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Update post with tactic-based proof.

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Danila Fedorin 1 year ago
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  1. 2
      code/aoc-2020
  2. 102
      code/aoc-coq/day1.v
  3. 315
      content/blog/01_aoc_coq.md

2
code/aoc-2020

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Subproject commit 661bcbb557bfd7c228341b2430efd5aa21de3878
Subproject commit 2b69cbd391398ed21ccf3b2b06d62dc46207fe85

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Require Import Coq.Lists.List.
Require Import Omega.
Definition has_pair (t : nat) (is : list nat) : Prop :=
exists n1 n2 : nat, n1 <> n2 /\ In n1 is /\ In n2 is /\ n1 + n2 = t.
Fixpoint find_matching (is : list nat) (total : nat) (x : nat) : option nat :=
match is with
| nil => None
| cons y ys =>
if Nat.eqb (x + y) total
then Some y
else find_matching ys total x
end.
Fixpoint find_sum (is : list nat) (total : nat) : option (nat * nat) :=
match is with
| nil => None
| cons x xs =>
match find_matching xs total x with
| None => find_sum xs total (* Was buggy! *)
| Some y => Some (x, y)
end
end.
Lemma find_matching_correct : forall is k x y,
find_matching is k x = Some y -> x + y = k.
Proof.
intros is. induction is;
intros k x y Hev.
- simpl in Hev. inversion Hev.
- simpl in Hev. destruct (Nat.eqb (x+a) k) eqn:Heq.
+ injection Hev as H; subst.
apply EqNat.beq_nat_eq. auto.
+ apply IHis. assumption.
Qed.
Lemma find_matching_skip : forall k x y i is,
find_matching is k x = Some y -> find_matching (cons i is) k x = Some y.
Proof.
intros k x y i is Hsmall.
simpl. destruct (Nat.eqb (x+i) k) eqn:Heq.
- apply find_matching_correct in Hsmall.
symmetry in Heq. apply EqNat.beq_nat_eq in Heq.
assert (i = y). { omega. } rewrite H. reflexivity.
- assumption.
Qed.
Lemma find_matching_works : forall is k x y, In y is /\ x + y = k ->
find_matching is k x = Some y.
Proof.
intros is. induction is;
intros k x y [Hin Heq].
- inversion Hin.
- inversion Hin.
+ subst a. simpl. Search Nat.eqb.
destruct (Nat.eqb_spec (x+y) k).
* reflexivity.
* exfalso. apply n. assumption.
+ apply find_matching_skip. apply IHis.
split; assumption.
Qed.
Theorem find_sum_works :
forall k is, has_pair k is ->
exists x y, (find_sum is k = Some (x, y) /\ x + y = k).
Proof.
intros k is. generalize dependent k.
induction is; intros k [x' [y' [Hneq [Hinx [Hiny Hsum]]]]].
- (* is is empty. But x is in is! *)
inversion Hinx.
- (* is is not empty. *)
inversion Hinx.
+ (* x is the first element. *)
subst a. inversion Hiny.
* (* y is also the first element; but this is impossible! *)
exfalso. apply Hneq. apply H.
* (* y is somewhere in the rest of the list.
We've proven that we will find it! *)
exists x'. simpl.
erewrite find_matching_works.
{ exists y'. split. reflexivity. assumption. }
{ split; assumption. }
+ (* x is not the first element. *)
inversion Hiny.
* (* y is the first element,
so x is somewhere in the rest of the list.
Again, we've proven that we can find it. *)
subst a. exists y'. simpl.
erewrite find_matching_works.
{ exists x'. split. reflexivity. rewrite plus_comm. assumption. }
{ split. assumption. rewrite plus_comm. assumption. }
* (* y is not the first element, either.
Of course, there could be another matching pair
starting with a. Otherwise, the inductive hypothesis applies. *)
simpl. destruct (find_matching is k a) eqn:Hf.
{ exists a. exists n. split.
reflexivity.
apply find_matching_correct with is. assumption. }
{ apply IHis. unfold has_pair. exists x'. exists y'.
repeat split; assumption. }
Qed.

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@ -397,10 +397,7 @@ Among those is `t`, the type an integer in such an arbitrary implementation. We
will not make an assumption about how the integers are implemented, and simply
use this generic `t` from now on.
#### Semantics in Coq
Now that we've seen finite sets and vectors, it's time to use them to
encode our semantics in Coq. Let's start with jumps. Suppose we wanted to write a function that _does_ return a valid program
Now, suppose we wanted to write a function that _does_ return a valid program
counter after adding the offset to it. Since it's possible for this function to fail
(for instance, if the offset is very negative), it has to return `option (fin (S n))`.
