Functional induction

Alternating lists, the manual way

We begin with a function that picks elements from two alternating input lists:

It seems to be doing what it should:

["red", "apple", "green", "banana", "blue", "yellow"]

Now let us try to prove a simple theorem about alternate, namely that the length of the resulting list is the sum of the argument lists' lengths.

The alternate function is not structurally recursive, so using induction xs is not going to be helpful (try it if you are curious how it fails). Instead, we can attempt to write a recursive theorem: we use the theorem we are in the process of proving in the proof, and then rely on Lean's termination checker to check that this is actually ok.

So a first attempt at the proof might start by unfolding the definition of alternate. This exposes a match expression, and our Lean muscle memory tells us to proceed using split. The first case is now trivial. In the second case we use the next tactic to name the new variables. Then we call the theorem we are currently proving on the smaller lists ys and xs, following the pattern of alternate itself, and after some simplification and arithmetic, this goal is closed. Lean will tell us that it cannot prove termination of the recursive call, so we help it by giving the same termination argument as we gave to alternate itself.

This works, but is not quite satisfactory. The split tactic is rather powerful, but having to name the variables with next is tedious. A more elegant variant is to use the same pattern matching in the proof as we do in the function definition:

Note that we no longer unfold the function definition, but let the simplifier reduce the function application in the respective cases.

This style brings us closer to the function to be defined, but we still have to repeat the (possibly complex) pattern match of the original function, and the (possibly complex) termination argument and proof. Moreover, we have to explicitly instantiate the recursive invocation of the theorem, or hope that simp will only use it at the arguments that we want it to.

This is not very satisfying. The repetition makes the proof brittle, as a change to the function definition may require applying the same change to every single proof. We already explained the structure of the function to Lean once, we should not have to do it again!

Alternating lists, with functional induction

This is where the functional induction theorem comes in. For every recursive function, Lean defines a functional induction principle; just append .induct to the name. In our example, it has the following type:

Seeing an induction principle in its full glory can be a bit intimidating, so let us spell it out in prose:

• Given a property of two lists that we want to prove (the motive),

• if it holds when the first list is empty (the minor premise case1, corresponding to the first branch of alternate),

• and if it holds when the first list is non-empty (the minor premise case2, corresponding to the second branch of alternate), where we may assume that it holds for the second list and the tail of the first list (the induction hypothesis, corresponding to the recursive call in alternate),

• then given two arbitrary lists (the targets xs and xy, which correspond to the varying parameters of the function)

• the property holds for these lists.

An induction principle like this is used every time you use the induction tactic; for example induction on natural numbers uses Nat.recAux.

With this bespoke induction principle, we can now use the induction tactic in our proof:

This looks still quite similar to the previous iteration, but note the absence of a termination argument. Furthermore, just like with normal induction, we were provided the induction hypothesis already specialized at the right argument, as you can see in the proof state after case2.

With all this boilerplate out of the way, chances are good that we can now solve both cases with just proof automation tactics, leading to the following two-line proof:

The * in simp's argument list indicates that simp should also consider local assumptions for rewriting; this way it picks up the induction hypothesis.

In terms of proof automation, this is almost as good as it can get: We state the theorem and the induction and leave the rest to Lean's automation. However, just in case you prefer explicit readable and educational proofs, we can of course also spell it out explicitly:

Functions with fixed parameters

The above example demonstrates how to use functional induction in the common simple case. We now look into some more tricky aspects of it.

The first arises if the the recursive function has fixed parameters, as in the following function, which cuts up a list into sublists of length at most n:

Note that in all the recursive calls, the parameter n remains the same, whereas the parameter xs varies. Lean distinguishes between a prefix of fixed parameters, which are identical in all recursive calls, and then the varying parameters, which can change.

