Library prosa.analysis.abstract.abstract_seq_rta

Abstract Response-Time Analysis with sequential tasks

In this section we propose the general framework for response-time analysis (RTA) of uni-processor scheduling of real-time tasks with arbitrary arrival models and sequential tasks.
We prove that the maximum among the solutions of the response-time bound recurrence for some set of parameters is a response-time bound for tsk. Note that in this section we do rely on the hypothesis about task sequentiality. This allows us to provide a more precise response-time bound function, since jobs of the same task will be executed strictly in the order they arrive.
Consider any type of tasks ...
  Context {Task : TaskType}.
  Context `{TaskCost Task}.
  Context `{TaskRunToCompletionThreshold Task}.

... and any type of jobs associated with these tasks.
  Context {Job : JobType}.
  Context `{JobTask Job Task}.
  Context `{JobArrival Job}.
  Context `{JobCost Job}.
  Context `{JobPreemptable Job}.

Consider any arrival sequence with consistent, non-duplicate arrivals...
Next, consider any ideal uniprocessor schedule of this arrival sequence...
... where jobs do not execute before their arrival nor after completion.
Assume that the job costs are no larger than the task costs.
Consider an arbitrary task set.
  Variable ts : list Task.

Let tsk be any task in ts that is to be analyzed.
  Variable tsk : Task.
  Hypothesis H_tsk_in_ts : tsk \in ts.

Consider a valid preemption model...
...and a valid task run-to-completion threshold function. That is, task_rtct tsk is (1) no bigger than tsk's cost, (2) for any job of task tsk job_rtct is bounded by task_rtct.
Let max_arrivals be a family of valid arrival curves, i.e., for any task tsk in ts max_arrival tsk is (1) an arrival bound of tsk, and (2) it is a monotonic function that equals 0 for the empty interval delta = 0.
Assume we are provided with abstract functions for interference and interfering workload.
Let's define some local names for clarity.
In this section, we introduce a few new definitions to make it easier to express the new bound of the worst-case execution time.
  Section Definitions.

When assuming sequential tasks, we can introduce an additional hypothesis that ensures that the values of interference and workload remain consistent. It states that any of tsk's job, that arrived before the busy interval, should be completed by the beginning of the busy interval.
    Definition interference_and_workload_consistent_with_sequential_tasks :=
       (j : Job) (t1 t2 : instant),
          arrives_in arr_seq j
          job_of_task tsk j
          job_cost j > 0
          busy_interval j t1 t2
          task_workload_between 0 t1 = task_service_of_jobs_in (arrivals_between 0 t1) 0 t1.

Next we introduce the notion of task interference. Intuitively, task tsk incurs interference when some of the jobs of task tsk incur interference. As a result, tsk cannot make any progress. More formally, task tsk experiences interference at a time instant time t, if at time t task tsk is not scheduled and there exists a job of tsk that (1) experiences interference and (2) has arrived before some time instant upper_bound.
It is important to note two subtle points: according to our semantics of the interference function, jobs from the same task can cause interference to each other. In the definition of interference of a task we want to avoid such situations. That is why we use the term ~~ task_scheduled_at tsk t.
Moreover, in order to make the definition constructive, we introduce an upper bound on the arrival time of jobs from task tsk. As a result, we need to consider only a finite number of jobs. For the function to produce the correct values it is enough to specify a sufficiently large upper_bound. Usually as upper_bound one can use the end of the corresponding busy interval.
Next we define the cumulative task interference.
We say that task interference is bounded by task_interference_bound_function (tIBF) iff for any job j of task tsk cumulative task interference within the interval t1, t1 + R) is bounded by function [tIBF(tsk, A, R)]. Note that this definition is almost the same as the definition of job_interference_is_bounded_by from the non-necessary-sequential case. However, in this case we ignore the interference that comes from jobs from the same task.
    Definition task_interference_is_bounded_by
               (task_interference_bound_function : Task duration duration duration) :=
       j R t1 t2,
        arrives_in arr_seq j
        job_of_task tsk j
        t1 + R < t2
        ~~ completed_by sched j (t1 + R)
        busy_interval j t1 t2
        let offset := job_arrival j - t1 in
        cumul_task_interference tsk t2 t1 (t1 + R) task_interference_bound_function tsk offset R.

  End Definitions.

In this section, we prove that the maximum among the solutions of the response-time bound recurrence is a response-time bound for tsk.
  Section ResponseTimeBound.

