
(**

{1 VerifyThis @ ETAPS 2016 competition, Challenge 2: Binary Tree Traversal}

{h

The following is the original description of the verification task,
reproduced verbatim from
<a href="http://http://etaps2016.verifythis.org/challenges">the competition web site</a>

<pre>

Challenge 2: Binary Tree Traversal

Consider a binary tree:

  class Tree {
    Tree left, right, parent;
    bool mark;
  }

We are given a binary tree with the following properties:
It is well formed, in the sense that following a child pointer (left or right) and then following a parent pointer brings us to the original node. Moreover, the parent pointer of the root is null.
It has at least one node, and each node has 0 or 2 children.
We do not know the initial value of the mark fields.

Our goal is to set all mark fields to true. The algorithm below (Morris 1979) works in time linear in the number of nodes, as usual, but uses only a constant amount of extra space.

  void markTree(Tree root) {
    Tree x, y;
    x = root;
    do {
          x.mark = true;
          if (x.left == null && x.right == null) {
                  y = x.parent;
          } else {
                y = x.left;
                x.left = x.right;
                x.right = x.parent;
                x.parent = y;
          }
          x = y;
    } while (x != null);
  }

Tasks. Prove that:
upon termination of the algorithm, all mark fields are set
the tree shape does not change
the code does not crash, and
the code terminates.

As a bonus, prove that the nodes are visited in depth-first order
</pre>}

The following is a solution by Martin Clochard (Université Paris-Sud)
who entered the competition.

*)

(** Component-as-array memory model with null pointers. *)
module Memory

  use map.Map
  (** `loc` is the type of memory locations (e.g pointers) *)
  type loc
  val predicate eq (x y:loc) ensures { result <-> x = y }
  (** Kinds of pointer fields. *)
  type kind = Parent | Left | Right
  (** Convenience alias for a pointer field table. *)
  type pmem = map kind (map loc loc)
  type memory = abstract {
    mutable accessor : pmem;
    mutable mark : map loc bool;
  }
  (** `memo` represent memory. *)
  val memo : memory
  (** the memory model has a `null` pointer. *)
  constant null : loc
  val null () : loc ensures { result = null }
  val is_null (l:loc) : bool ensures { result <-> l = null }
  (** Memory getter & setters. non-null preconditions to
      check absence of null pointer access. *)
  val get_mark (l:loc) : bool
    requires { l <> null }
    reads { memo }
    ensures { result = memo.mark[l] }
  val set_mark (l:loc) (b:bool) : unit
    requires { l <> null }
    writes { memo.mark }
    ensures { memo.mark = (old memo).mark[l <- b] }
  val get_acc (p:kind) (l:loc) : loc
    requires { l <> null }
    reads { memo }
    ensures { result = memo.accessor[p][l] }
  val set_acc (p:kind) (l d:loc) : unit
    requires { l <> null }
    writes { memo.accessor }
    ensures { memo.accessor =
      (old memo).accessor[p <- (old memo).accessor[p][l <- d]] }
  (** Ghost global accessors. Those are technical tools
      to create ghost witnesses of past states. *)
  val ghost get_all_accs () : pmem
    reads { memo }
    ensures { result = memo.accessor }
  val ghost get_all_marks () : map loc bool
    reads { memo }
    ensures { result = memo.mark }

  (** Non-deterministic initialization. In the original alorithm,
      the variable y starts uninitialized. *)
  type non_det_magic
  type non_det = abstract { mutable non_det_field : non_det_magic }
  val non_det : non_det
  val anyloc () : loc writes { non_det }

end

(** In module `TreeShape` we describe the correlation between
    the memory and the binary tree model. We also gives
    elements of the algorithm specification, and prove
    some frame/separation properties. *)
module TreeShape

  use int.Int
  use set.Set
  use map.Map
  use bintree.Tree
  use bintree.Size
  use Memory

  (** The described structure can be modeled as a tree of locations *)
  type treel = tree loc

  (** `is_tree memo t c p` describe the fact that `c` points to
      a well-formed tree modeled by `t`, whose root parent node is `p`.  *)
  predicate is_tree (memo:pmem) (t:treel) (c p:loc) = match t with
    | Empty -> c = null
    | Node l m r -> c <> null /\ c = m /\ memo[Parent][c] = p /\
      let cl = memo[Left][c] in
      let cr = memo[Right][c] in
      (cl = null <-> cr = null) /\
      is_tree memo l cl c /\ is_tree memo r cr c
    end

