In Interactive Theorem Proving and Program Development the authors explain constraints on constructors of inductive types in Coq.

For inductive type $T$, a constructor must have the form $t_1 \rightarrow t_2 \rightarrow ... \rightarrow t_l \rightarrow T a_1 ... a_k$ where $k$ is the number of arguments of the inductive type $T$.

There are some restrictions placed on the $t_i$s, which I cannot understand.

If $t_i$ is the type of a constant, then $t_i$ can have the form "$g(Tb_{1,1} ... b_{1,k})...(Tb_{l,1} ... b_{l,k}),$" provided the expressions $b_{i,j}$ also satisfy the typing rules, and $T$ does not occur in these expressions or in $g$.


If $t_i$ is the type of a function, this type can have the form $t_1'\rightarrow ... \rightarrow t'_m \rightarrow g(Tb_{1,1} ... b_{1,k})...(Tb_{l,1} ... b_{l,k})$ but the type $T$ cannot occur in the expressions $t'_1, ... t'_m$ or in the expressions $b_{i,j}$

I am confused on what it means for a type $t_i$ "to have the form of $g(Tb_{1,1} ... b_{1,k})...(Tb_{l,1} ... b_{l,k})$".

For instance, does nat (the type of natural numbers) have the form of $g(Tb_{1,1} ... b_{1,k})...(Tb_{l,1} ... b_{l,k})$"?

If so, how does nat obtain such a elaborate form? What is $g$ and what are the terms in the $(l,k)$ matrix $b_{i,j}$?

If not, am I prohibited from using nat in type constructors of inductive types? (That would seem absurd...)

Below I have uploaded a screenshot of the page in question.

Interactive Theorem Proving and Program Development pg 379

Update: I was able to construct a counterexample to the textbook:

Definition g := fun x y z : Type => nat.

Inductive T : Set :=
  t : (g T T T) -> T.

Here $l = 1, k = 0$, and $t_1$ can only have the form $g(T)$ but I was able to use $T$ 3 times inside an actual Coq environment.

Edit 2: I think what you guys are saying about the reduced form makes sense so I tried to make another "counterexample"

Inductive g (x y z: Type) : Set.

Inductive T : Set :=
  t : (g T T T) -> T.
  • 1
    $\begingroup$ I think Coq reduces g T T T to its normal form nat, and then checks the resulting t: nat -> T which complies with the rules. The book probably does not mention this normalization step. $\endgroup$
    – chi
    Oct 17, 2018 at 13:16

1 Answer 1


Take $l = k = 0$, the matrix of $b_{ij}$'s has size $0 \times 0$, hence we do not have to defined any $b_{ij}$, and take $g = \mathbf{nat}$.

  • $\begingroup$ Thanks for your answer @AndrejBauer but I don't think we can choose $l, k$ aribitrarily. In fact, they are related to the terms in the constructor. So $l = 1$ since the constructor would have one argument of type nat and $k$ is decided by the number of arguments to the inductive type $T$. $\endgroup$
    – Mark
    Oct 16, 2018 at 6:34
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    $\begingroup$ You misunderstand the textbook. The $g$ and the $b_{ij}$ are schematic, i.e., they show you the form of syntax that can be used, and are not meant to be actual Coq code. Your counter-example reduces to Inductive T : Set := t : nat -> T by the definition of g and shows nothing. You need to compare the reduced definition of T (i.e., with the definition of g unfolded) to the indicated forms in the textbook. But I doubt I can explain this point in a comment, unless you understand the difference between a meta-variable and a variable. $\endgroup$ Oct 17, 2018 at 15:03
  • $\begingroup$ To address your comment further: I did not "choose $l$, $k$ arbitrarily". You might be confused about how to quantify $l$ and $k$. If we want to verify that some particular expression $E$ has the form $g (T b_{1,1} \ldots b_{1,k}) \cdots (T b_{l,1} \ldots b_{l,k})$ then we need to find $l, k$, the $b_{ij}$'s and the $g$ so that we end up getting $E$. Example: take $l = 1$, $k = 2$, $b_{11} = a$, $b_{12} = a$, and $g(x) = \sqrt{1+x}$. (I purposely use strange things to emphasize that these are purely syntactic manipulations of expressions.) Then $g(T b_{11} b_{12}) = \sqrt{1 + T a a}$. $\endgroup$ Oct 17, 2018 at 15:09
  • $\begingroup$ Consider the case $l = k = 0$: then $g (T b_{1,1} \ldots b_{1,k}) \cdots (T b_{l,1} \ldots b_{l,k})$ is just $g$. Furthermore, consider $g = \mathbf{nat}$. Then we get just $\mathbf{nat}$. This is precisely what you asked for: what should we take for $l$, $k$, $b_{ij}$ and $g$ so that $g (T b_{1,1} \ldots b_{1,k}) \cdots (T b_{l,1} \ldots b_{l,k})$ becomes equal to $\mathbf{nat}$. I just answered your question. $\endgroup$ Oct 17, 2018 at 15:09
  • $\begingroup$ I think what you are saying about reduced definition makes sense. But I'm still confused about $l =0$. $g$ is the schematic of a type $t_i$ inside one of the arguments of a constructor for inductive type $T$. But this means that the constructor has at least one argument (namely $t_i$) and therefore $l$ (the number of $i$'s is greater than or equal to 1. $\endgroup$
    – Mark
    Oct 17, 2018 at 15:37

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