For simplicity, let's define some local names.
We assume that the schedule is work-conserving.
Unlike the previous theorem uniprocessor_response_time_bound, we assume that (1) tasks are sequential, moreover (2) functions interference and interfering_workload are consistent with the hypothesis of sequential tasks.
Assume we have a constant L which bounds the busy interval of any of tsk's jobs.
Next, we assume that task_interference_bound_function is a bound on interference incurred by the task.
Given any job j of task tsk that arrives exactly A units after the beginning of the busy interval, the bound on the total interference incurred by j within an interval of length Δ is no greater than task_rbf (A + ε) - task_cost tsk + task's IBF Δ. Note that in case of sequential tasks the bound consists of two parts: (1) the part that bounds the interference received from other jobs of task tsk -- task_rbf (A + ε) - task_cost tsk and (2) any other interference that is bounded by task_IBF(tsk, A, Δ).
Note that since we consider the modified interference bound function, the search space has also changed. One can see that the new search space is guaranteed to include any A for which task_rbf (A) task_rbf (A + ε), since this implies the fact that total_interference_bound (tsk, A, Δ) total_interference_bound (tsk, A + ε, Δ).
Consider any value R, and assume that for any relative arrival time A from the search space there is a solution F of the response-time recurrence that is bounded by R. In contrast to the formula in "non-sequential" Abstract RTA, assuming that tasks are sequential leads to a more precise response-time bound. Now we can explicitly express the interference caused by other jobs of the task under consideration.
To understand the right part of the fix-point in the equation it is helpful to note that the bound on the total interference (bound_of_total_interference) is equal to task_rbf (A + ε) - task_cost tsk + tIBF tsk A Δ. Besides, a job must receive enough service to become non-preemptive task_lock_in_service tsk. The sum of these two quantities is exactly the right-hand side of the equation.
    Variable R : nat.
    Hypothesis H_R_is_maximum_seq:
       (A : duration),
        is_in_search_space_seq A
         (F : duration),
          A + F (task_rbf (A + ε) - (task_cost tsk - task_rtct tsk))
                  + task_interference_bound_function tsk A (A + F)
          R F + (task_cost tsk - task_rtct tsk).

In this section we prove a few simple lemmas about the completion of jobs from the task considering the busy interval of the job under consideration.
    Section CompletionOfJobsFromSameTask.

Consider any two jobs j1 j2 of tsk.
      Variable j1 j2 : Job.
      Hypothesis H_j1_arrives: arrives_in arr_seq j1.
      Hypothesis H_j2_arrives: arrives_in arr_seq j2.
      Hypothesis H_j1_from_tsk: job_of_task tsk j1.
      Hypothesis H_j2_from_tsk: job_of_task tsk j2.
      Hypothesis H_j1_cost_positive: job_cost_positive j1.

Consider the busy interval [t1, t2) of job j1.
      Variable t1 t2 : instant.
      Hypothesis H_busy_interval : busy_interval j1 t1 t2.

We prove that if a job from task tsk arrived before the beginning of the busy interval, then it must be completed before the beginning of the busy interval
      Lemma completed_before_beginning_of_busy_interval:
        job_arrival j2 < t1
        completed_by sched j2 t1.
      Proof.
        moveJA; move: (H_j2_from_tsk) ⇒ /eqP TSK2eq.
        rewrite /completed_by.
        move: (posnP (@job_cost _ H3 j2)) ⇒ [-> | POS]; first by done.
        move: (H_interference_and_workload_consistent_with_sequential_tasks j1 t1 t2) ⇒ SWEQ.
        feed_n 4 SWEQ; try by done.
        apply all_jobs_have_completed_equiv_workload_eq_service with (j := j2) in SWEQ ⇒ //.
        - by apply ideal_proc_model_provides_unit_service.
        - by apply arrived_between_implies_in_arrivals.
      Qed.

Next we prove that if a job is pending after the beginning of the busy interval [t1, t2) then it arrives after t1.
      Lemma arrives_after_beginning_of_busy_interval:
         t,
          t1 t
          pending sched j2 t
          arrived_between j2 t1 t.+1.
      Proof.
        intros t GE PEND.
        rewrite /arrived_between; apply/andP; split; last first.
        { by move: PEND ⇒ /andP [ARR _]; rewrite ltnS. }
        rewrite leqNgt; apply/negP; intros LT.
        move: (H_busy_interval) ⇒ [[/andP [AFR1 AFR2] [QT _]] _].
        have L12 := completed_before_beginning_of_busy_interval LT.
        apply completion_monotonic with (t' := t) in L12; try done.
        by move: PEND ⇒ /andP [_ /negP T2].
      Qed.