  (** `footprint t` is the memory footprint of `t`, e.g the set of
      locations occuring in the tree. *)
  function footprint (t:treel) : set loc = match t with
    | Empty -> empty
    | Node l m r -> add m (union (footprint l) (footprint r))
    end

  (** `ext s memo1 memo2` mean that the pointer fields associated to
      locations in set `s` are identical in memo1 and memo2. *)
  predicate ext (s:set loc) (memo1 memo2:pmem) =
    forall k x. mem x s -> memo1[k][x] = memo2[k][x]
  (** `unchanged memo1 memo2` mean that all pointer fields in
      `memo1` and `memo2` are identical. *)
  predicate unchanged (memo1 memo2:pmem) =
    forall k x. memo1[k][x] = memo2[k][x]
  (** `was_marked t memo1 memo2` mean that `memo2` is the update
      of `memo1` obtained by marking all elements in `t`. *)
  predicate was_marked (t:treel) (memo1 memo2:map loc bool) =
    (forall l. mem l (footprint t) -> memo2[l]) /\
    (forall l. not mem l (footprint t) -> memo2[l] = memo1[l])

  (** General extensionality property. *)
  let rec ghost ext_set (s:set loc) (memo1 memo2:pmem)
                                    (t:treel) (c p:loc) : unit
    requires { ext s memo1 memo2 }
    requires { subset (footprint t) s }
    requires { is_tree memo1 t c p }
    ensures { is_tree memo2 t c p }
    variant { t }
  = let ghost rc = ext_set s memo1 memo2 in
    match t with
    | Empty -> ()
    | Node l _ r -> rc l (memo1[Left][c]) c; rc r (memo1[Right][c]) c
    end

  (** Specialized for our use case. *)
  let ghost ext_cur (memo1:pmem) (t:treel) (c p:loc) : unit
    reads { memo }
    requires { ext (footprint t) memo1 memo.accessor }
    requires { is_tree memo1 t c p }
    ensures { is_tree memo.accessor t c p }
  = ext_set (footprint t) memo1 (get_all_accs ()) t c p

  (** The tree model corresponding to a given pointer is unique. *)
  let rec ghost unicity (memo:pmem) (t1 t2:treel) (c p1 p2:loc) : unit
    requires { is_tree memo t1 c p1 /\ is_tree memo t2 c p2 }
    ensures { t1 = t2 }
    variant { t1 }
  = match t1, t2 with
    | Empty, Empty -> ()
    | Empty, _ | _, Empty -> absurd
    | Node l1 _ r1, Node l2 _ r2 ->
        unicity memo l1 l2 (memo[Left][c]) c c;
        unicity memo r1 r2 (memo[Right][c]) c c
    end

  (** In a non-empty well formed tree, the top pointer
      cannot occur in the child subtree. Otherwise,
      their would be an infinite branch in the tree,
      which is impossible in our inductive tree setting. *)
  let ghost not_below (memo:pmem) (lf rg:treel) (c p:loc) : unit
    requires { is_tree memo (Node lf c rg) c p }
    ensures { not mem c (footprint lf) /\ not mem c (footprint rg) }
  = let t0 = Node lf c rg in
    let rec aux (t:treel) (c2 p2:loc) : unit
      requires { size t < size t0 }
      requires { is_tree memo t c2 p2 }
      ensures { not mem c (footprint t) }
      variant { t }
    = match t with
      | Empty -> ()
      | Node l _ r ->
        if eq c2 c then (unicity memo t t0 c p2 p;absurd);
        aux l (memo[Left][c2]) c2; aux r (memo[Right][c2]) c2
      end in
    aux lf (memo[Left][c]) c; aux rg (memo[Right][c]) c

  (** Algorithm phases.
      `GoLeft` mean that the algorithm will try to explore the left
        subtree
      `GoRight` mean that the algorithm will explore the right subtree
      `GoBack` mean that the algorithm will go back to the parent node
      `Finish` mean that the alogrithm will exit *)
  type phase =
    | GoLeft
    | GoRight
    | GoBack
    | Finish

