Merge

Invariant.v

@@ -3,28 +3,42 @@ Require Import Relations.
 Set Implicit Arguments.
 
 
-Section Invariant.
-  Variable state : Type.
-  Variable step : state -> state -> Prop.
-  Variable invariant : state -> Prop.
+Record trsys state := {
+  Initial : state -> Prop;
+  Step : state -> state -> Prop
+}.
 
-  Hint Constructors trc.
+Definition invariantFor {state} (sys : trsys state) (invariant : state -> Prop) :=
+  forall s, sys.(Initial) s
+            -> forall s', sys.(Step)^* s s'
+                          -> invariant s'.
 
-  Definition safe (s : state) :=
-    forall s', step^* s s' -> invariant s'.
+Theorem use_invariant : forall {state} (sys : trsys state) (invariant : state -> Prop) s s',
+  sys.(Step)^* s s'
+  -> sys.(Initial) s
+  -> invariantFor sys invariant
+  -> invariant s'.
+Proof.
+  firstorder.
+Qed.
 
-  Variable s0 : state.
+Theorem invariantFor_monotone : forall {state} (sys : trsys state)
+  (invariant1 invariant2 : state -> Prop),
+  (forall s, invariant1 s -> invariant2 s)
+  -> invariantFor sys invariant1
+  -> invariantFor sys invariant2.
+Proof.
+  unfold invariantFor; intuition eauto.
+Qed.
 
-  Hypothesis Hinitial : invariant s0.
-
-  Hypothesis Hstep : forall s s', invariant s -> step s s' -> invariant s'.
-
-  Lemma safety : safe s0.
-  Proof.
-    generalize dependent s0.
-    unfold safe.
-    induction 2; eauto.
-  Qed.
-End Invariant.
-
-Hint Resolve safety.
+Theorem invariant_induction : forall {state} (sys : trsys state)
+  (invariant : state -> Prop),
+  (forall s, sys.(Initial) s -> invariant s)
+  -> (forall s, invariant s -> forall s', sys.(Step) s s' -> invariant s')
+  -> invariantFor sys invariant.
+Proof.
+  unfold invariantFor; intros.
+  assert (invariant s) by eauto.
+  clear H1.
+  induction H2; eauto.
+Qed.