    End CompletionOfJobsFromSameTask.

Since we are going to use the uniprocessor_response_time_bound theorem to prove the theorem of this section, we have to show that all the hypotheses are satisfied. Namely, we need to show that hypotheses H_sequential_tasks, H_i_w_are_task_consistent and H_task_interference_is_bounded_by imply H_job_interference_is_bounded, and the fact that H_R_is_maximum_seq implies H_R_is_maximum.
In this section we show that there exists a bound for cumulative interference for any job of task tsk, i.e., the hypothesis H_job_interference_is_bounded holds.
Consider any job j of tsk.
      Variable j : Job.
      Hypothesis H_j_arrives : arrives_in arr_seq j.
      Hypothesis H_job_of_tsk : job_of_task tsk j.
      Hypothesis H_job_cost_positive : job_cost_positive j.

Consider the busy interval [t1, t2) of job j.
      Variable t1 t2 : instant.
      Hypothesis H_busy_interval : busy_interval j t1 t2.

Let's define A as a relative arrival time of job j (with respect to time t1).
      Let A : duration := job_arrival j - t1.

Consider an arbitrary time x ...
      Variable x : duration.
... such that (t1 + x) is inside the busy interval...
      Hypothesis H_inside_busy_interval : t1 + x < t2.
... and job j is not completed by time (t1 + x).
      Hypothesis H_job_j_is_not_completed : ~~ completed_by sched j (t1 + x).

In this section, we show that the cumulative interference of job j in the interval [t1, t1 + x) is bounded by the sum of the task workload in the interval t1, t1 + A + ε) and the cumulative interference of [j]'s task in the interval [t1, t1 + x). Note that the task workload is computed only on the interval [t1, t1 + A + ε). Thanks to the hypothesis about sequential tasks, jobs of task [tsk] that arrive after [t1 + A + ε] cannot interfere with j.
      Section TaskInterferenceBoundsInterference.

We start by proving a simpler analog of the lemma which states that at any time instant t ∈ [t1, t1 + x) the sum of interference j t and scheduled_at j t is no larger than the sum of the service received by jobs of task tsk at time t and task_iterference tsk t.
Next we consider 4 cases.
        Section CaseAnalysis.

Consider an arbitrary time instant t ∈ [t1, t1 + x).
          Variable t : instant.
          Hypothesis H_t_in_interval : t1 t < t1 + x.

          Section Case1.

Assume the processor is idle at time t.
            Hypothesis H_idle : sched t = None.

In case when the processor is idle, one can show that interference j t = 1, scheduled_at j t = 0. But since interference doesn't come from a job of task tsk task_interference tsk = 1. Which reduces to 1 1.
            Lemma interference_plus_sched_le_serv_of_task_plus_task_interference_idle:
              interference j t + scheduled_at sched j t
              service_of_jobs_at (job_of_task tsk) (arrivals_between t1 (t1 + A + ε)) t +
              task_interference_received_before tsk t2 t.
            Proof.
              move: (H_busy_interval) ⇒ [[/andP [BUS LT] _] _].
              rewrite /cumul_task_interference /definitions.cumul_interference
                      /Sequential_Abstract_RTA.cumul_task_interference /task_interference_received_before
                      /task_scheduled_at /task_schedule.task_scheduled_at /service_of_jobs_at
                      /service_of_jobs.service_of_jobs_at/= scheduled_at_def.
              erewrite eq_bigr; last by movei j' /=; rewrite service_at_def H_idle /=.
              rewrite /= big1_eq add0n H_idle addn0.
              case INT: (interference j t); last by done.
              rewrite /= lt0b.
              apply/hasP; j; last by done.
              rewrite mem_filter; apply/andP; split; first by done.
              by apply arrived_between_implies_in_arrivals.
            Qed.

          End Case1.

          Section Case2.

Assume a job j' from another task is scheduled at time t.
            Variable j' : Job.
            Hypothesis H_sched : sched t = Some j'.
            Hypothesis H_not_job_of_tsk : ~~ job_of_task tsk j'.