That is, this function may either fail (returning `None`) or succeed, returning
@ -409,11 +406,93 @@ the function in Coq (again, don't worry too much about the definition):
{{< codelines "Coq" "aoc-2020/day8.v" 61 61 >}}
But earlier, didn't we say:
We will make use of this function when we define and verify our semantics.
Let's take a look at that next.
#### Semantics in Coq
Now that we've seen finite sets and vectors, it's time to use them to
encode our semantics in Coq. Before we do anything else, we need
to provide Coq definitions for the various components of our
language, much like what we did with `tinylang`. We can start with opcodes:
{{< codelines "Coq" "aoc-2020/day8.v" 20 23 >}}
Now we can define a few other parts of our language and semantics, namely
states, instructions and programs (which I called "inputs" since, we'll, they're
our puzzle input). A state is simply the 3-tuple of the program counter, the set
of valid program counters, and the accumulator. We write it as follows:
{{< codelines "Coq" "aoc-2020/day8.v" 33 33 >}}
The star `*` is used here to represent a [product type](https://en.wikipedia.org/wiki/Product_type)
rather than arithmetic multiplication. Our state type accepts an argument,
`n`, much like a finite natural number or a vector. In fact, this `n` is passed on
to the state's program counter and set types. Rightly, a state for a program
of length \\(n\\) will not be of the same type as a state for a program of length \\(n+1\\).
An instruction is also a tuple, but this time containing only two elements: the opcode and
the number. We write this as follows:
{{< codelines "Coq" "aoc-2020/day8.v" 36 36 >}}
Finally, we have to define the type of a program. This type will also be
indexed by `n`, the program's length. A program of length `n` is simply a
vector of instructions `inst` of length `n`. This leads to the following
definition:
{{< codelines "Coq" "aoc-2020/day8.v" 38 38 >}}
So far, so good! Finally, it's time to get started on the semantics themselves.
We begin with the inductive definition of \\((\\rightarrow_i)\\).
I think this is fairly straightforward. However, we do use
`t` instead of \\(n\\) from the rules, and we use `FS`
instead of \\(+1\\). Also, we make the formerly implicit
assumption that \\(c+n\\) is valid explicit, by
providing a proof that `valid_jump_t pc t = Some pc'`.
{{< codelines "Coq" "aoc-2020/day8.v" 103 110 >}}
Next, it will help us to combine the premises for a
"failed" and "ok" terminations into Coq data types.
This will make it easier for us to formulate a lemma later on.
Here are the definitions:
{{< codelines "Coq" "aoc-2020/day8.v" 112 117 >}}
Since all of out "termination" rules start and
end in the same state, there's no reason to
write that state twice. Thus, both `done`
and `stuck` only take the input `inp`,
and the state, which includes the accumulator
`acc`, set of allowed program counters `v`, and
the program counter at which the program came to an end.
When the program terminates successfully, this program
counter will be equal to the length of the program `n`,
so we use `nat_to_fin n`. On the other hand, if the program
terminates in as stuck state, it must be that it terminated
at a program counter that points to an instruction. Thus, this
program counter is actually a \\(\\text{Fin} \\; n\\), and not
a \\(\\text{Fin} \\ (n+1)\\), and is not in the set of allowed program counters.
We use the same "weakening" trick we saw earlier to represent
this.
Finally, we encode the three inference rules we came up with:
{{< codelines "Coq" "aoc-2020/day8.v" 119 126 >}}
Notice that we fused two of the premises in the last rule.
Instead of naming the instruction at the current program
counter and using it in another premise, we simply use
`nth inp pc`, which corresponds to \\(p[c]\\) in our
"paper" semantics.
Before we go on writing some actual proofs, we have
one more thing we have to address. Earlier, we said:
> All jumps will lead either to an instruction, or right to the end of a program.
To make Coq aware of this constraint, we'll have to formalize this notion. To
To make Coq aware of this constraint, we'll have to formalize it. To
start off, we'll define the notion of a "valid instruction", which is guaranteed
to keep the program counter in the correct range.
There are a couple of ways to do this, but we'll use yet another definition based
@ -423,7 +502,7 @@ is perfectly valid for a program with thousands of instructions, but if it occur
in a program with only 3 instructions, it will certainly lead to disaster. Specifically,
the validity of an instruction depends on the length of the program in which it resides,
and the program counter at which it's encountered.
Thus, we refine our idea of validity to "being valid for a program of length n at program counter f".
Thus, we refine our idea of validity to "being valid for a program of length \\(n\\) at program counter \\(f\\)".
For this, we can use the following two inference rules:
{{< latex >}}
@ -444,11 +523,12 @@ which, as we discussed above, is perfectly valid.
