Coinduction Notes

Prepared for the WG-traits design meeting on 2019.11.08.

The problem

See chalk#248 for details. The short version is that we fail to handle a case like this correctly, where Ci are all co-inductive goals:

C1 :- C2, C3.
C2 :- C1.

What happens is that we

• start to prove C1
• start to prove C2
• see a recursive attempt to prove C1, assume it is succesful
• consider C2 proved and cache this
• start to prove C3, fail
• consider C1 unproven

Now we incorrectly have a result that C2 is true – but that result was made on the assumption that C1 was true, and it was not.

Some other tricky cases to consider

Unification failures

One thing to consider is that even when we have "coinduction obligations" to prove, we have to remember their substitutions too:

C1(X) :- C2(Y), X = 22.
C2(X) :- C3(X), X = 44.
C3(X) :- C1(X), C2(X).

None of these predicates should be provable, because C1(X) and C2(X) don't hold for the same X.

If we're not careful, we might:

• start to prove C1
• start to prove C2
• start to prove C3, see the recursive calls to C1 and C2
  • maybe we wait to consider it proven until C1 and C2 complete

In this case, it's not enough that C1 and C2 are provable at all, they have to be provable for the same X.

Non-trivial self-cycles

C1(A) :- C1(B), B = 22, C2(A).
C2(44).

This case is not provable, even though the only cycle is C1(X) :- C1(Y) – but it turns out that X must not be 22. The catch is that while this might appear to be a trivial self-cycle, it is not!

Actually I have to think about the best way to handle this case, as my proposed solution doesn't quite cut it. It wouldn't be wrong but it seems not ideal. – Niko

Delayed trivial cycles

C1(A, B) :- C2(A, B), A = 22, B = 22.
C2(A, B) :- C1(B, A).

This should be provable, but the cycle from C2 to C1 is not immediately visible as a trivial cycle, at least if subgoals are solved in order.

Delayed trivial cycles, variant 2

C1(A, B) :- C2(A, B), A = 22.
C2(A, B) :- C1(B, A).

As above, here the only complete answer is C1(22, 22). This is because the C1, C2 cycle effectively guarantees equality.

Delayed trivial cycles, variant 3

C1(A, B) :- C1(B, A).

This is true for all A, B

Other cases?

Approach in existing PR

High-level idea

• When we encounter a co-inductive subgoal, we delay them in the current Strand
• When all subgoals have been tested, and there are remaining delayed co-inductive subgoals, this is propogated up, marking the current Strand as co-inductive
• When the co-inductive Strands reach the root table, we only then pursue an answer

Failing tests

• Tests were added for the tricky cases above:
  • First
  • Second
• Both are provable to have no solutions, but are returned as ambiguous

Niko's proposed solution

High-level idea

• We only consider a co-induction subgoal proven for trivial recursion – i.e., self-recursion where you have C1 :- C1.
• For non-trivial recursion, we propagate the co-inductive subgoal to the parent. This continues until it becomes trivial.

Implementation steps

Extend Answer in two ways.

Currently Answer has a "constrained substitution" that includes values for the table's substitution + region constraints:

struct Answer {
    constrained_subst: Canonical<ConstrainedSubst>,
    is_ambiguous: bool
}

struct ConstrainedSubst {
    substitution: Substitution,
    region_constraints: Vec<RegionConstraint>,
}

we would first extend ConstrainedSubst to include as yet unproven co-inductive subgoals (this might actually be better done as a new type):

struct ConstrainedSubst {
    substitution: Substitution,
    delayed_subgoals: Vec<Literal>,
    region_constraints: Vec<RegionConstraint>,
}

then we would extend Answer slightly as well so it can be "ok" or ambiguous, as today, but also an error case

enum AnswerMode {
    OK,
    Ambiguous,
    Error,
}

struct Answer {
    constrained_subst: Canonical<ConstrainedSubst>,
    mode: AnswerMode
}

We won't need this error case till later, so let's ignore it for now. (And in a way, we never need it.)