Let's first see the function in action:

[["😏", "😄", "😊"], ["😎", "😍", "😘"], ["🥰", "😃", "😉"], ["😋", "😆"]]

Joining these lists back together should give the original list. If we try to use functional induction in the same way as before, we find that we have more goals than we expect. In addition to the expected case1 and case2, we also get a subgoal called n:

unsolved goals
case n
α : Type u_1
n : Nat
⊢ Nat

case case1
α : Type u_1
n : Nat
⊢ (cutN n []).join = []

case case2
α : Type u_1
n : Nat
x✝ : α
xs✝ : List α
ih1✝ : (cutN n (List.drop (?n - 1) xs✝)).join = List.drop (?n - 1) xs✝
⊢ (cutN n (x✝ :: xs✝)).join = x✝ :: xs✝

This is not really a “case”, but the induction tactic does not know that. If we look at the type of cutN.induct, we see that – just like cutN itself - it has a parameter called n, which is hardly surprising:

Note that the motive depends only on the varying parameter of cutN, namely xs, whereas the fixed parameter of cutN, namely n, did not turn into a target of the induction principle, but merely a regular parameter.

So because n is not a target, we cannot write induction n, xs. Instead, we have to instantiate this parameter in the expression passed to using, and now the proof turns out rather nice:

We also could have used the syntax cutN.induct (n := n) if we prefer named arguments.

Overlapping patterns

So far we only defined functions with non-overlapping patterns. Lean also supports defining functions with overlapping patterns, and here functional induction can be particularly helpful.

Our (contrived) example function is one that removes repeated elements from a list of Boolean values:

The first three patterns of the function definition do not overlap, so they can be read as theorems that hold unconditionally. Lean generates such theorems, for use with rw and simp:

But the fourth equation is not a theorem. In general, it is not true that destutter (x :: xs) = x :: destutter xs holds. This equation is only true when the previous cases did not apply. Lean expresses this by preconditions on the equational lemma:

One rather annoying consequence is that if we try to construct a proof about destutter by repeating the same pattern matches, we cannot make progress:

In the last case of the match, Lean does not know that the previous cases did not apply, so it cannot use destutter.eq_4 and thus cannot simplify the goal.

One way out is to resort to the unfold destutter; split idiom. Functional induction also covers this case. The induction theorem produced by Lean provides the assumptions needed by the equational theorem in the appropriate case:

With these assumptions in the context, simp makes progress, and the proof becomes quite short:

You might be surprised that simp [destutter] suffices here, and we did not have to explicitly write simp [destutter, *] to tell simp that it should also use facts in the local context. This is due to special handling in simp: if rewrite rules have premises that look like like the premise of a function’s equational lemma, simp will consult the local context anyways.

This special handling does not exist in rewrite and rw, so if we use these tactics to unfold the function, we will get new goals with these side conditions, and have to resolve them explicitly, for example using assumption.

Mutual recursion

Proofs about mutually defined functions can become quite tricky: following the rule that the proof should follow the structure of the function, the usual approach is to prove two mutually recursive theorems, replicating the case splits of the original functions by matching or using split. Again, functional induction can help us avoid this repetition.

We will use a minimal example of mutually defined functions, even and odd:

The functional induction principle generated for one of these functions looks as follows:

This theorem now has two motives, one for each function in the recursive group, and four cases, corresponding to the two cases of even and then two cases of odd.

Recall that the role of the motive is to specify the statement that we want to prove about the parameters of the recursive function. If we have two mutually recursive functions, we typically have two such statements in mind that we want to prove together, one for each of the two functions.

One of the main benefits of the induction tactic, over simply trying to apply an induction principle, is that the induction tactic constructs the motive from the current proof goal. If the proof goal is P x y then induction x, y sets (motive := fun x y => P x y). (There is a bit more to it when there are assumptions about x or y in the local context, or the generalizing clause is used.)

So if the induction principle expects multiple motives, only of them can be inferred from the goal, and we will have to instantiate the other motive explicitly. So to prove that all numbers of the form 2n are even, were we want to use that all numbers of the form 2n + 1 are odd, we can write:

We can of course also prove the statement about odd. Now the induction tactic infers motive2 and we have to give motive1 explicitly:

If you compare even.induct and odd.induct, or inspect the proof state after induction, you will notice that we get the same proof obligations in both proofs, and we had to perform the same proofs twice to get both theorems. In the example above, where the proofs are rather trivial, this is not so bad, but in general we would be quite unsatisfied by that repetition. In that case, we can use the mutual functional induction principle:

This induction principle proves both motive1 and motive2. Unfortunately, this kind of induction principle can no longer be used by induction, so we have to manually prepare the goal to be of the right shape to be able to apply the theorem, essentially constructing both motives manually, and then project out the components:

We plan to provide commands to perform such a joint declaration of two theorems with less repetition and more convenience in the future.

