Library prosa.analysis.facts.tdma

From mathcomp Require Import div.
Require Export prosa.model.schedule.tdma.
Require Import prosa.util.all.

In this section, we define the properties of TDMA and prove some basic lemmas.

Section TDMAFacts.
Context {Task:eqType}.

Consider any task set ts...

Variable ts: {set Task }.

... with a TDMA policy

Context `{TDMAPolicy Task}.

Section TimeSlotFacts.

Consider any task in ts

Variable task: Task.
Hypothesis H_task_in_ts: task \in ts.

Assume task_time_slot is valid time slot

Hypothesis time_slot_positive:
valid_time_slot ts.

Obviously, the TDMA cycle is greater or equal than any task time slot which is in TDMA cycle

    Lemma TDMA_cycle_ge_each_time_slot:
      TDMA_cycle ts ≥ task_time_slot task.
    Proof.
      rewrite /TDMA_cycle (big_rem task) //.
      apply:leq_trans; last by exact: leq_addr.
        by apply leqnn.
    Qed.

Thus, a TDMA cycle is always positive

    Lemma TDMA_cycle_positive:
      TDMA_cycle ts > 0.
    Proof.
      move:time_slot_positive=>/(_ task H_task_in_ts)/leq_trans;apply;apply TDMA_cycle_ge_each_time_slot.
    Qed.

Slot offset is less then cycle

    Lemma Offset_lt_cycle:
      task_slot_offset ts task < TDMA_cycle ts.
    Proof.
      rewrite /task_slot_offset /TDMA_cycle big_mkcond.
      apply leq_ltn_trans with (n:=\sum_(prev_task <- ts )if prev_task !=task then task_time_slot prev_task else 0).
      - apply leq_sum. intros× T. case (slot_order i task);auto.
      - rewrite -big_mkcond (bigD1_seq task)?set_uniq//=.
        rewrite -subn_gt0 -addnBA. rewrite subnn addn0 //.
        exact: time_slot_positive.
        easy.
    Qed.

For a task, the sum of its slot offset and its time slot is less then or equal to cycle.

    Lemma Offset_add_slot_leq_cycle:
      task_slot_offset ts task + task_time_slot task ≤ TDMA_cycle ts.
    Proof.
      rewrite /task_slot_offset /TDMA_cycle.
      rewrite addnC (bigD1_seq task) //=. rewrite leq_add2l.
      rewrite big_mkcond.
      replace (\sum_(i <- ts | i != task ) task_time_slot i)
        with (\sum_(i <- ts ) if i != task then task_time_slot i else 0).
      apply leq_sum. intros×T. case (slot_order i task);auto.
        by rewrite -big_mkcond. apply (set_uniq ts).
    Qed.
  End TimeSlotFacts.

Now we prove that no two tasks share the same time slot at any time.

Section TimeSlotOrderFacts.

Consider any task in ts

Variable task: Task.
Hypothesis H_task_in_ts: task \in ts.

Assume task_time_slot is valid time slot

Hypothesis time_slot_positive:
valid_time_slot ts.

Assume that slot order is total...

    Hypothesis slot_order_total:
      total_slot_order ts.

    (*..., antisymmetric... *)
    Hypothesis slot_order_antisymmetric:
      antisymmetric_slot_order ts.

    (*... and transitive. *)
    Hypothesis slot_order_transitive:
      transitive_slot_order.

Then, we can prove that the difference value between two offsets is at least a slot

    Lemma relation_offset:
      ∀ tsk1 tsk2, tsk1 \in ts →
                                 tsk2 \in ts →
                                          slot_order tsk1 tsk2 →
                                          tsk1 != tsk2 →
                                          task_slot_offset ts tsk2 ≥
                                          task_slot_offset ts tsk1 + task_time_slot tsk1 .
    Proof.
      intros× IN1 IN2 ORDER NEQ.
      rewrite /task_slot_offset big_mkcond addnC/=.
      replace (\sum_(tsk <- ts | slot_order tsk tsk2 && (tsk != tsk2)) task_time_slot tsk)
        with (task_time_slot tsk1 + \sum_(tsk <- ts )if slot_order tsk tsk2 && (tsk != tsk1) && (tsk !=tsk2) then task_time_slot tsk else O).
      rewrite leq_add2l. apply leq_sum_seq. intros× IN T.
      case (slot_order i tsk1)eqn:SI2;auto. case (i==tsk1)eqn:IT2;auto;simpl.
      case (i==tsk2)eqn:IT1;simpl;auto.
      - by move/eqP in IT1;rewrite IT1 in SI2;apply slot_order_antisymmetric in ORDER;auto;apply ORDER in SI2;move/eqP in NEQ.
      - by rewrite (slot_order_transitive _ _ _ SI2 ORDER).
      - symmetry. rewrite big_mkcond /=. rewrite→bigD1_seq with (j:=tsk1);auto;last by apply (set_uniq ts).
        move/eqP /eqP in ORDER. move/eqP in NEQ. rewrite ORDER //=. apply /eqP.
        have TS2: (tsk1 != tsk2) = true . apply /eqP;auto. rewrite TS2.
        rewrite eqn_add2l. rewrite big_mkcond. apply/eqP. apply eq_bigr;auto.
        intros× T. case(i!=tsk1);case (slot_order i tsk2);case (i!=tsk2) ;auto.
    Qed.

Then, we proved that no two tasks share the same time slot at any time.

    Lemma task_in_time_slot_uniq:
      ∀ tsk1 tsk2 t,
        tsk1 \in ts → task_time_slot tsk1 > 0 →
        tsk2 \in ts → task_time_slot tsk2 > 0 →
        task_in_time_slot ts tsk1 t →
        task_in_time_slot ts tsk2 t →
        tsk1 = tsk2.
    Proof.
      intros × IN1 SLOT1 IN2 SLOT2.
      rewrite /task_in_time_slot.
      set cycle := TDMA_cycle ts.
      set O1 := task_slot_offset ts tsk1.
      set O2 := task_slot_offset ts tsk2.
      have CO1 : O1 < cycle by apply Offset_lt_cycle.
      have CO2 : O2 < cycle by apply Offset_lt_cycle.
      have C : cycle > 0 by apply (TDMA_cycle_positive tsk1).
      have → : O1 %% cycle = O1 by apply modn_small, Offset_lt_cycle.
      have → : O2 %% cycle = O2 by apply modn_small, Offset_lt_cycle.
      have SO1 : O1 + task_time_slot tsk1 ≤ cycle by apply (Offset_add_slot_leq_cycle tsk1).
      have SO2 : O2 + task_time_slot tsk2 ≤ cycle by apply (Offset_add_slot_leq_cycle tsk2).
      repeat rewrite mod_elim; auto.
      case (O1 ≤ t %% cycle) eqn:O1T; case (O2 ≤ t %% cycle) eqn:O2T; intros G1 G2; try lia.
      rewrite ltn_subLR // in G1; rewrite ltn_subLR // in G2.
      case (tsk1 == tsk2) eqn:NEQ; move/eqP in NEQ; auto.
      destruct (slot_order_total tsk1 tsk2) as [order | order]; auto.
      all: by apply relation_offset in order; fold O1 O2 in order; try lia; auto; apply/eqP; auto.
    Qed.

  End TimeSlotOrderFacts.

End TDMAFacts.