If a job j' from another task is scheduled at time t, then interference j t = 1, scheduled_at j t = 0. But since interference doesn't come from a job of task tsk task_interference tsk = 1. Which reduces to 1 1.
            Lemma interference_plus_sched_le_serv_of_task_plus_task_interference_task:
              interference j t + scheduled_at sched j t
              service_of_jobs_at (job_of_task tsk) (arrivals_between t1 (t1 + A + ε)) t +
              task_interference_received_before tsk t2 t.
            Proof.
              move: (H_busy_interval) ⇒ [[/andP [BUS LT] _] _].
              rewrite /cumul_task_interference /definitions.cumul_interference
                      /Sequential_Abstract_RTA.cumul_task_interference /task_interference_received_before
                      /task_scheduled_at /task_schedule.task_scheduled_at /service_of_jobs_at
                      /service_of_jobs.service_of_jobs_at scheduled_at_def/=.
              have ARRs: arrives_in arr_seq j'; first by apply H_jobs_come_from_arrival_sequence with t; rewrite scheduled_at_def; apply/eqP.
              rewrite H_sched H_not_job_of_tsk; simpl.
              have ->: Some j' == Some j = false; last rewrite addn0.
              { apply/negP ⇒ /eqP CONTR; inversion CONTR; subst j'.
                by move: (H_not_job_of_tsk); rewrite H_job_of_tsk. }
              erewrite eq_bigr; last by movei _; rewrite service_at_def H_sched.
              have ZERO: \sum_(i <- arrivals_between t1 (t1 + A + ε) | job_task i == tsk) (Some j' == Some i) = 0.
              { apply big1j2 TSK.
                apply/eqP; rewrite eqb0; apply/negP ⇒ /eqP EQ; inversion EQ; subst j'.
                by move: (H_not_job_of_tsk); rewrite / job_of_task TSK.
              }
              rewrite {}ZERO ?addn0 add0n /=.
              case INT: (interference j t); last by done.
              rewrite /= lt0b.
              apply/hasP; j; last by done.
              rewrite mem_filter; apply/andP; split; first by done.
              by eapply arrived_between_implies_in_arrivals.
            Qed.

          End Case2.

          Section Case3.

Assume a job j' (different from j) of task tsk is scheduled at time t.
            Variable j' : Job.
            Hypothesis H_sched : sched t = Some j'.
            Hypothesis H_not_job_of_tsk : job_of_task tsk j'.
            Hypothesis H_j_neq_j' : j != j'.

