Add draft of the first portion of day 8 Coq writeup.
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title: "Advent of Code in Coq - Day 8"
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date: 2020-12-31T17:55:25-08:00
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tags: ["Advent of Code", "Coq"]
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draft: true
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---
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Huh? We're on day 8? What happened to days 2 through 7?
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Well, for the most part, I didn't think they were that interesting from the Coq point of view.
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Day 7 got close, but not close enough to inspire me to create a formalization. Day 8, on the other
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hand, is
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{{< sidenote "right" "pl-note" "quite interesting," >}}
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Especially to someone like me who's interested in programming languages!
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{{< /sidenote >}} and took quite some time to formalize.
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As before, here's an (abridged) description of the problem:
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> Given a tiny assembly-like language, determine the state of its accumulator
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> when the same instruction is executed twice.
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Before we start on the Coq formalization, let's talk about an idea from
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Programming Language Theory (PLT), _big step operational semantics_.
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### Big Step Operational Semantics
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What we have in Advent of Code's Day 8 is, undeniably, a small programming language.
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We are tasked with executing this language, or, in PLT lingo, defining its _semantics_.
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There are many ways of doing this - at university, I've been taught of [denotational](https://en.wikipedia.org/wiki/Denotational_semantics), [axiomatic](https://en.wikipedia.org/wiki/Axiomatic_semantics),
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and [operational](https://en.wikipedia.org/wiki/Operational_semantics) semantics.
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I believe that Coq's mechanism of inductive definitions lends itself very well
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to operational semantics, so we'll take that route. But even "operational semantics"
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doesn't refer to a concrete technique - we have a choice between small-step (structural) and
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big-step (natural) operational semantics. The former describe the minimal "steps" a program
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takes as it's being evaluated, while the latter define the final results of evaluating a program.
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I decided to go with big-step operational semantics, since they're more intutive (natural!).
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So, how does one go about "[defining] the final results of evaluating a program?" Most commonly,
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we go about using _inference rules_. Let's talk about those next.
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#### Inference Rules
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Inference rules are a very general notion. The describe how we can determine (infer) a conclusion
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from a set of assumption. It helps to look at an example. Here's a silly little inference rule:
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{{< latex >}}
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\frac
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{\text{I'm allergic to cats} \quad \text{My friend has a cat}}
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{\text{I will not visit my friend very much}}
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{{< /latex >}}
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It reads, "if I'm allergic to cats, and if my friend has a cat, then I will not visit my friend very much".
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Here, "I'm allergic to cats" and "my friend has a cat" are _premises_, and "I will not visit my friend very much" is
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a _conclusion_. An inference rule states that if all its premises are true, then its conclusion must be true.
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Here's another inference rule, this time with some mathematical notation instead of words:
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{{< latex >}}
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\frac
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{n < m}
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{n + 1 < m + 1}
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{{< /latex >}}
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This one reads, "if \\(n\\) is less than \\(m\\), then \\(n+1\\) is less than \\(m+1\\)". We can use inference
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rules to define various constructs. As an example, let's define what it means for a natural number to be even.
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It takes two rules:
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{{< latex >}}
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\frac
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{}
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{0 \; \text{is even}}
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\quad
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\frac
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{n \; \text{is even}}
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{n+2 \; \text{is even}}
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{{< /latex >}}
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First of all, zero is even. We take this as fact - there are no premises for the first rule, so they
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are all trivially true. Next, if a number is even, then adding 2 to that number results in another
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even number. Using the two of these rules together, we can correctly determine whether any number
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is or isn't even. We start knowing that 0 is even. Adding 2 we learn that 2 is even, and adding 2
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again we see that 4 is even, as well. We can continue this to determine that 6, 8, 10, and so on
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are even too. Never in this process will we visit the numbers 1 or 3 or 5, and that's good - they're not even!
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Let's now extend this notion to programming languages, starting with a simple arithmetic language.
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This language is made up of natural numbers and the \\(\square\\) operation, which represents the addition
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of two numbers. Again, we need two rules:
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{{< latex >}}
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\frac
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{n \in \mathbb{N}}
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{n \; \text{evaluates to} \; n}
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\quad
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\frac
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{e_1 \; \text{evaluates to} \; n_1 \quad e_2 \; \text{evaluates to} \; n_2}
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{e_1 \square e_2 \; \text{evaluates to} \; n_1 + n_2}
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{{< /latex >}}
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First, let me explain myself. I used \\(\square\\) to demonstrate two important points. First, languages can be made of
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any kind of characters we want; it's the rules that we define that give these languages meaning.
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Second, while \\(\square\\) is the addition operation _in our language_, \\(+\\) is the _mathematical addition operator_.
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They are not the same - we use the latter to define how the former works.