  function next_phase (ph:phase) : phase =
    match ph with
    | GoLeft -> GoRight
    | GoRight -> GoBack
    | GoBack -> GoLeft
    | Finish -> Finish
    end
  (** `rotated memo1 memo2 ph c` describe how `c` is rotated in
      `memo2` with respect to its initial position in `memo1` for
      current phase `ph`. In final phase, it is not rotated but
      null instead. *)
  predicate rotated (memo1 memo2:pmem) (ph:phase) (c:loc) =
    (forall k x. x <> c -> memo1[k][x] = memo2[k][x]) /\
    (ph <> Finish -> c <> null) /\
    match ph with
    | GoLeft -> unchanged memo1 memo2
    | GoRight -> memo2[Left][c] = memo1[Right][c] /\
        memo2[Right][c] = memo1[Parent][c] /\
        memo2[Parent][c] = memo1[Left][c]
    | GoBack -> memo1[Left][c] = memo2[Right][c] /\
        memo1[Right][c] = memo2[Parent][c] /\
        memo1[Parent][c] = memo2[Left][c]
    | Finish -> c = null
    end

end

(** In this module, we prove the algorithm by modifying
    the code to put it in recursive form. We justify that
    our limited usage of recursion makes the code equivalent
    to the one proposed in the challenge. *)
module Recursive

  use map.Map
  use bintree.Tree
  use ref.Ref
  use Memory
  use TreeShape

  (** Recursion-based proof of the algorithm.
      The principal idea is the following: in its recursive
      fashion, the algorithm is a standard tree traversal
      (which is easy to prove correct).
      Hence we nest the algorithm inside a recursive procedure
      to link it with its recursive version. However,
      the obtained algorithm does not really need the call stack
      to run. Here is how we achieve this:
      - We create a non-recursive sub-routine `bloc` corresponding to
        one turn of the while loop in the challenge's code.
        It uses an exception to simulate exit.
        Also, this sub-routine does not have any non-ghost argument,
        so calling `bloc` really amounts to advancing in the loop execution.
      - We forbids use of any side-effect in the recursive procedure
        except those obtained by calling `bloc`. Hence calling
        the recursive procedure is equivalent to run a certain
        amounts of turns in the loop execution.
        Note that since we will prove that the recursive procedure
        terminates, it might not run a finite amount of turns and diverges
        afterward.
      - After the topmost call to the recursive procedure,
        we add an infinite loop who only calls `bloc` at every turns.
        This call and the loop are enclosed in an exception catching
        construction to detect loop termination.
      Hence it is justifiable that the algorithm with
      the recursive procedure is equivalent (in the sense that
      they have the same side-effects) as an infinite loop of `bloc` calls,
      encapsulated in an exception catching expression. And this
      algorithm is evidently equivalent to the original challenge code.
    *)

