From iris.algebra Require Export sts.
From iris.base_logic Require Import lib.own.
From stdpp Require Export gmap.
Set Default Proof Using "Type".
(** The STS describing the main barrier protocol. Every state has an index-set
associated with it. These indices are actually [gname], because we use them
with saved propositions. *)
Inductive phase := Low | High.
Record state := State { state_phase : phase; state_I : gset gname }.
Add Printing Constructor state.
Inductive token := Change (i : gname) | Send.
Global Instance stateT_inhabited: Inhabited state := populate (State Low ∅).
Global Instance Change_inj : Inj (=) (=) Change.
Proof. by injection 1. Qed.
Inductive prim_step : relation state :=
| ChangeI p I2 I1 : prim_step (State p I1) (State p I2)
| ChangePhase I : prim_step (State Low I) (State High I).
Definition tok (s : state) : set token :=
{[ t | ∃ i, t = Change i ∧ i ∉ state_I s ]} ∪
(if state_phase s is High then {[ Send ]} else ∅).
Global Arguments tok !_ /.
Canonical Structure sts := sts.STS prim_step tok.
(* The set of states containing some particular i *)
Definition i_states (i : gname) : set state := {[ s | i ∈ state_I s ]}.
(* The set of low states *)
Definition low_states : set state := {[ s | state_phase s = Low ]}.
Lemma i_states_closed i : sts.closed (i_states i) {[ Change i ]}.
Proof.
split; first (intros [[] I]; set_solver).
(* If we do the destruct of the states early, and then inversion
on the proof of a transition, it doesn't work - we do not obtain
the equalities we need. So we destruct the states late, because this
means we can use "destruct" instead of "inversion". *)
intros s1 s2 Hs1 [T1 T2 Hdisj Hstep'].
inversion_clear Hstep' as [? ? ? ? Htrans _ _ Htok].
destruct Htrans as [[] ??|]; done || set_solver.
Qed.
Lemma low_states_closed : sts.closed low_states {[ Send ]}.
Proof.
split; first (intros [??]; set_solver).
intros s1 s2 Hs1 [T1 T2 Hdisj Hstep'].
inversion_clear Hstep' as [? ? ? ? Htrans _ _ Htok].
destruct Htrans as [[] ??|]; done || set_solver.
Qed.
(* Proof that we can take the steps we need. *)
Lemma signal_step I : sts.steps (State Low I, {[Send]}) (State High I, ∅).
Proof. apply rtc_once. constructor; first constructor; set_solver. Qed.
Lemma wait_step i I :
i ∈ I →
sts.steps (State High I, {[ Change i ]}) (State High (I ∖ {[ i ]}), ∅).
Proof.
intros. apply rtc_once.
constructor; first constructor; [set_solver..|].
apply elem_of_equiv=>-[j|]; last set_solver.
destruct (decide (i = j)); set_solver.
Qed.
Lemma split_step p i i1 i2 I :
i ∈ I → i1 ∉ I → i2 ∉ I → i1 ≠ i2 →
sts.steps
(State p I, {[ Change i ]})
(State p ({[i1; i2]} ∪ I ∖ {[i]}), {[ Change i1; Change i2 ]}).
Proof.
intros. apply rtc_once. constructor; first constructor.
- destruct p; set_solver.
- destruct p; set_solver.
- apply elem_of_equiv=> /= -[j|]; last set_solver.
set_unfold; rewrite !(inj_iff Change).
assert (Change j ∈ match p with Low => ∅ | High => {[Send]} end ↔ False)
as -> by (destruct p; set_solver).
destruct (decide (i1 = j)) as [->|]; first naive_solver.
destruct (decide (i2 = j)) as [->|]; first naive_solver.
destruct (decide (i = j)) as [->|]; naive_solver.
Qed.