If a job j' (different from j) of task tsk is scheduled at time t, then interference j t = 1, scheduled_at j t = 0. Moreover, since interference comes from a job of the same task task_interference tsk = 0. However, in this case service_of_jobs of tsk = 1. Which reduces to 1 1.
            Lemma interference_plus_sched_le_serv_of_task_plus_task_interference_job:
              interference j t + scheduled_at sched j t
              service_of_jobs_at (job_of_task tsk) (arrivals_between t1 (t1 + A + ε)) t +
              task_interference_received_before tsk t2 t.
            Proof.
              move: (H_busy_interval) ⇒ [[/andP [BUS LT] _] _].
              rewrite /cumul_task_interference /definitions.cumul_interference
                      /Sequential_Abstract_RTA.cumul_task_interference /task_interference_received_before
                      /task_scheduled_at /task_schedule.task_scheduled_at /service_of_jobs_at
                      /service_of_jobs.service_of_jobs_at scheduled_at_def/=.
              have ARRs: arrives_in arr_seq j'; first by apply H_jobs_come_from_arrival_sequence with t; rewrite scheduled_at_def; apply/eqP.
              move: (H_not_job_of_tsk) ⇒ /eqP; rewrite H_sched ⇒ ->; rewrite eq_refl addn0; simpl.
              have ->: Some j' == Some j = false by
                  apply/negP ⇒ /eqP EQ; inversion EQ; subst j'; move:H_j_neq_j' ⇒ /negP NEQ; apply: NEQ.
              replace (interference j t) with true; last first.
              { have NEQT: t1 t < t2.
                { by move: H_t_in_interval ⇒ /andP [NEQ1 NEQ2]; apply/andP; split; last apply ltn_trans with (t1 + x). }
                move: (H_work_conserving j t1 t2 t H_j_arrives H_job_of_tsk H_job_cost_positive H_busy_interval NEQT) ⇒ [Hn _].
                apply/eqP;rewrite eq_sym eqb_id; apply/negPn/negP; intros CONTR; move: CONTR ⇒ /negP CONTR.
                apply Hn in CONTR; rewrite scheduled_at_def in CONTR; simpl in CONTR.
                by move: CONTR; rewrite H_sched ⇒ /eqP EQ; inversion EQ; subst; move: H_j_neq_j' ⇒ /eqP.
              }
              rewrite big_mkcond; apply/sum_seq_gt0P; j'; split; last first.
              { move: H_not_job_of_tsk ⇒ /eqP TSK.
                by rewrite /job_of_task TSK eq_refl service_at_def H_sched eq_refl. }
              { intros. have ARR:= arrives_after_beginning_of_busy_interval j j' _ _ _ _ _ t1 t2 _ t.
                feed_n 8 ARR; try (done || by move: H_t_in_interval ⇒ /andP [T1 T2]).
                { move: H_sched ⇒ /eqP SCHEDt.
                  apply scheduled_implies_pending;
                    auto using ideal_proc_model_ensures_ideal_progress.
                  by rewrite scheduled_at_def. }
                case_eq (job_arrival j' job_arrival j) ⇒ ARRNEQ.
                { move: ARR ⇒ /andP [РР _].
                  eapply arrived_between_implies_in_arrivals; eauto 2.
                  by apply/andP; split; last rewrite /A subnKC // addn1 ltnS. }
                { exfalso.
                  apply negbT in ARRNEQ; rewrite -ltnNge in ARRNEQ.
                  move: (H_sequential_tasks j j' t) ⇒ CONTR.
                  feed_n 5 CONTR; try done.
                  { by rewrite /same_task eq_sym; move: (H_job_of_tsk) ⇒ /eqP →. }
                  { by move: H_sched ⇒ /eqP SCHEDt; rewrite scheduled_at_def. }
                  move: H_job_j_is_not_completed ⇒ /negP T; apply: T.
                  apply completion_monotonic with t; try done.
                  by apply ltnW; move: H_t_in_interval ⇒ /andP [_ NEQ]. } }
            Qed.

          End Case3.

          Section Case4.

Assume that job j is scheduled at time t.
            Hypothesis H_sched : sched t = Some j.

If job j is scheduled at time t, then interference = 0, scheduled_at = 1, but note that service_of_jobs of tsk = 1, therefore inequality reduces to 1 1.
            Lemma interference_plus_sched_le_serv_of_task_plus_task_interference_j:
              interference j t + scheduled_at sched j t
              service_of_jobs_at (job_of_task tsk) (arrivals_between t1 (t1 + A + ε)) t +
              task_interference_received_before tsk t2 t.
            Proof.
              have j_is_in_arrivals_between: j \in arrivals_between t1 (t1 + A + ε).
              { eapply arrived_between_implies_in_arrivals; eauto 2.
                move: (H_busy_interval) ⇒ [[/andP [GE _] [_ _]] _].
                by apply/andP; split; last rewrite /A subnKC // addn1.
              } intros.
              rewrite /cumul_task_interference /definitions.cumul_interference
                      /Sequential_Abstract_RTA.cumul_task_interference /task_interference_received_before
                      /task_scheduled_at /task_schedule.task_scheduled_at /service_of_jobs_at
                      /service_of_jobs.service_of_jobs_at scheduled_at_def.
              rewrite H_sched; move: H_job_of_tsk ⇒ /eqP ->; rewrite eq_refl eq_refl addn0 //=.
              move: (H_work_conserving j _ _ t H_j_arrives H_job_of_tsk H_job_cost_positive H_busy_interval) ⇒ WORK.
              feed WORK.
              { move: H_t_in_interval ⇒ /andP [NEQ1 NEQ2].
                by apply/andP; split; last apply ltn_trans with (t1 + x). }
              move: WORK ⇒ [_ ZIJT].
              feed ZIJT; first by rewrite scheduled_at_def H_sched; simpl.
              move: ZIJT ⇒ /negP /eqP; rewrite eqb_negLR; simpl; move ⇒ /eqP ZIJT; rewrite ZIJT; simpl; rewrite add0n.
              rewrite big_mkcond //=; apply/sum_seq_gt0P.
               j; split; first by apply j_is_in_arrivals_between.
              by move: (H_job_of_tsk) ⇒ ->; rewrite service_at_def H_sched eq_refl.
            Qed.

          End Case4.