Unrelated to mutual induction, but worth pointing out at this point: in this section we were able to solve all subgoals using simp_all [even, odd]. This high level of automation means that our proofs are quite stable under changes to the function definitions, and indeed, if we changed the definition of odd in

all of our proofs still go through, even though the number and shape of the cases changed!

What is a branch?

When Lean defines the functional induction principle, it has to decide what constitutes a branch of the function, and should therefore become a case of the induction principle. This is necessarily a heuristic, and is guided by the notion of tail position: Lean splits match statements and if-then-else statements that occur in the body of the function or in a case of such statements, but not when they appear, for example, in arguments to functions.

As a slight variant of cutN, consider a function that inserts a separator before, between and after groups of n:

["🐕", "😏", "😄", "😊", "🐕", "😎", "😍", "😘", "🐕", "🥰", "😃", "😉", "🐕", "😋", "😆", "🐕"]

Instead of using pattern matching as with cutN, let us try to use if-then-else:

Because we use if-then-else in a tail position the induction principle has two goals, one for each of the two branches:

We could also have defined the function slightly differently, moving the if-then-else into the (sep :: ·), as follows:

Now we get an induction principle with just one case:

Also note that the induction hypothesis in this case now has a condition ¬x.length = 0. This is to be expected: The corresponding recursive call is in the else-branch, and if x.length = 0 we may not assume that the induction hypothesis holds.

Hurdle: Looping simp

At this point we should discuss an issue with simp and certain recursive function definitions that can lead to simp [foo] looping. This is relevant because unfolding the function’s definition with simp is often the first thing we do in the case of a functional induction.

Consider the following proof that relates sepWith and cutN

We have two recursive functions, each with their own induction principle. We chose to use cutN's induction principle, because those generated from pattern matching tend to be easier to work with than those generated from if-then-else, and indeed we can conclude the proof.

You might wonder why we wrote

    rw [sepWith]; simp_all [cutN, List.join_cons]

and not

    simp_all [sepWith, cutN, List.join_cons]

The latter would lead to an error:

tactic 'simp' failed, nested error:
maximum recursion depth has been reached (use `set_option maxRecDepth <num>` to increase limit)

The problem is that the equational theorem generated for sepWith matches any call to sepWith, including the one on its own right hand side, and that sends simp into a loop:

The solution is to unfold such functions more carefully using rw (or possibly unfold). We hope that we can improve Lean's behaviour in this regard in the future.

Hurdle: Complex parameters

We saw that when a function like unstutter is defined with overlapping patterns, its equational theorems have preconditions, such as

These side-conditions are sometimes hard for simp to discharge. Consider the following theorem that says that destutter commutes with List.map not. This is obviously true, and we'd hope that the straight-forward proof

would work, but it fails with

unsolved goals
case case4
x✝² : Bool
xs✝ : List Bool
x✝¹ : ∀ (xs_1 : List Bool), x✝² = true → xs✝ = true :: xs_1 → False
x✝ : ∀ (xs_1 : List Bool), x✝² = false → xs✝ = false :: xs_1 → False
ih1✝ : List.map not (destutter xs✝) = destutter (List.map not xs✝)
⊢ (!x✝²) :: destutter (List.map not xs✝) = destutter ((!x✝²) :: List.map not xs✝)

It is stuck in the fourth case, and we can see that the destutter call on the left hand side has been simplified, but not the one on the right-hand side. The simplifier is not able to prove the preconditions of destutter.eq_4 for the arguments !x✝² and List.map not xs✝ from the information present in the goal. More powerful proof automation might handle that in the future, but for now we have to work around this issue.

One solution is to use rw to apply the equational theorem and manually discharge the side-conditions: