We consider the system from.

  Alphabet:

    chain : [N -> N * list] --> list 
    cons : [N * list] --> list 
    false : [] --> B 
    from : [N * list] --> list 
    if : [B * list * list] --> list 
    incch : [list] --> list 
    lteq : [N * N] --> B 
    nil : [] --> list 
    o : [] --> N 
    s : [N] --> N 
    true : [] --> B 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    lteq(s(x), o) => false 
    lteq(o, x) => true 
    lteq(s(x), s(y)) => lteq(x, y) 
    from(x, nil) => nil 
    from(x, cons(y, z)) => if(lteq(x, y), cons(y, z), from(x, z)) 
    chain(f, nil) => nil 
    chain(f, cons(x, y)) => cons(f x, chain(f, from(f x, y))) 
    incch(x) => chain(/\y.s(y), x) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] lteq#(s(X), s(Y)) =#> lteq#(X, Y)   
    1] from#(X, cons(Y, Z)) =#> if#(lteq(X, Y), cons(Y, Z), from(X, Z))   
    2] from#(X, cons(Y, Z)) =#> lteq#(X, Y)   
    3] from#(X, cons(Y, Z)) =#> from#(X, Z)   
    4] chain#(F, cons(X, Y)) =#> chain#(F, from(F X, Y))   
    5] chain#(F, cons(X, Y)) =#> from#(F X, Y)   
    6] incch#(X) =#> chain#(/\x.s(x), X)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    lteq(s(X), o) => false 
    lteq(o, X) => true 
    lteq(s(X), s(Y)) => lteq(X, Y) 
    from(X, nil) => nil 
    from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
    chain(F, nil) => nil 
    chain(F, cons(X, Y)) => cons(F X, chain(F, from(F X, Y))) 
    incch(X) => chain(/\x.s(x), X) 

Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 :  
    * 2 : 0 
    * 3 : 1, 2, 3 
    * 4 : 4, 5 
    * 5 : 1, 2, 3 
    * 6 : 4, 5 

This graph has the following strongly connected components:

  P_1:

    lteq#(s(X), s(Y)) =#> lteq#(X, Y)   

  P_2:

    from#(X, cons(Y, Z)) =#> from#(X, Z)   

  P_3:

    chain#(F, cons(X, Y)) =#> chain#(F, from(F X, Y))   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_3, R_0, static, formative) is finite.

We consider the dependency pair problem (P_3, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[#argfun-incch#(x_1)]] = x_1 
  [[chain#(x_1, x_2)]] = x_2 
  [[false]] = _|_ 
  [[from(x_1, x_2)]] = from(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[incch(x_1)]] = x_1 
  [[lteq(x_1, x_2)]] = lteq 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, chain, cons, from, if, lteq, nil, o, s}, and the following precedence: lteq > nil > o > s > chain > cons > from > if > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  cons(X, Y) > from(Y) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  lteq >= _|_ 
  lteq >= _|_ 
  lteq >= lteq 
  from(nil) >= nil 
  from(cons(X, Y)) >= if(cons(X, Y), from(Y)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y))) 
  chain(/\x.s(x), X) >= chain(/\x.s(x), X) 

With these choices, we have:

  1] cons(X, Y) > from(Y)  because [2], by definition 
  2] cons*(X, Y) >= from(Y)  because cons > from and [3], by (Copy) 
  3] cons*(X, Y) >= Y  because [4], by (Select) 
  4] Y >= Y  by (Meta) 

  5] if(X, Y) >= X  because [6], by (Star) 
  6] if*(X, Y) >= X  because [7], by (Select) 
  7] X >= X  by (Meta) 

  8] if(X, Y) >= Y  because [9], by (Star) 
  9] if*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 

  11] lteq >= _|_  by (Bot) 

  12] lteq >= _|_  by (Bot) 

  13] lteq >= lteq  because lteq in Mul, by (Fun) 

  14] from(nil) >= nil  because [15], by (Star) 
  15] from*(nil) >= nil  because [16], by (Select) 
  16] nil >= nil  by (Fun) 

  17] from(cons(X, Y)) >= if(cons(X, Y), from(Y))  because [18], by (Star) 
  18] from*(cons(X, Y)) >= if(cons(X, Y), from(Y))  because from > if, [19] and [23], by (Copy) 
  19] from*(cons(X, Y)) >= cons(X, Y)  because [20], by (Select) 
  20] cons(X, Y) >= cons(X, Y)  because cons in Mul, [21] and [22], by (Fun) 
  21] X >= X  by (Meta) 
  22] Y >= Y  by (Meta) 
  23] from*(cons(X, Y)) >= from(Y)  because [24], by (Select) 
  24] cons(X, Y) >= from(Y)  because [25], by (Star) 
  25] cons*(X, Y) >= from(Y)  because cons > from and [26], by (Copy) 
  26] cons*(X, Y) >= Y  because [22], by (Select) 

  27] chain(F, nil) >= nil  because [28], by (Star) 
  28] chain*(F, nil) >= nil  because [16], by (Select) 

  29] chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because [30], by (Star) 
  30] chain*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because chain > cons, [31] and [38], by (Copy) 
  31] chain*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because chain > @_{o -> o}, [32] and [34], by (Copy) 
  32] chain*(F, cons(X, Y)) >= F  because [33], by (Select) 
  33] F >= F  by (Meta) 
  34] chain*(F, cons(X, Y)) >= X  because [35], by (Select) 
  35] cons(X, Y) >= X  because [36], by (Star) 
  36] cons*(X, Y) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] chain*(F, cons(X, Y)) >= chain(F, from(Y))  because chain in Mul, [39] and [40], by (Stat) 
  39] F >= F  by (Meta) 
  40] cons(X, Y) > from(Y)  because [2], by definition 

  41] chain(/\x.s(x), X) >= chain(/\x.s(x), X)  because chain in Mul, [42] and [45], by (Fun) 
  42] /\y.s(y) >= /\y.s(y)  because [43], by (Abs) 
  43] s(x) >= s(x)  because s in Mul and [44], by (Fun) 
  44] x >= x  by (Var) 
  45] X >= X  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  from#(X, cons(Y, Z)) >? from#(X, Z) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + 2y1 
  from# = \y0y1.y1 

Using this interpretation, the requirements translate to:

  [[from#(_x0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[from#(_x0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.

We consider the dependency pair problem (P_1, R_0, static, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  lteq#(s(X), s(Y)) >? lteq#(X, Y) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  lteq# = \y0y1.y0 + y1 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[lteq#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[lteq#(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.