We use the above case analysis to prove that any time instant t ∈ [t1, t1 + x) the sum of interference j t and scheduled_at j t is no larger than the sum of the service received by jobs of task tsk at time t and task_iterference tsk t.
          Lemma interference_plus_sched_le_serv_of_task_plus_task_interference:
            interference j t + scheduled_at sched j t
             service_of_jobs_at (job_of_task tsk) (arrivals_between t1 (t1 + A + ε)) t
              + task_interference_received_before tsk t2 t.
          Proof.
            move: (H_busy_interval) ⇒ [[/andP [BUS LT] _] _].
            case SCHEDt: (sched t) ⇒ [j1 | ].
            2: by apply interference_plus_sched_le_serv_of_task_plus_task_interference_idle.
            have ARRs: arrives_in arr_seq j1;
              first by apply H_jobs_come_from_arrival_sequence with t; rewrite scheduled_at_def; apply/eqP.
            case_eq (job_task j1 == tsk) ⇒ TSK.
            2: by eapply interference_plus_sched_le_serv_of_task_plus_task_interference_task; [eassumption| apply/negbT].
            case EQ: (j == j1); [move: EQ ⇒ /eqP EQ; subst j1 | ].
            1: by apply interference_plus_sched_le_serv_of_task_plus_task_interference_j.
            eapply interference_plus_sched_le_serv_of_task_plus_task_interference_job;
              auto; repeat split; eauto; apply/eqP; move: EQ ⇒ /eqP EQ; auto.
          Qed.

        End CaseAnalysis.

Next we prove cumulative version of the lemma above.
        Lemma cumul_interference_plus_sched_le_serv_of_task_plus_cumul_task_interference:
          cumul_interference j t1 (t1 + x)
           (task_service_of_jobs_in (arrivals_between t1 (t1 + A + ε)) t1 (t1 + x)
              - service_during sched j t1 (t1 + x)) + cumul_task_interference t2 t1 (t1 + x).
        Proof.
          have j_is_in_arrivals_between: j \in arrivals_between t1 (t1 + A + ε).
          { eapply arrived_between_implies_in_arrivals; eauto 2.
            move: (H_busy_interval) ⇒ [[/andP [GE _] [_ _]] _].
            by apply/andP; split; last rewrite /A subnKC // addn1.
          }
          rewrite /cumul_interference /cumul_interference /task_service_of_jobs_in
          /service_of_jobs.task_service_of_jobs_in
          /service_of_jobs exchange_big //=.
          rewrite -(leq_add2r (\sum_(t1 t < (t1 + x)) service_at sched j t)).
          rewrite [X in _ X]addnC addnA subnKC; last first.
          { rewrite (exchange_big _ _ (arrivals_between _ _)) /= (big_rem j) //=.
            by rewrite H_job_of_tsk leq_addr. }
          rewrite -big_split -big_split //=.
          rewrite big_nat_cond [X in _ X]big_nat_cond leq_sum //; movet /andP [NEQ _].
          rewrite {1}service_at_def -scheduled_at_def.
          by apply interference_plus_sched_le_serv_of_task_plus_task_interference.
        Qed.

On the other hand, the service terms in the inequality above can be upper-bound by the workload terms.
        Lemma serv_of_task_le_workload_of_task_plus:
          task_service_of_jobs_in (arrivals_between t1 (t1 + A + ε)) t1 (t1 + x)
          - service_during sched j t1 (t1 + x) + cumul_task_interference t2 t1 (t1 + x)
           (task_workload_between t1 (t1 + A + ε) - job_cost j)
            + cumul_task_interference t2 t1 (t1 + x).
        Proof.
          have j_is_in_arrivals_between: j \in arrivals_between t1 (t1 + A + ε).
          { eapply arrived_between_implies_in_arrivals; eauto 2.
            move: (H_busy_interval) ⇒ [[/andP [GE _] [_ _]] _].
            by apply/andP; split; last rewrite /A subnKC // addn1.
          }
          rewrite leq_add2r.
          rewrite /task_workload /task_service_of_jobs_in
                  /service_of_jobs.task_service_of_jobs_in/service_of_jobs /workload_of_jobs.
          rewrite (big_rem j) ?[X in _ X - _](big_rem j) //=; auto using j_is_in_arrivals_between.
          rewrite H_job_of_tsk addnC -addnBA; last by done.
          rewrite [X in _ X - _]addnC -addnBA; last by done.
          rewrite !subnn !addn0.
          by apply service_of_jobs_le_workload; auto using ideal_proc_model_provides_unit_service.
        Qed.