The second rule works for the other two instructions. It has an extra premise:
the result of `jump_valid_t` (written as \\(J_v\\)) has to be \\(\\text{Some} \\; c'\\),
that is, `jump_valid_t` must succeed. Now, if an instruction satisfies these validity
rules for a given program at a given program counter, evaluating it will always
result in a program counter that has a proper value.
that is, `jump_valid_t` must succeed. Note that we require this even for no-ops,
since it later turns out of the them may be a jump after all.
We encode this in Coq as follows:
We now have our validity rules. If an instruction satisfies them for a given program
and at a given program counter, evaluating it will always result in a program counter that has a proper value.
We encode the rules in Coq as follows:
{{< codelines "Coq" "aoc-2020/day8.v" 152 157 >}}
@ -470,44 +550,191 @@ encode this in Coq, too:
{{< codelines "Coq" "aoc-2020/day8.v" 160 161 >}}
In the above, we use `input n` to mean "a program of length `n`".
This is just an alias for `vect inst n`, a vector of instructions
of length `n`. In the above, `n` is made implicit where possible.
In the above, `n` is made implicit where possible.
Since \\(c\\) (called `pc` in the code) is of type \\(\\text{Fin} \\; n\\), there's no
need to write \\(n\\) _again_.
need to write \\(n\\) _again_. The curly braces tell Coq to infer that
argument where possible.
Finally, it's time to get started on the semantics themselves.
We start with the inductive definition of \\((\\rightarrow_i)\\).
I think this is fairly straightforward. We use
`t` instead of \\(n\\) from the rules, and we use `FS`
instead of \\(+1\\). Also, we make the formerly implicit
assumption that \\(c+n\\) is valid explicit, by
providing a proof that `valid_jump_t pc t = Some pc'`.
### Proving Termination
Here we go! It's finally time to make some claims about our
definitions. Who knows - maybe we wrote down total garbage!
We will be creating several related lemmas and theorems.
All of them share two common assumptions:
{{< codelines "Coq" "aoc-2020/day8.v" 103 110 >}}
* We have some valid program `inp` of length `n`.
* This program is a valid input, that is, `valid_input` holds for it.
There's no sense in arguing based on an invalid input program.
Next, it will help us to combine the premises for a
"failed" and "ok" terminations into Coq data types.
This will help us formulate a lemma later on. Here they are:
We represent these grouped assumptions by opening a Coq
`Section`, which we call `ValidInput`, and listing our assumptions:
{{< codelines "Coq" "aoc-2020/day8.v" 112 117 >}}
{{< codelines "Coq" "aoc-2020/day8.v" 163 166 >}}
Since all of out "termination" rules start and
end in the same state, there's no reason to
write that state twice. Thus, both `done`
and `stuck` only take the input `inp`,
and the state, which includes the accumulator
`acc`, set of allowed program counters `v`, and
the program counter at which the program came to an end.
When the program terminates successfully, this program
counter will be equal to the length of the program `n`,
so we use `nat_to_fin n`. On the other hand, if the program
terminates in as stuck state, it must be that it terminated
at a program counter that points to an instruction. Thus, this
program counter is actually a \\(\\text{Fin} \\; n\\), and not
a \\(\\text{Fin} \\ (n+1)\\) (we use the same "weakening" trick
we saw earlier), and is not in the set of allowed program counters.
We had to also explicitly mention the length `n` of our program.
From now on, the variables `n`, `inp`, and `Hv` will be
available to all of the proofs we write in this section.
The first proof is rather simple. The claim is:
Finally, we encode the three inference rules we came up with:
> For our valid program, at any program counter `pc`
and accumulator `acc`, there must exists another program
counter `pc'` and accumulator `acc'` such that the
instruction evaluation relation \\((\rightarrow_i)\\)
connects the two. That is, valid addresses aside,
we can always make a step.
{{< codelines "Coq" "aoc-2020/day8.v" 119 126 >}}
Here is this claim encoded in Coq:
{{< codelines "Coq" "aoc-2020/day8.v" 168 169 >}}
We start our proof by introducing all the relevant variables into
the global context. I've mentioned this when I wrote about
day 1, but here's the gist: the `intros` keyword takes
variables from a `forall`, and makes them concrete.
In short, `intros x` is very much like saying "suppose
we have an `x`", and going on with the proof.
