Library prosa.analysis.facts.transform.replace_at

In this file, we make a few simple observations about schedules with replacements.
Section ReplaceAtFacts.

For any given type of jobs...
Context {Job : JobType}.

... and any given type of processor states, ...
Context {PState: eqType}.
Context `{ProcessorState Job PState}.

...consider any given reference schedule.
Variable sched: schedule PState.

Suppose we are given a specific time t' ...
Variable t': instant.

...and a replacement state state.
Variable nstate: PState.

In the following, let sched' denote the schedule with replacement at time t'.
We begin with the trivial observation that the schedule doesn't change at other times.
Lemma rest_of_schedule_invariant:
t, t t' sched' t = sched t.
Proof.
movet DIFF.
rewrite /sched' /replace_at.
case (t' == t) eqn:TT; last by done.
by move/eqP in TT; rewrite TT in DIFF; contradiction.
Qed.

As a result, the service in intervals that do not intersect with t' is invariant, too.
Lemma service_at_other_times_invariant:
t1 t2,
t2 t' t' < t1
j,
service_during sched j t1 t2
=
service_during sched' j t1 t2.
Proof.
movet1 t2 SWAP_EXCLUDED j.
rewrite /service_during /service_at.
apply eq_big_natt /andP [le_t1t lt_tt2].
rewrite rest_of_schedule_invariant//.
eapply point_not_in_interval; eauto.
apply /andP; by split.
Qed.

Next, we consider the amount of service received in intervals that do span across the replacement point. We can relate the service received in the original and the new schedules by adding the service in the respective "missing" state...
Lemma service_delta:
t1 t2,
t1 t' < t2
j,
service_during sched j t1 t2 + service_at sched' j t'
=
service_during sched' j t1 t2 + service_at sched j t'.
Proof.
movet1 t2 TIMES j.
rewrite -(service_split_at_point sched _ _ t' _) //
-(service_split_at_point sched' _ _ t' _) //.
by rewrite !service_at_other_times_invariant -/sched'; [ring | right | left].
Qed.

...which we can also rewrite as follows.
Corollary service_in_replaced:
t1 t2,
t1 t' < t2
j,
service_during sched' j t1 t2
=
service_during sched j t1 t2 + service_at sched' j t' - service_at sched j t'.
Proof. movet1 t2 ORDER j. by rewrite service_delta// addnK. Qed.

As another simple invariant (useful for case analysis), we observe that if a job is scheduled neither in the replaced nor in the new state, then at any time it receives exactly the same amount of service in the new schedule with replacements as in the original schedule.
Lemma service_at_of_others_invariant (j: Job):
~~ scheduled_in j (sched' t')
~~ scheduled_in j (sched t')
t,
service_at sched j t = service_at sched' j t.
Proof.
moveNOT_IN_NEW NOT_IN_OLD t.
rewrite /service_at.
case: (boolP (t == t')) ⇒ /eqP TT.
- rewrite !TT !service_implies_scheduled //; by apply negbTE.
- rewrite rest_of_schedule_invariant//.
Qed.

Based on the previous observation, we can trivially lift the invariant that jobs not involved in the replacement receive equal service to the cumulative service received in any interval.
Corollary service_during_of_others_invariant (j: Job):
~~ scheduled_in j (sched' t')
~~ scheduled_in j (sched t')
t1 t2,
service_during sched j t1 t2 = service_during sched' j t1 t2.
Proof.
moveNOT_IN_NEW NOT_IN_OLD t1 t2.
rewrite /service_during.
apply eq_big_natt /andP [le_t1t lt_tt2].
rewrite service_at_of_others_invariant //.
Qed.

End ReplaceAtFacts.