Finally, we show that the cumulative interference of job j in the interval [t1, t1 + x) is bounded by the sum of the task workload in the interval t1, t1 + A + ε) and the cumulative interference of [j]'s task in the interval [t1, t1 + x).
        Lemma cumulative_job_interference_le_task_interference_bound:
          cumul_interference j t1 (t1 + x)
           (task_workload_between t1 (t1 + A + ε) - job_cost j)
            + cumul_task_interference t2 t1 (t1 + x).
        Proof.
          apply leq_trans with
              (task_service_of_jobs_in (arrivals_between t1 (t1 + A + ε)) t1 (t1 + x)
               - service_during sched j t1 (t1 + x)
               + cumul_task_interference t2 t1 (t1 + x));
            [ apply cumul_interference_plus_sched_le_serv_of_task_plus_cumul_task_interference
            | apply serv_of_task_le_workload_of_task_plus].
        Qed.

      End TaskInterferenceBoundsInterference.

In order to obtain a more convenient bound of the cumulative interference, we need to abandon the actual workload in favor of a bound which depends on task parameters only. So, we show that actual workload of the task excluding workload of any job j is no greater than bound of workload excluding the cost of job j's task.
      Lemma task_rbf_excl_tsk_bounds_task_workload_excl_j:
        task_workload_between t1 (t1 + A + ε) - job_cost j task_rbf (A + ε) - task_cost tsk.
      Proof.
        move: H_j_arrives H_job_of_tsk H_busy_intervalARR TSK [[/andP [JAGET1 JALTT2] _] _].
        apply leq_trans with
            (task_cost tsk × number_of_task_arrivals arr_seq tsk t1 (t1 + A + ε) - task_cost tsk); last first.
        { rewrite leq_sub2r // leq_mul2l; apply/orP; right.
          rewrite -addnA -{2}[(A+1)](addKn t1).
          by apply H_is_arrival_curve; auto using leq_addr. }
        have Fact6: j \in arrivals_between (t1 + A) (t1 + A + ε).
        { apply arrived_between_implies_in_arrivals; try done.
          apply/andP; split; rewrite /A subnKC //.
          by rewrite addn1 ltnSn //. }
        have Fact4: j \in arrivals_at arr_seq (t1 + A).
        { by move: ARR ⇒ [t ARR]; rewrite subnKC //; feed (H_arrival_times_are_consistent j t); try (subst t). }
        have Fact1: 1 number_of_task_arrivals arr_seq tsk (t1 + A) (t1 + A + ε).
        { rewrite /number_of_task_arrivals /task_arrivals_between /arrival_sequence.arrivals_between.
          by rewrite size_filter -has_count; apply/hasP; j; last rewrite TSK.
        }
        rewrite (@num_arrivals_of_task_cat _ _ _ _ _ (t1 + A)); last by apply/andP; split; rewrite leq_addr //.
        rewrite mulnDr.
        have Step1: task_workload_between t1 (t1 + A + ε)
                    = task_workload_between t1 (t1 + A) + task_workload_between (t1 + A) (t1 + A + ε).
        { by apply workload_of_jobs_cat; apply/andP; split; rewrite leq_addr. } rewrite Step1; clear Step1.
        rewrite -!addnBA; first last.
        { by rewrite /task_workload_between /workload.task_workload_between /task_workload
             /workload_of_jobs (big_rem j) //= TSK leq_addr. }
        { apply leq_trans with (task_cost tsk); first by done.
          by rewrite -{1}[task_cost tsk]muln1 leq_mul2l; apply/orP; right. }
        rewrite leq_add; [by done | by eapply task_workload_le_num_of_arrivals_times_cost; eauto | ].
        rewrite /task_workload_between /workload.task_workload_between /task_workload /workload_of_jobs
                /arrival_sequence.arrivals_between /number_of_task_arrivals /task_arrivals_between
                /arrival_sequence.arrivals_between.
        rewrite {1}addn1 big_nat1 addn1 big_nat1.
        rewrite (big_rem j) //= TSK //= addnC -addnBA // subnn addn0.
        rewrite (filter_size_rem j); [ | by done | by rewrite TSK].
        rewrite mulnDr mulnC muln1 -addnBA // subnn addn0 mulnC.
        apply sum_majorant_constant.
        movej' ARR' /eqP TSK2.
        by rewrite -TSK2; apply H_valid_job_cost; (t1 + A); apply rem_in in ARR'.
      Qed.