Deferring coinductive subgoals

When we encounter a co-inductive subgoal, we check if it is trivial cycle or not. A trivial cycle is a case like C1 :- C1. We can simply consider such cycles to be true (but note the distinction between a trivial cycle and a self-cycle – see the "non-trivial self-cycle" example above).

For non-trivial cycles, we will want to store the cycle to be validated later. To accommodate that, we extend ExClause to include a delayed_subgoals list as well. We can write this the same way SLG does, so Goal :- DelayedSubgoals | Subgoals

In our example, proving C2 :- C1 would result in adding C1 to the list of delayed subgoals.

When we reach the end of the list of subgoals, we can create an answer that contains the delayed subgoals. We don't have to add all the goals – we can check for those that are trivial self-cycles again and remove them (in some cases, something which was not trivial to start may have become trivial through later unifications, see Delayed Trivial Self-Cycle case). Note that we do have to add all non-trivial cycles, including non-trivial self-cycles – see the walkthrough of Non-trivial self-cycle variant 3.

So the answer to C2 would be

substitution: [] // no variables
delayed_subgoals: ["C1"]
region_constraints: []

We can denote this as C2 :- C1 |, to use SLG notation.

Incorporating an answer with deferred subgoals.

When a table gets back an answer that has deferred sub-goals, they get added to the current list of subgoals.

So e.g. in our case, we had a ExClause like:

C1 :- | C2, C3

and we get the answer C2 :- C1 |, so we would convert it to

C1 :- | C3, C1

i.e., we have added C1 to the list of goals to prove. When we go to prove C3, of course, we will fail – but it had succeeded, we would go on to prove C1 but encounter a trivial cycle and hence succeed.

Extending root answer

So we failed to prove C1, but we do have a (conditional) answer to C2 – C2 :- C1 |, even though C2 is unprovable. What happens if ensure_root_answer is invoked on C2?

What we have here is a conditional answer. We know that C1 must have ultimately resolved itself somehow (although it might not yet be proven). What we can do is create a strand in C2 to evaluate C1 again – if this strand succeeds, it can actually overwrite the C2 :- C1 | answer in place with C2 :- (i.e., an unconditional answer). This is just a refinement of what we had. If the strand fails, though, we'll want to remember the error.

I think when we get a new answer, we want it to overwrite the old answer in place, rather than create a new answer. This is valid because it's not a new answer, it's just a more refined form of the old answer (although note that it might have different substitutions and other details, see the "delayed trivial cycle" case).

In particular, it could be that the table already has a "complete" set of answers – i.e., somebody invoked ensure_answer(N) and got back None. We don't want to be adding new answers which would change the result of that call. It is a bit strange that we are changing the result of ensure_answer(i) for the current i, but then the result is the same answer, just a bit more elaborated.

The idea then would be to create a strand associated with this answer somehow (it doesn't, I don't think, live in the normal strand table; we probably have a separate "refinment strand" table). This strand has as its subgoals the delayed subgoals. It pursues them. This either results in an answer (which replaces the existing answer) or an error (in which case the existing answer is marked as error). This may require extending strand with an optional answer index that it should overwrite, or perhaps we thread it down as an argument to pursue_strand (optional because, in the normal mode, we are just appending a new answer).

(Question: What distinguishes root answer? Nothing – we could actually do this process for any answer, so long as the delayed subgoals are not to tables actively on the stack. This just happens to be trivially true for root answers. The key part though is that the answer must be registered in the table first before the refinement strand is created, see Delayed Self-Cycle Variant 3.)

This is complex, so let's walk through an example or two.

The original problem. When we finish solving C1, we leave C2 with a single answer C2 :- C1 |. If someone invokes ensure_root_answer(C2, 0), we would see the delayed literal and create a refinement strand for the answer: C2 :- | C1. We would pursue C1 and get back the successful answer. So the refinement strand would terminate and we can overwrite with the answer C2 :- |.

Delayed trivial self-cycle. Similar to above, but the answer is C2(?A, ?B) :- C1(?B, ?A) |. In other words, in the canonical answer, we have a (identity) substitution of [^0, ^1] and a delayed goal of C1(^1, ^0). The strand we create will find only one answer to C1, C1(22, 22), so we wind up with an answer C2(22, 22).