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Finally, writing "evaluates to" gets quite tedious, especially for complex languages. Instead,
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PLT people use notation to make their semantics more concise. The symbol \\(\Downarrow\\) is commonly
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used to mean "evaluates to"; thus, \\(e \Downarrow v\\) reads "the expression \\(e\\) evaluates to the value \\(v\\).
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Using this notation, our rules start to look like the following:
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{{< latex >}}
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\frac
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{n \in \mathbb{N}}
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{n \Downarrow n}
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\quad
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\frac
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{e_1 \Downarrow n_1 \quad e_2 \Downarrow n_2}
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{e_1 \square e_2 \Downarrow n_1 + n_2}
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{{< /latex >}}
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If nothing else, these are way more compact! Though these may look intimidating at first, it helps to
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simply read each symbol as its English meaning.
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#### Encoding Inference Rules in Coq
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Now that we've seen what inference rules are, we can take a look at how they can be represented in Coq.
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We can use Coq's `Inductive` mechanism to define the rules. Let's start with our "is even" property.
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```Coq
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Inductive is_even : nat -> Prop :=
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| zero_even : is_even 0
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| plustwo_even : is_even n -> is_even (n+2).
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```
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The first line declares the property `is_even`, which, given a natural number, returns proposition.
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This means that `is_even` is not a proposition itself, but `is_even 0`, `is_even 1`, and `is_even 2`
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are all propositions.
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The following two lines each encode one of our aforementioned inference rules. The first rule, `zero_even`,
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is of type `is_even 0`. The `zero_even` rule doesn't require any arguments, and we can use it to create
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a proof that 0 is even. On the other hand, the `plustwo_even` rule _does_ require an argument, `is_even n`.
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To construct a proof that a number `n+2` is even using `plustwo_even`, we need to provide a proof
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that `n` itself is even. From this definition we can see a general principle: we encode each inference
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rule as constructor of an inductive Coq type. Each rule encoded in this manner takes as arguments
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the proofs of its premises, and returns a proof of its conclusion.
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For another example, let's encode our simple addition language. First, we have to define the language
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itself:
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```Coq
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Inductive tinylang : Type :=
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| number (n : nat) : tinylang
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| box (e1 e2 : tinylang) : tinylang.
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```
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This defines the two elements of our example language: `number n` corresponds to \\(n\\), and `box e1 e2` corresponds
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to \\(e_1 \square e_2\\). Finally, we define the inference rules:
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```Coq {linenos=true}
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Inductive tinylang_sem : tinylang -> nat -> Prop :=
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| number_sem : forall (n : nat), tinylang_sem (number n) n
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| box_sem : forall (e1 e2 : tinylang) (n1 n2 : nat),
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tinylang_sem e1 n1 -> tinylang_sem e2 n2 ->
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tinylang_sem (box e1 e2) (n1 + n2).
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```
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When we wrote our rules earlier, by using arbitrary variables like \\(e_1\\) and \\(n_1\\), we implicitly meant
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that our rules work for _any_ number or expression. When writing Coq we have to make this assumption explicit
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by using `forall`. For instance, the rule on line 2 reads, "for any number `n`, the expression `n` evaluates to `n`".
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#### Semantics of Our Language
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We've now written some example big-step operational semantics, both "on paper" and in Coq. Now, it's time to take a look at
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the specific semantics of the language from Day 8! Our language consists of a few parts.
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First, there are three opcodes: \\(\texttt{jmp}\\), \\(\\texttt{nop}\\), and \\(\\texttt{add}\\). Opcodes, combined
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with an integer, make up an instruction. For example, the instruction \\(\\texttt{add} \\; 3\\) will increase the
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content of the accumulator by three. Finally, a program consists of a sequence of instructions; They're separated
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by newlines in the puzzle input, but we'll instead separate them by semicolons. For example, here's a complete program.
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{{< latex >}}
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\texttt{add} \; 0; \; \texttt{nop} \; 2; \; \texttt{jmp} \; -2
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{{< /latex >}}
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Now, let's try evaluating this program. Starting at the beginning and with 0 in the accumulator,
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it will add 0 to the accumulator (keeping it the same),
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do nothing, and finally jump back to the beginning. At this point, it will try to run the addition instruction again,
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which is not allowed; thus, the program will terminate.
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Did you catch that? The semantics of this language will require more information than just our program itself (which we'll denote \\(p\\)).
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* First, to evaluate the program we will need a program counter, \\(\\textit{c}\\). This program counter
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will tell us the position of the instruction to be executed next. It can also point past the last instruction,
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which means our program terminated successfully.
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* Next, we'll need the accumulator \\(a\\). Addition instructions can change the accumulator, and we will be interested
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in the number that ends up in the accumulator when our program finishes executing.