  exception Stop

  let markTree (ghost t:treel) (root:loc) : unit
    (* Requires a well-formed non-empty tree *)
    requires { is_tree memo.accessor t root null /\ root <> null }
    writes { memo, non_det }
    (* Tree shape is unchanged *)
    ensures { is_tree memo.accessor t root null }
    ensures { unchanged (old memo).accessor memo.accessor }
    (* All nodes are marked. *)
    ensures { was_marked t (old memo).mark memo.mark }
  = let x = ref (anyloc ()) in (* ==> Tree x, y *)
    let y = ref (anyloc ()) in
    x := root; (* ==> x = root *)
    let entered = ref false in
    (* ^ Used to encode the do { .. } while() loop as a while() {..}.
       Iterations will set the flag. *)
    let ghost z = ref (null ()) in (* Store previously explored node. *)
    (* Loop body. `mem0` is memory state at start of recursive call,
       `ph` is phase. *)
    let bloc (ghost mem0:pmem) (ghost ph:phase) : unit
      (* current node `!x` is rotated correctly with
         respect to the phase. *)
      requires { rotated mem0 memo.accessor ph !x }
      requires { ph = Finish -> !entered }
      writes { memo, x, y, z, entered }
      (* Caracterise `x` and `z` reference updates. *)
      ensures { !z = !(old x) /\ !x = memo.accessor[Parent][!z] }
      (* `Finish` phase never ends normally *)
      ensures { ph <> Finish /\ !entered }
      (* Node is marked, and other nodes do not change. *)
      ensures { memo.mark[!z] }
      ensures { forall l. l <> !z -> memo.mark[l] = (old memo).mark[l] }
      (* Describe phase shift. *)
      ensures {
        if (old memo).accessor[Left][!z] = null
           = (old memo).accessor[Right][!z]
        then memo.accessor = (old memo).accessor
        else rotated mem0 memo.accessor (next_phase ph) !z }
      (* In `Finish` phase, the code throws `Stop` without changing
         anything. *)
      raises { Stop -> ph = Finish /\ memo = old memo }
    = if !entered && is_null !x then raise Stop; (* ==> do { BODY         *)
      entered := true;                           (*  } while (x != null); *)
      (ghost z := !x);
      set_mark !x true; (* ==> x.mark = true; *)
      if is_null (get_acc Left !x) && is_null (get_acc Right !x)
         (* ==> if (x.left = null && x.right == null) { *)
      then y := get_acc Parent !x (* ==> y = x.parent *)
      else begin (* ==> } else { *)
        y := get_acc Left !x; (* ==> y = x.left; *)
        set_acc Left !x (get_acc Right !x); (* ==> x.left = x.right; *)
        set_acc Right !x (get_acc Parent !x); (* ==> x.right = x.parent; *)
        set_acc Parent !x !y; (* ==> x.parent = y; *)
      end; (* ==> } *)
      x := !y (* ==> x = y; *)
    in
    (* 'Recursive proof', corresponding to the
        expected depth-first traversal. *)
    let rec aux (ghost t:treel) : unit
      requires { !x <> null }
      requires { is_tree memo.accessor t !x !z }
      writes { memo, x, y, z, entered }
      ensures { unchanged (old memo).accessor memo.accessor }
      ensures { !x = old !z /\ !z = old !x }
      ensures { was_marked t (old memo).mark memo.mark }
      ensures { !entered }
      raises { Stop -> false }
      variant { t }
    = let ghost mem0 = get_all_accs () in
      let ghost _c = !x in
      let _lf = get_acc Left !x in
      let _rg = get_acc Right !x in
      let lf, rg = ghost match t with
        | Empty -> absurd
        | Node l _ r -> l, r
        end in
      (ghost not_below mem0 lf rg _c !z);
      bloc mem0 GoLeft;
      let b = begin ensures { result <-> _lf = null /\ _rg = null }
        ensures { result -> lf = Empty /\ rg = Empty }
        is_null _lf && is_null _rg end in
      if not b then begin
        (ghost ext_cur mem0 lf _lf _c);
        aux lf;
        bloc mem0 GoRight;
        (ghost ext_cur mem0 rg _rg _c);
        aux rg;
        bloc mem0 GoBack
      end
    in
    let ghost mem0 = get_all_accs () in
    try aux t;
      label I in
      while true do
        invariant { memo = (memo at I) }
        invariant { rotated mem0 memo.accessor Finish !x /\ !entered }
        variant { 0 }
        bloc mem0 Finish
      done
    with Stop -> (ghost ext_cur mem0 t root (null ())); () end

end

(** In this module we provide a proof of the initial algorithm by
    derecursiving the previous one *)
module Iterative

  use int.Int
  use map.Map
  use map.Const
  use bintree.Tree
  use bintree.Size
  use option.Option
  use ref.Ref
  use Memory
  use TreeShape

  (** Snapshot of all relevant memory values in the program *)
  type snap = {
    pointers : pmem;
    cursor : loc;
    parent : loc;
    marks : map loc bool;
  }

  (** Stack frame in the recursive version. Fields are
      assigned as the code pointer increase. *)
  type frame = {
    (* Memory & ghost argument at call site (pc>=0). *)
    memo0 : snap;
    tree : treel;
    (* right & left trees/pointers (pc>=1) *)
    tleft : treel;
    pleft : loc;
    tright : treel;
    pright : loc;
    (* Memory after first bloc call (pc>=2) *)
    memo1 : snap;
    (* Memory after first recursive call (pc >= 4) *)
    memo2 : snap;
    (* Memory after second bloc call (pc >= 5) *)
    memo3 : snap;
    (* Memory after second recursive call (pc >= 6) *)
    memo4 : snap;
  }

  (** Find out current memory state with respect to code pointer. *)
  function frame_memo (f:frame) (pc:int) : snap =
    if pc <= 0 then f.memo0
    else if pc <= 1 then f.memo1
    else if pc <= 2 then f.memo2
    else if pc <= 3 then f.memo3
    else f.memo4

  (** Postcondition relation for `bloc` calls. *)
  predicate bloc_rel (mem0:pmem) (ph:phase) (s1 s2:snap) =
    s2.parent = s1.cursor /\ s2.cursor = s2.pointers[Parent][s2.parent] /\
    s2.marks[s2.parent] /\
    (forall l. l <> s2.parent -> s2.marks[l] = s1.marks[l]) /\
    if s1.pointers[Left][s1.cursor] = null = s1.pointers[Right][s1.cursor]
    then s2.pointers = s1.pointers
    else rotated mem0 s2.pointers (next_phase ph) s2.parent