{{< codelines "Coq" "aoc-2020/day8.v" 170 171 >}}
Here, we said "take any program counter `pc` and any
accumulator `acc`". Now what? Well, first of all,
we want to take a look at the instruction at the current
`pc`. We know that this instruction is a combination
of an opcode and a number, so we use `destruct` to get
access to both of these parts:
{{< codelines "Coq" "aoc-2020/day8.v" 172 172 >}}
Now, Coq reports the following proof state:
```
1 subgoal
n : nat
inp : input n
Hv : valid_input inp
pc : Fin.t n
acc : t
o : opcode
t0 : t
Hop : nth inp pc = (o, t0)
========================= (1 / 1)
exists (pc' : fin (S n)) (acc' : t),
step_noswap (o, t0) (pc, acc) (pc', acc')
```
We have some opcode `o`, and some associated number
`t0`, and we must show that there exist a `pc'`
and `acc'` to which we can move on. To prove
that something exists in Coq, we must provide
an instance of that "something". If we claim
that there exists a dog that's not a good boy,
we better have this elusive creature in hand.
In other words, proofs in Coq are [constructive](https://en.wikipedia.org/wiki/Constructive_proof).
Without knowing the kind of operation we're dealing with, we can't
say for sure how the step will proceed. Thus, we proceed by
case analysis on `o`.
{{< codelines "Coq" "aoc-2020/day8.v" 173 173 >}}
There are three possible cases we have to consider,
one for each type of instruction.
* If the instruction is \\(\\texttt{add}\\), we know
that `pc' = pc + 1` and `acc' = acc + t0`. That is,
the program counter is simply incremented, and the accumulator
is modified with the number part of the instruction.
* If the instruction is \\(\\texttt{nop}\\), the program
coutner will again be incremented (`pc' = pc + 1`),
but the accumulator will stay the same, so `acc' = acc`.
* If the instruction is \\(\\texttt{jmp}\\), things are
more complicated. We must rely on the assumption
that our input is valid, which tells us that adding
`t0` to our `pc` will result in `Some f`, and not `None`.
Given this, we have `pc' = f`, and `acc' = acc`.
This is how these three cases are translated to Coq:
{{< codelines "Coq" "aoc-2020/day8.v" 174 177 >}}
For the first two cases, we simply provide the
values we expect for `pc'` and `acc'`, and
apply the corresponding inference rule that
is satisfied by these values. For the third case, we have
to invoke `Hv`, the hypothesis that our input is valid.
In particular, we care about the instruction at `pc`,
so we use `specialize` to plug `pc` into the more general
hypothesis. We then replace `nth inp pc` with its known
value, `(jmp, t0)`. This tells us the following, in Coq's words:
```
Hv : valid_inst (jmp, t0) pc
```
That is, `(jmp, t0)` is a valid instruction at `pc`. Then, using
Coq's `inversion` tactic, we ask: how is this possible? There is
only one inference rule that gives us such a conclusion, and it is named `valid_inst_jmp`
in our Coq code. Since we have a proof that our `jmp` is valid,
it must mean that this rule was used. Furthermore, sicne this
rule requires that `valid_jump_t` evaluates to `Some f'`, we know
that this must be the case here! Coq now has adds the following
two lines to our proof state:
```
f' : fin (S n)
H0 : valid_jump_t pc t0 = Some f'
```
Finally, we specify, as mentioned earlier, that `pc' = f'` and `acc' = acc`.
As before, we apply the corresponding step rule for `jmp`. When it asks
for a proof that `valid_jump_t` produces a valid program counter,
we hand it `H0` using `apply H0`. And with that, Coq is happy!
Next, we prove a claim that a valid program can always do _something_,
and that something is one of three things:
* It can terminate in the "ok" state if the program counter
reaches the programs' end.
* It can terminate with an error if it's currently at a program
counter that is not included in the valid set.
* Otherwise, it can run the current instruction and advance
to a "next" state.
Alternatively, we could say that one of the inference rules
for \\((\\Rightarrow_p)\\) must apply. This is not the case if the input
is not valid, since, as I said
before, an arbitrary input program can lead us to jump
to a negative address (or to an address _way_ past the end of the program).
Here's the claim, translated to Coq:
{{< codelines "Coq" "aoc-2020/day8.v" 181 186 >}}
Informally, we can prove this as follows:
* If the current program counter is equal to the length
of the program, we've reached the end. Thus, the program
can terminate in the "ok" state.
* Otherwise, the current program counter must be
less than the length of the program.
* If we've already encountered this program counter (that is,
if it's gone from the set of valid program counters),
then the program will terminate in the "error" state.
* Otherwise, the program counter is in the set of
valid instructions. By our earlier theorem, in a valid
program, the instruction at any program counter can be correctly
executed, taking us to the next state. Now too
our program can move to this next state.
Below is the Coq translation of the above.
{{< codelines "Coq" "aoc-2020/day8.v" 187 203 >}}
It doesn't seem like we're that far from being done now.
A program can always take a step, and each time it does,
the set of valid program counters decreases in size. Eventually,
this set will become empty, so if nothing else, our program will
eventually terminate in an "error" state. Thus, it will stop
running no matter what.
The problem is: how do we express that?

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