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* Finally, and more subtly, we'll need to keep track of the states we visited. For instance,
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in the course of evaluating our program above, we encounter the \\((c, a)\\) pair of \\((0, 0)\\) twice: once
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at the beginning, and once at the end. However, whereas at the beginning we have not yet encountered the addition
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instruction, at the end we have, so the evaluation behaves differently. To make the proofs work better in Coq,
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we'll use a set \\(v\\) of
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{{< sidenote "right" "allowed-note" "allowed (valid) program counters (as opposed to visited program counters)." >}}
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Whereas the set of "visited" program counters keeps growing as our evaluation continues,
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the set of "allowed" program counters keeps shrinking. Because the "allowed" set never stops shrinking,
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assuming we're starting with a finite set, our execution will eventually terminate.
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{{< /sidenote >}}
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Now we have all the elements of our evaluation. Let's define some notation. A program starts at some state,
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and terminates in another, possibly different state. In the course of a regular evaluation, the program
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never changes; only the state does. So I propose this (rather unorthodox) notation:
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{{< latex >}}
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(c, a, v) \Rightarrow_p (c', a', v')
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{{< /latex >}}
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This reads, "after starting at program counter \\(c\\), accumulator \\(a\\), and set of valid addresses \\(v\\),
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the program \\(p\\) terminates with program counter \\(c'\\), accumulator \\(a'\\), and set of valid addresses \\(v'\\)".
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Before creating the inference rules for this evaluation relation, let's define the effect of evaluating a single
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instruction, using notation \\((c, a) \rightarrow_i (c', a')\\). An addition instruction changes the accumulator,
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and increases the program counter by 1.
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{{< latex >}}
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\frac{}
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{(c, a) \rightarrow_{\texttt{add} \; n} (c+1, a+n)}
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{{< /latex >}}
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A no-op instruction does even less. All it does is increment the program counter.
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{{< latex >}}
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\frac{}
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{(c, a) \rightarrow_{\texttt{nop} \; n} (c+1, a)}
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{{< /latex >}}
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Finally, a jump instruction leaves the accumulator intact, but adds a number to the program counter itself!
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{{< latex >}}
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\frac{}
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{(c, a) \rightarrow_{\texttt{jmp} \; n} (c+n, a)}
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{{< /latex >}}
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None of these rules have any premises, and they really are quite simple. Now, let's define the rules
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for evaluating a program. First of all, a program starting in a state that is not considered "valid"
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is done evaluating, and is in a "failed" state.
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{{< latex >}}
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\frac{c \not \in v \quad c \not= \text{length}(p)}
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{(c, a, v) \Rightarrow_{p} (c, a, v)}
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{{< /latex >}}
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We use \\(\\text{length}(p)\\) to represent the number of instructions in \\(p\\). Note the second premise:
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even if our program counter \\(c\\) is not included in the valid set, if it's "past the end of the program",
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the program terminates in an "ok" state. Here's a rule for terminating in the "ok" state:
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{{< latex >}}
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\frac{c = \text{length}(p)}
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{(c, a, v) \Rightarrow_{p} (c, a, v)}
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{{< /latex >}}
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When our program counter reaches the end of the program, we are also done evaluating it. Even though
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both rules {{< sidenote "right" "redundant-note" "lead to the same conclusion," >}}
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In fact, if the end of the program is never included in the valid set, the second rule is completely redundant.
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{{< /sidenote >}}
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it helps to distinguish the two possible outcomes. Finally, if neither of the termination conditions are met,
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our program can take a step, and continue evaluating from there.
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{{< latex >}}
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\frac{c \in v \quad p[c] = i \quad (c, a) \rightarrow_i (c', a') \quad (c', a', v - \{c\}) \Rightarrow_p (c'', a'', v'')}
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{(c, a, v) \Rightarrow_{p} (c'', a'', v'')}
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{{< /latex >}}
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This is quite a rule. A lot of things need to work out for a program to evauate from a state that isn't
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currently the final state:
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* The current program counter \\(c\\) must be valid. That is, it must be an element of \\(v\\).
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* This program counter must correspond to an instruction \\(i\\) in \\(p\\), which we write as \\(p[c] = i\\).
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* This instruction must be executed, changing our program counter from \\(c\\) to \\(c'\\) and our
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accumulator from \\(a\\) to \\(a'\\). The set of valid instructions will no longer include \\(c\\),
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and will become \\(v - \\{c\\}\\).
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* Our program must then finish executing, starting at state
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\\((c', a', v - \\{c\\})\\), and ending in some (unknown) state \\((c'', a'', v'')\\).
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If all of these conditions are met, our program, starting at \\((c, a, v)\\), will terminate in the state \\((c'', a'', v'')\\). This third rule completes our semantics; a program being executed will keep running instructions using the third rule, until it finally
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hits an invalid program counter (terminating with the first rule) or gets to the end of the program (terminating with the second rule).
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