Handling error answers

We introduced the idea of an "error answer"…how do we handle that? It's fairly simple. If a strand encounters an error answer, it simply fails. Done. The outer search however needs to treat an error answer as basically a no-op – so e.g. the answer iterator should simply increment the error counter and move to the next answer.

Walk through: delayed trivial self cycle, variant 2

C1(A, B) :- C2(A, B), A = 22.
C2(A, B) :- C1(B, A).

• ensure_root_answer(C1(?A, ?B)) is invoked
  • We start solving C1(?A, ?B) with the ex-clause C1(?A, ?B) :- | C2(?A, ?B), ?A = 22
    • That starts solving C2(?A, ?B)
      • This gets an answer C2(?A, ?B) :- C1(?B, ?A) |
      • When answer is incorporated, we get C1(?A, ?B) :- | C1(?B, ?A), ?A = 22
    • C1(?B, ?A) is a non-trivial cycle, and so we get
      • C1(?A, ?B) :- C1(?B, ?A) | ?A = 22
    • Unification completes, leaving us with
      • C1(22, ?B) :- C1(?B, 22) |
    • This is a complete answer
  • ensure root answer attempts to refine this answer, creating a strand for C1(22, ?B) :- | C1(?B, 22)
    • This creates a table for C1(?B, 22) with ex-clause C1(?B, 22) :- | C2(?B, 22), ?B = 22
      • We start solving C2(?B, 22), which has ex-clause C2(?B, 22) :- C1(22, ?B)
        • This creates a table for C1(22, ?B), with ex-clause C1(22, ?B) :- C2(22, ?B), 22 = 22
          • This starts solving C2(22, ?B), which is a fresh table with ex-clause C2(22, ?B) :- C1(?B, 22)
            • This is a co-inductive cycle
            • So our answer is C2(22, ?B) :- C1(?B, 22) |
          • Incorporating this answer yields C1(22, ?B) :- 22 = 22, C1(?B, 22)
          • The unification constraint succeeds, leaving C1(22, ?B) :- C1(?B, 22)
          • Co-inductive cycle detected, so answer is
            • C1(22, ?B) :- C1(?B, 22) |
      • This answer is incorporated into C2, yielding the ex-clause
        • C2(?B, 22) :- C1(?B, 22)
      • Pursuing that sub-goal gives a co-inductive cycle, so our final answer is
        • C2(?B, 22) :- C1(?B, 22) |
    • This answer is incorporated, yielding ex-clause C1(?B, 22) :- | ?B = 22, C1(?B, 22)
    • Unification yields C1(22, 22) :- C1(22, 22)
    • Trivial self-cycle detected, so final answer is
      • C1(22, 22)
  • the answer for C1(?A, ?B) is thus updated to C1(22, 22)

Walk through: delayed trivial self cycle, variant 3

C1(A, B) :- C1(B, A).

This example is interesting because it shows that we have to incorporate non-trivial self cycles into an answer so they can recursively build on one another.

• we get an initial answer of
  • C1(?A, ?B) :- C1(?B, ?A) |
• if we attempt to refine this, we will get a strand C1(?X, ?Y) :- C1(?Y, ?X)
  • pursuing the first subgoal C1(?Y, ?X) leads us to our own table, but at answer 0
    • (the very answer we are refining)
    • the answer is C1(?Y, ?X) :- C1(?X, ?Y) |
  • this strand incorporates its own answer, yielding
    • C1(?X, ?Y) :- C1(?X, ?Y)
  • next subgoal is a trivial self-cycle, discard, yielding
    • C1(?X, ?Y) :-
• result: true

Walk through: non-trivial self cycle

Let's walk through one more case, the non-trivial self cycle.

C1(A) :- C1(B), B = 22, C2(A).
C2(44).

What happens here is that we get an initial answer from C1 that looks like:

C1(44) :- C1(22) |

Ensure root answer will thus try to refine by trying to solve C1(22). Interestingly, this is going to go to a distinct table, because the canonical form is not the same, but that table will just fail.