Finally, we use the lemmas above to obtain the bound on interference in terms of task_rbf and task_interference.
      Lemma cumulative_job_interference_bound:
        cumul_interference j t1 (t1 + x)
         (task_rbf (A + ε) - task_cost tsk) + cumul_task_interference t2 t1 (t1 + x).
      Proof.
        set (y := t1 + x) in ×.
        have IN: j \in arrivals_between t1 (t1 + A + ε).
        { eapply arrived_between_implies_in_arrivals; eauto 2.
          move: (H_busy_interval) ⇒ [[/andP [GE _] _] _].
          by apply/andP; split; last rewrite /A subnKC // addn1.
        }
        apply leq_trans with (task_workload_between t1 (t1+A+ε) - job_cost j + cumul_task_interference t2 t1 y).
        - by apply cumulative_job_interference_le_task_interference_bound.
        - rewrite leq_add2r.
          eapply task_rbf_excl_tsk_bounds_task_workload_excl_j; eauto 2.
      Qed.

    End BoundOfCumulativeJobInterference.

In this section, we prove that H_R_is_maximum_seq implies H_R_is_maximum.
Consider any job j of tsk.
      Variable j : Job.
      Hypothesis H_j_arrives : arrives_in arr_seq j.
      Hypothesis H_job_of_tsk : job_of_task tsk j.

For simplicity, let's define a local name for the search space.
      Let is_in_search_space A :=
        is_in_search_space tsk L total_interference_bound A.

We prove that H_R_is_maximum holds.
      Lemma max_in_seq_hypothesis_implies_max_in_nonseq_hypothesis:
         (A : duration),
          is_in_search_space A
           (F : duration),
            A + F task_rtct tsk +
                    (task_rbf (A + ε) - task_cost tsk + task_interference_bound_function tsk A (A + F))
            R F + (task_cost tsk - task_rtct tsk).
      Proof.
        move: H_valid_run_to_completion_threshold ⇒ [PRT1 PRT2].
        intros A INSP.
        clear H_sequential_tasks H_interference_and_workload_consistent_with_sequential_tasks.
        move: (H_R_is_maximum_seq _ INSP) ⇒ [F [FIX LE]].
         F; split; last by done.
        rewrite -{2}(leqRW FIX).
        rewrite addnA leq_add2r.
        rewrite addnBA; last first.
        { apply leq_trans with (task_rbf 1).
          eapply task_rbf_1_ge_task_cost; eauto 2.
          eapply task_rbf_monotone; eauto 2.
          by rewrite addn1.
        }
        by rewrite subnBA; auto; rewrite addnC.
      Qed.

    End MaxInSeqHypothesisImpMaxInNonseqHypothesis.

Finally, we apply the uniprocessor_response_time_bound theorem, and using the lemmas above, we prove that all the requirements are satisfied. So, R is a response time bound.
    Theorem uniprocessor_response_time_bound_seq:
      response_time_bounded_by tsk R.
    Proof.
      intros j ARR TSK.
      eapply uniprocessor_response_time_bound with
          (interference_bound_function :=
             fun tsk A Rtask_rbf (A + ε) - task_cost tsk + task_interference_bound_function tsk A R); eauto 2.
      apply ideal_proc_model_ensures_ideal_progress.
      apply ideal_proc_model_provides_unit_service.
      { clear ARR TSK H_R_is_maximum_seq R j.
        intros t1 t2 R j BUSY NEQ ARR TSK COMPL.
        move: (posnP (@job_cost _ H3 j)) ⇒ [ZERO|POS].
        { exfalso.
          move: COMPL ⇒ /negP COMPL; apply: COMPL.
          by rewrite /service.completed_by /completed_by ZERO.
        }
        set (A := job_arrival j - t1) in ×.
        apply leq_trans with
            (task_rbf (A + ε) - task_cost tsk + cumul_task_interference t2 t1 (t1 + R)).
        - by eapply cumulative_job_interference_bound; eauto 2.
        - by rewrite leq_add2l; apply H_task_interference_is_bounded.
      }
      { by eapply max_in_seq_hypothesis_implies_max_in_nonseq_hypothesis; eauto. }
    Qed.

  End ResponseTimeBound.

End Sequential_Abstract_RTA.