### Update post with tactic-based proof.

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Danila Fedorin 1 year ago
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3 changed files with 272 additions and 147 deletions
1. 2
code/aoc-2020
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code/aoc-coq/day1.v
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content/blog/01_aoc_coq.md

#### 2 code/aoc-2020

 @ -1 +1 @@ Subproject commit 661bcbb557bfd7c228341b2430efd5aa21de3878 Subproject commit 2b69cbd391398ed21ccf3b2b06d62dc46207fe85

#### 102 code/aoc-coq/day1.v View File

 @ -1,102 +0,0 @@ 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.

#### 315 content/blog/01_aoc_coq.md View File

 @ -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?