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Tree substitutions with respect to a cover

  • Key definitions: subst_pred
  • Initial author: Laurent.Thery@inria.fr (2003)

From Huffman Require Import HeightPred.

Set Default Proof Using "Type".

Section SubstPred.
Variable A : Type.

Take two covers and substitute the elements of one by the element of the other
Inductive subst_pred : list (btree A) -> list (btree A) -> btree A -> btree A -> Prop :=
  | subst_pred_id :
      forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
      subst_pred (t1 :: []) (t2 :: []) t1 t2
  | subst_pred_node :
      forall (t1 t2 t3 t4 : btree A) (l1 l2 l3 l4 l5 l6 : list (btree A)),
      subst_pred l1 l2 t1 t2 ->
      subst_pred l3 l4 t3 t4 ->
      subst_pred (l1 ++ l3) (l2 ++ l4) (node t1 t3) (node t2 t4).
Local Hint Resolve subst_pred_id subst_pred_node : core.

The first cover of the substitution is an ordered cover
Theorem subst_pred_ordered_cover_l :
 forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
 subst_pred l1 l2 t1 t2 -> ordered_cover l1 t1.
Proof.
intros t1 t2 l1 l2 H; elim H; auto.
Qed.

The second cover of the substitution is an ordered cover
Theorem subst_pred_ordered_cover_r :
 forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
 subst_pred l1 l2 t1 t2 -> ordered_cover l2 t2.
Proof.
intros t1 t2 l1 l2 H; elim H; auto.
Qed.

The two covers have same length
Theorem subst_pred_length :
 forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
 subst_pred l1 l2 t1 t2 -> length l1 = length l2.
Proof.
intros t1 t2 l1 l2 H; elim H; auto.
intros t0 t3 t4 t5 l0 l3 l4 l5 l6 l7 H0 H1 H2 H3; repeat rewrite app_length; auto.
Qed.

An ordered cover can be completed in a substitution
Theorem ordered_cover_subst_pred :
 forall (t1 : btree A) (l1 l2 : list (btree A)),
 ordered_cover l1 t1 ->
 length l1 = length l2 -> exists t2 : btree A, subst_pred l1 l2 t1 t2.
Proof.
intros t1 l1 l2 H; generalize l2; elim H; clear t1 l1 l2 H.
intros t l l2; case l2.
simpl in |- *; intros; discriminate.
intros b l0; case l0; simpl in |- *; auto.
intros H; exists b; auto.
intros; discriminate.
intros t1 t2 l1 l2 l3 H H0 H1 H2 l0 H3.
case (H0 (firstn (length l1) l0)); auto.
rewrite firstn_le_length_eq; auto;
 (rewrite <- H3; rewrite app_length; auto with arith).
intros t4 HH1.
case (H2 (skipn (length l1) l0)); auto.
rewrite skipn_length; auto;
 (rewrite <- H3; rewrite app_length; rewrite <- Nat.add_comm; rewrite Nat.add_sub; auto with arith).
intros t5 HH2.
exists (node t4 t5); auto.
rewrite <- (firstn_skipn (length l1) l0); auto.
Qed.

A height predicate can be completed in a substitution
Theorem height_pred_subst_pred :
 forall (n : nat) (ln : list nat) (t1 : btree A) (l1 l2 : list (btree A)),
 height_pred n ln l1 t1 ->
 length l1 = length l2 ->
 exists t2 : btree A, height_pred n ln l2 t2 /\ subst_pred l1 l2 t1 t2.
Proof.
intros n ln t1 l1 l2 H; generalize l2; elim H; clear H n ln t1 l1 l2; auto.
intros n t l2; case l2.
simpl in |- *; intros; discriminate.
intros b l0; case l0; intros; try discriminate; exists b; auto.
intros n ln1 ln2 t1 t2 l1 l2 H H0 H1 H2 l0 H3.
case (H0 (firstn (length l1) l0)); auto.
rewrite firstn_le_length_eq; auto;
 (rewrite <- H3; rewrite app_length; auto with arith).
intros t4 (HH1, HH2).
case (H2 (skipn (length l1) l0)); auto.
rewrite skipn_length; auto;
 (rewrite <- H3; rewrite app_length; rewrite <- Nat.add_comm; rewrite Nat.add_sub; auto with arith).
intros t5 (HH3, HH4).
exists (node t4 t5); rewrite <- (firstn_skipn (length l1) l0); auto.
Qed.

End SubstPred.
Arguments subst_pred [A].
#[export] Hint Resolve subst_pred_id : core.