TransitionSystems.v

@@ -0,0 +1,237 @@
+(** Formal Reasoning About Programs <http://adam.chlipala.net/frap/>
+  * Chapter 4: Transition Systems
+  * Author: Adam Chlipala
+  * License: https://creativecommons.org/licenses/by-nc-nd/4.0/ *)
+
+Require Import Frap.
+
+
+(* Here's a classic recursive, functional program for factorial. *)
+Fixpoint fact (n : nat) : nat :=
+  match n with
+  | O => 1
+  | S n' => fact n' * S n'
+  end.
+
+(* But let's reformulate factorial relationally, as an example to explore
+ * treatment of inductive relations in Coq.  First, these are the states of our
+ * state machine. *)
+Inductive fact_state :=
+| AnswerIs (answer : nat)
+| WithAccumulator (input accumulator : nat).
+
+(* This *predicate* captures which states are starting states.
+ * Before the main colon of [Inductive], we list *parameters*, which stay fixed
+ * throughout recursive invocations of a predicate (though this definition does
+ * not use recursion).  After the colon, we give a type that expresses which
+ * additional arguments exist, followed by [Prop] for "proposition."
+ * Putting this inductive definition in [Prop] is what marks at as a predicate.
+ * Our prior definitions have implicitly been in [Set], the normal universe
+ * of mathematical objects. *)
+Inductive fact_init (original_input : nat) : fact_state -> Prop :=
+| FactInit : fact_init original_input (WithAccumulator original_input 1).
+
+(** And here are the states where we declare execution complete. *)
+Inductive fact_final : fact_state -> Prop :=
+| FactFinal : forall ans, fact_final (AnswerIs ans).
+
+(** The most important part: the relation to step between states *)
+Inductive fact_step : fact_state -> fact_state -> Prop :=
+| FactDone : forall acc,
+  fact_step (WithAccumulator O acc) (AnswerIs acc)
+| FactStep : forall n acc,
+  fact_step (WithAccumulator (S n) acc) (WithAccumulator n (acc * S n)).
+
+(* We care about more than just single steps.  We want to run factorial to
+ * completion, for which it is handy to define a general relation of
+ * *transitive-reflexive closure*, like so. *)
+Inductive trc {A} (R : A -> A -> Prop) : A -> A -> Prop :=
+| TrcRefl : forall x, trc R x x
+| TrcFront : forall x y z,
+  R x y
+  -> trc R y z
+  -> trc R x z.
+
+(* Ironically, this definition is not obviously transitive!
+ * Let's prove transitivity as a lemma. *)
+Theorem trc_trans : forall {A} (R : A -> A -> Prop) x y, trc R x y
+  -> forall z, trc R y z
+    -> trc R x z.
+Proof.
+  induct 1; simplify.
+
+  assumption.
+  (* [assumption]: prove a conclusion that matches some hypothesis exactly. *)
+
+  eapply TrcFront.
+  (* [eapply H]: like [apply], but works when it is not obvious how to
+   *   instantiate the quantifiers of theorem/hypothesis [H].  Instead,
+   *   placeholders are inserted for those quantifiers, to be determined
+   *   later. *)
+  eassumption.
+  (* [eassumption]: prove a conclusion that matches some hypothesis, when we
+   *   choose the right clever instantiation of placeholders.  Those placehoders
+   *   are then replaced everywhere with their new values. *)
+  apply IHtrc.
+  assumption.
+  (* [assumption]: like [eassumption], but never figures out placeholder
+   *   values. *)
+Qed.
+
+(* Transitive-reflexive closure is so common that it deserves a shorthand notation! *)
+Notation "R ^*" := (trc R) (at level 0).
+
+(* Now let's use it to execute the factorial program. *)
+Example factorial_3 : fact_step^* (WithAccumulator 3 1) (AnswerIs 6).
+Proof.
+  eapply TrcFront.
+  apply FactStep.
+  simplify.
+  eapply TrcFront.
+  apply FactStep.
+  simplify.
+  eapply TrcFront.
+  apply FactStep.
+  simplify.
+  eapply TrcFront.
+  apply FactDone.
+  apply TrcRefl.
+Qed.
+
+(* That was exhausting yet uninformative.  We can use a different tactic to blow
+ * through such obvious proof trees. *)
+Example factorial_3_auto : fact_step^* (WithAccumulator 3 1) (AnswerIs 6).
+Proof.
+  repeat econstructor.
+  (* [econstructor]: tries all declared rules of the predicate in the
+   *   conclusion, attempting each with [eapply] until one works. *)
+
+  (* Note that here [econstructor] is doing double duty, applying the rules of
+   * both [trc] and [fact_step]. *)
+Qed.
+
+(* It will be useful to give state machines more first-class status, as
+ * *transition systems*, formalized by this record type.  It has one type
