->B) -> formula.
(* How should I write this *)
Definition compute_formula {A B} (f: formula) (extraArg:A) : B :=
match f with
|Formula {A B} part1 part2 part3=>
if (A isof type2 && B isof type3) then add part1 part2+extraArg
if (A isof type4 && B isof type5) then mul part1 part2+extraArg
end.
coq
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
add a comment |
Following my question here, I have several functions with different types of arguments which I defined the Inductive type formula on them. Is there anyway to use Inductive formula in compute_formula. I am doing this to make proving easier by decreasing the number of constructors that I have to handle in proofs. Thank you.
Fixpoint add (n:type1) (m:type2): type3 :=
match n with
(*body for add*)
end.
Fixpoint mul (n:type1) (m:type4): type5 :=
match n with
(*body for mul*)
end.
Inductive formula : Type :=
| Formula {A B}: type1-> A -> (type1->A->B) -> formula.
(* How should I write this *)
Definition compute_formula {A B} (f: formula) (extraArg:A) : B :=
match f with
|Formula {A B} part1 part2 part3=>
if (A isof type2 && B isof type3) then add part1 part2+extraArg
if (A isof type4 && B isof type5) then mul part1 part2+extraArg
end.
coq
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
Following my question here, I have several functions with different types of arguments which I defined the Inductive type formula on them. Is there anyway to use Inductive formula in compute_formula. I am doing this to make proving easier by decreasing the number of constructors that I have to handle in proofs. Thank you.
Fixpoint add (n:type1) (m:type2): type3 :=
match n with
(*body for add*)
end.
Fixpoint mul (n:type1) (m:type4): type5 :=
match n with
(*body for mul*)
end.
Inductive formula : Type :=
| Formula {A B}: type1-> A -> (type1->A->B) -> formula.
(* How should I write this *)
Definition compute_formula {A B} (f: formula) (extraArg:A) : B :=
match f with
|Formula {A B} part1 part2 part3=>
if (A isof type2 && B isof type3) then add part1 part2+extraArg
if (A isof type4 && B isof type5) then mul part1 part2+extraArg
end.
coq
coq
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
edited Nov 22 '18 at 20:12
Tom And.
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
asked Nov 22 '18 at 19:52
Tom And.Tom And.
736
736
add a comment |
add a comment |
1 Answer
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What do you want the output type of compute_formula to be? The way the signature is written, the function would have to be able to compute an element of B no matter what B is. Since this is obviously impossible (what if B is Empty?), I'll show you a different approach.
The idea is to use the formula to get the output type.
Definition output_type (f: formula) :=
match f with
| @Formula _ B _ _ _ => B
end.
Then we can define compute_formula as
Definition compute_formula (f: formula): output_type f :=
match f with
| @Formula _ _ t a func => func t a
end.
A few other things. I'm not sure what you mean with the extraArg part. If you're more specific about what that means I might be able to help you. Also, there isn't (at least outside of tactics) a way to do what you want with A isof type2.
answered Nov 23 '18 at 23:42
UserUser
61558
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
What do you want the output type of compute_formula to be? The way the signature is written, the function would have to be able to compute an element of B no matter what B is. Since this is obviously impossible (what if B is Empty?), I'll show you a different approach.
The idea is to use the formula to get the output type.
Definition output_type (f: formula) :=
match f with
| @Formula _ B _ _ _ => B
end.
Then we can define compute_formula as
Definition compute_formula (f: formula): output_type f :=
match f with
| @Formula _ _ t a func => func t a
end.
A few other things. I'm not sure what you mean with the extraArg part. If you're more specific about what that means I might be able to help you. Also, there isn't (at least outside of tactics) a way to do what you want with A isof type2.
answered Nov 23 '18 at 23:42
UserUser
61558
add a comment |
What do you want the output type of compute_formula to be? The way the signature is written, the function would have to be able to compute an element of B no matter what B is. Since this is obviously impossible (what if B is Empty?), I'll show you a different approach.
The idea is to use the formula to get the output type.
Definition output_type (f: formula) :=
match f with
| @Formula _ B _ _ _ => B
end.
Then we can define compute_formula as
Definition compute_formula (f: formula): output_type f :=
match f with
| @Formula _ _ t a func => func t a
end.
A few other things. I'm not sure what you mean with the extraArg part. If you're more specific about what that means I might be able to help you. Also, there isn't (at least outside of tactics) a way to do what you want with A isof type2.
answered Nov 23 '18 at 23:42
UserUser
61558
add a comment |
What do you want the output type of compute_formula to be? The way the signature is written, the function would have to be able to compute an element of B no matter what B is. Since this is obviously impossible (what if B is Empty?), I'll show you a different approach.
The idea is to use the formula to get the output type.
Definition output_type (f: formula) :=
match f with
| @Formula _ B _ _ _ => B
end.
Then we can define compute_formula as
Definition compute_formula (f: formula): output_type f :=
match f with
| @Formula _ _ t a func => func t a
end.
A few other things. I'm not sure what you mean with the extraArg part. If you're more specific about what that means I might be able to help you. Also, there isn't (at least outside of tactics) a way to do what you want with A isof type2.
answered Nov 23 '18 at 23:42
UserUser
61558
What do you want the output type of compute_formula to be? The way the signature is written, the function would have to be able to compute an element of B no matter what B is. Since this is obviously impossible (what if B is Empty?), I'll show you a different approach.
The idea is to use the formula to get the output type.
Definition output_type (f: formula) :=
match f with
| @Formula _ B _ _ _ => B
end.
Then we can define compute_formula as
Definition compute_formula (f: formula): output_type f :=
match f with
| @Formula _ _ t a func => func t a
end.
A few other things. I'm not sure what you mean with the extraArg part. If you're more specific about what that means I might be able to help you. Also, there isn't (at least outside of tactics) a way to do what you want with A isof type2.
answered Nov 23 '18 at 23:42
UserUser
61558
answered Nov 23 '18 at 23:42
UserUser
61558
answered Nov 23 '18 at 23:42
UserUser
61558
answered Nov 23 '18 at 23:42
UserUser
61558
61558
add a comment |
add a comment |
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