  (** postcondition relation for recursive (`aux`) calls. *)
  predicate rec_rel (t:treel) (s1 s2:snap) =
    unchanged s1.pointers s2.pointers /\
    s2.cursor = s1.parent /\ s2.parent = s1.cursor /\
    was_marked t s1.marks s2.marks

  (** Call stack *)
  type stack =
    | Bottom
    | Running stack int frame
    | Done

  (** Describe a valid call stack. Mostly precise which relation
      between states holds. Note that in the previous version,
      Why3 did that part for us via the weakest precondition calculus. *)
  predicate is_stack (t:treel) (stop scur:snap)
                     (s:stack) (calls:option treel) =
    match s with
    | Bottom -> stop = scur /\ calls = Some t
    | Running s pc f ->
        0 <= pc <= 4 /\
        (* Correctness of caller stack. *)
        let m0 = f.memo0 in
        is_stack t stop m0 s (Some f.tree) /\
        (* Precondition for recursive call. *)
        m0.cursor <> null /\ is_tree m0.pointers f.tree m0.cursor m0.parent /\
        (* Initially obtained pointers & subtrees. *)
        f.tree = Node f.tleft m0.cursor f.tright /\
        f.pleft = m0.pointers[Left][m0.cursor] /\
        f.pright = m0.pointers[Right][m0.cursor] /\
        (* First bloc run, if completed normally. *)
        (pc >= 1 -> bloc_rel m0.pointers GoLeft m0 f.memo1 /\
                    f.pleft <> null /\ f.pright <> null) /\
        (* First recursive call. *)
        (pc >= 2 -> rec_rel f.tleft f.memo1 f.memo2) /\
        (* Second bloc run. *)
        (pc >= 3 -> bloc_rel m0.pointers GoRight f.memo2 f.memo3) /\
        (* Second recursive call. *)
        (pc >= 4 -> rec_rel f.tright f.memo3 f.memo4) /\
        (* Current memory state. *)
        frame_memo f pc = scur /\
        match calls with
        | None -> pc <> 1 /\ pc <> 3
        | Some u ->
            if pc = 1 then u = f.tleft else pc = 3 /\ u = f.tright
        end
    | Done -> rec_rel t stop scur /\ calls = None
    end

  (** Termination argument. In fully general cases, we would need
     a lexicographic ordering but here an integer size suffice. *)
  constant large_enough : int = 100
  function stack_size (st:stack) : int = match st with
    | Bottom -> 1
    | Done -> 0
    | Running s pc f ->
        stack_size s + (large_enough - pc) + if pc = 0
        then large_enough * (size f.tleft + size f.tright)
        else if pc <= 2
        then large_enough * size f.tright
        else 0
    end

  let rec lemma stack_size_pos (st:stack) : unit
    requires { exists t stop scur calls. is_stack t stop scur st calls }
    ensures { stack_size st >= 0 }
    variant { st }
  = match st with Running s _ _ -> stack_size_pos s | _ -> () end

  (** Create a stack frame for a recursive call *)
  let ghost opening (t ct:treel) (stop scur:snap) (ghost st:ref stack)
    requires { is_stack t stop scur !st (Some ct) }
    requires { is_tree scur.pointers ct scur.cursor scur.parent }
    requires { scur.cursor <> null }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st <= stack_size !(old st) +
                                large_enough * size ct }
  = match ct with
    | Empty -> absurd
    | Node lf _ rg ->
      let f = {
        memo0 = scur; tree = ct; tleft = lf; tright = rg;
        pleft = scur.pointers[Left][scur.cursor];
        pright = scur.pointers[Right][scur.cursor];
        memo1 = scur; memo2 = scur; memo3 = scur; memo4 = scur;
      } in
      st := Running !st 0 f
    end

  (** Remove a stack frame at end of recursive call simulation *)
  let ghost closing (t ct:treel) (stop sprev scur:snap)
                    (ghost st:ref stack) : unit
    requires { is_stack t stop sprev !st (Some ct) }
    requires { rec_rel ct sprev scur }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st < stack_size !(old st) }
  = match !st with
    | Bottom -> st := Done
    | Done -> absurd
    | Running s pc f ->
      let f = if pc = 1 then { f with memo2 = scur }
      else { f with memo4 = scur } in
      st := Running s (pc+1) f
    end