+ * parameter, [state], which records the type of states. *)
+Record trsys state := {
+  Initial : state -> Prop;
+  Step : state -> state -> Prop
+}.
+
+Definition invariantFor {state} (sys : trsys state) (invariant : state -> Prop) :=
+  forall s, sys.(Initial) s
+            -> forall s', sys.(Step)^* s s'
+                          -> invariant s'.
+
+Theorem use_invariant : forall {state} (sys : trsys state) (invariant : state -> Prop) s s',
+  sys.(Step)^* s s'
+  -> sys.(Initial) s
+  -> invariantFor sys invariant
+  -> invariant s'.
+Proof.
+  firstorder.
+Qed.
+
+Theorem invariantFor_monotone : forall {state} (sys : trsys state)
+  (invariant1 invariant2 : state -> Prop),
+  (forall s, invariant1 s -> invariant2 s)
+  -> invariantFor sys invariant1
+  -> invariantFor sys invariant2.
+Proof.
+  unfold invariantFor; intuition eauto.
+Qed.
+
+Theorem invariant_induction : forall {state} (sys : trsys state)
+  (invariant : state -> Prop),
+  (forall s, sys.(Initial) s -> invariant s)
+  -> (forall s, invariant s -> forall s', sys.(Step) s s' -> invariant s')
+  -> invariantFor sys invariant.
+Proof.
+  unfold invariantFor; intros.
+  assert (invariant s) by eauto.
+  clear H1.
+  induction H2; eauto.
+Qed.
+
+
+(* To prove that our state machine is correct, we rely on the crucial technique
+ * of *invariants*.  What is an invariant?  Here's a general definition, in
+ * terms of an arbitrary *transition system* defined by a set of states,
+ * an initial-state relation, and a step relation. *)
+Definition invariantFor {state} (initial : state -> Prop) (step : state -> state -> Prop) (invariant : state -> Prop) :=
+  forall s, initial s
+            -> forall s', step^* s s'
+                          -> invariant s'.
+(* That is, when we begin in an initial state and take any number of steps, the
+ * place we wind up always satisfied the invariant. *)
+
+(* Here's a simple lemma to help us apply an invariant usefully,
+ * really just restating the definition. *)
+Theorem use_invariant : forall {state} (initial : state -> Prop)
+  (step : state -> state -> Prop) (invariant : state -> Prop) s s',
+  step^* s s'
+  -> initial s
+  -> invariantFor initial step invariant
+  -> invariant s'.
+Proof.
+  unfold invariantFor.
+  simplify.
+  eapply H1.
+  eassumption.
+  assumption.
+Qed.
+
+(* What's the most fundamental way to establish an invariant?  Induction! *)
+Theorem invariant_induction : forall {state} (initial : state -> Prop)
+  (step : state -> state -> Prop) (invariant : state -> Prop),
+  (forall s, initial s -> invariant s)
+  -> (forall s, invariant s -> forall s', step s s' -> invariant s')
+  -> invariantFor initial step invariant.
+Proof.
+  unfold invariantFor; intros.
+  assert (invariant s) by eauto.
+  clear H1.
+  induction H2; eauto.
+Qed.
+
+(* Here's a good invariant for factorial, parameterized on the original input
+ * to the program. *)
+Definition fact_invariant (original_input : nat) (st : fact_state) : Prop :=
+  match st with
+  | AnswerIs ans => fact original_input = ans
+  | WithAccumulator n acc => fact original_input = fact n * acc
+  end.
+
+(* We can use [invariant_induction] to prove that it really is a good
+ * invariant. *)
+Theorem fact_invariant_ok : forall original_input,
+  invariantFor (fact_init original_input) fact_step (fact_invariant original_input).
+Proof.
+  simplify.
+  apply invariant_induction; simplify.
+
+  (* Step 1: invariant holds at the start. (base case) *)
+  (* We have a hypothesis establishing [fact_init original_input s].
+   * By inspecting the definition of [fact_init], we can draw conclusions about
+   * what [s] must be.  The [invert] tactic formalizes that intuition,
+   * replacing a hypothesis with certain "obvious inferences" from the original.
+   * In general, when multiple different rules may have been used to conclude a
+   * fact, [invert] may generate one new subgoal per eligible rule, but here the
+   * predicate is only defined with one rule. *)
+  invert H.
+  (* We magically learn [s = WithAccumulator original_input 1]! *)
+  simplify.
+  ring.
+
+  (* Step 2: steps preserve the invariant. (induction step) *)
+  invert H0.
+  (* This time, [invert] is used on a predicate with two rules, neither of which
+   * can be ruled out for this case, so we get two subgoals from one. *)
+
+  simplify.
+  linear_arithmetic.
+
+  simplify.
+  rewrite H.
+  ring.
+Qed.