  (** Advance code pointer when a bloc is run. Takes care to
      open/close new frame as needed. *)
  let ghost continuing (t:treel) (stop sprev scur:snap)
                       (ghost st:ref stack)
    requires { is_stack t stop sprev !st None }
    requires { match !st with
      | Bottom | Done -> false
      | Running _ pc f ->
          let ph =
            if pc = 0 then GoLeft else if pc = 2 then GoRight else GoBack in
          bloc_rel f.memo0.pointers ph sprev scur
      end }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st < stack_size !(old st) }
  = match !st with
    | Bottom | Done -> absurd
    | Running s pc f ->
      not_below f.memo0.pointers f.tleft f.tright
                f.memo0.cursor f.memo0.parent;
      if pc = 0 then
        if is_null f.pleft && is_null f.pright
        then begin
          st := s;
          assert { match f.tleft, f.tright with
            Empty, Empty  -> true | _ -> false end };
          closing t f.tree stop f.memo0 scur st
        end else begin
          let f = { f with memo1 = scur } in
          st := Running s (pc+1) f;
          ext_set (footprint f.tleft) f.memo0.pointers scur.pointers
                  f.tleft f.pleft f.memo0.cursor;
          opening t f.tleft stop scur st;
        end
      else if pc = 2 then begin
        let f = { f with memo3 = scur } in
        st := Running s (pc+1) f;
        ext_set (footprint f.tright) f.memo0.pointers scur.pointers
                f.tright f.pright f.memo0.cursor;
        opening t f.tright stop scur st;
      end else if pc = 4 then begin
        st := s;
        assert { unchanged scur.pointers f.memo0.pointers };
        closing t f.tree stop f.memo0 scur st
      end
    end

  (** The main algorithm. *)
  let markTree (ghost t:treel) (root:loc) : unit
    (* Requires a well-formed non-empty tree *)
    requires { is_tree memo.accessor t root null /\ root <> null }
    writes { memo, non_det }
    (* Tree shape is unchanged *)
    ensures { is_tree memo.accessor t root null }
    ensures { unchanged (old memo).accessor memo.accessor }
    (* All nodes are marked. *)
    ensures { was_marked t (old memo).mark memo.mark }
  = let x = ref (anyloc ()) in (* ==> Tree x, y *)
    let y = ref (anyloc ()) in
    let ghost z = ref (null ()) in
    let ghost snapshot () : snap
      ensures { result.pointers = memo.accessor }
      ensures { result.cursor = !x }
      ensures { result.parent = !z }
      ensures { result.marks = memo.mark }
    = { pointers = get_all_accs ();
        cursor = !x; parent = !z;
        marks = get_all_marks () } in
    x := root; (* ==> x = root; *)
    let ghost stop = snapshot () in
    let ghost scur = ref stop in
    let ghost st = ref Bottom in
    ghost opening t t stop stop st;
    let entered = ref false in (* encode do-while loop. *)
    while not (begin ensures { result <-> !x = null }
        !entered && is_null !x end)
    do (* ==> do { BODY } while(x != null); *)
      invariant { !x = null -> !entered }
      invariant { !scur.pointers = memo.accessor }
      invariant { !scur.cursor = !x }
      invariant { !scur.parent = !z }
      invariant { !scur.marks = memo.mark }
      invariant { is_stack t stop !scur !st None }
      variant { stack_size !st }
      entered := true;
      (ghost z := !x);
      set_mark !x true; (* ==> x.mark = true; *)
      if is_null (get_acc Left !x) && is_null (get_acc Right !x)
        (* ==> if (x.left == null && x.right == null) { *)
      then begin
        y := get_acc Parent !x; (* ==> y = x.parent; *)
      end
      else begin (* ==> } else { *)
        y := get_acc Left !x; (* ==> y = x.left; *)
        set_acc Left !x (get_acc Right !x); (* ==> x.left = x.right; *)
        set_acc Right !x (get_acc Parent !x); (* ==> x.right = x.parent; *)
        set_acc Parent !x !y; (* ==> x.parent = y; *)
      end; (* ==> } *)
      x := !y; (* ==> x = y; *)
      let ghost sprev = !scur in
      ghost scur := snapshot ();
      ghost continuing t stop sprev !scur st;
    done; (* ==> end 'do while' *)
    ghost (match !st with Done -> () | _ -> absurd end);
    ghost ext_cur stop.pointers t root (null ())

end
