We consider the system length.

  Alphabet:

    0 : [] --> nat 
    cons : [nat * list] --> list 
    foldr : [nat -> nat -> nat * nat * list] --> nat 
    length : [list] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 

  Rules:

    foldr(f, x, nil) => x 
    foldr(f, x, cons(y, z)) => f y foldr(f, x, z) 
    length(x) => foldr(/\y./\z.s(z), 0, 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] foldr#(F, X, cons(Y, Z)) =#> foldr#(F, X, Z)   
    1] length#(X) =#> foldr#(/\x./\y.s(y), 0, X)   

  Rules R_0:

    foldr(F, X, nil) => X 
    foldr(F, X, cons(Y, Z)) => F Y foldr(F, X, Z) 
    length(X) => foldr(/\x./\y.s(y), 0, 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 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:

  foldr#(F, X, cons(Y, Z)) >? foldr#(F, X, Z) 
  length#(X) >? foldr#(/\x./\y.s(y), 0, X) 
  foldr(F, X, nil) >= X 
  foldr(F, X, cons(Y, Z)) >= F Y foldr(F, X, Z) 
  length(X) >= foldr(/\x./\y.s(y), 0, X) 

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

  pi( length(X) ) = #argfun-length#(foldr(/\x./\y.s(y), 0, X)) 
  pi( length#(X) ) = #argfun-length##(foldr#(/\x./\y.s(y), 0, 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-length#(x_1)]] = x_1 
  [[0]] = _|_ 
  [[foldr#(x_1, x_2, x_3)]] = foldr#(x_1) 
  [[length(x_1)]] = x_1 
  [[length#(x_1)]] = length# 

We choose Lex = {} and Mul = {#argfun-length##, @_{o -> o -> o}, @_{o -> o}, cons, foldr, foldr#, length#, nil, s}, and the following precedence: foldr# > cons > s > foldr > @_{o -> o -> o} > nil > @_{o -> o} > length# > #argfun-length##

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

  foldr#(F) >= foldr#(F) 
  #argfun-length##(foldr#(/\x./\y.s(y))) > foldr#(/\x./\y.s(y)) 
  foldr(F, X, nil) >= X 
  foldr(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) 
  foldr(/\x./\y.s(y), _|_, X) >= foldr(/\x./\y.s(y), _|_, X) 

With these choices, we have:

  1] foldr#(F) >= foldr#(F)  because foldr# in Mul and [2], by (Fun) 
  2] F >= F  by (Meta) 

  3] #argfun-length##(foldr#(/\x./\y.s(y))) > foldr#(/\x./\y.s(y))  because [4], by definition 
  4] #argfun-length##*(foldr#(/\x./\y.s(y))) >= foldr#(/\x./\y.s(y))  because [5], by (Select) 
  5] foldr#(/\x./\y.s(y)) >= foldr#(/\x./\y.s(y))  because foldr# in Mul and [6], by (Fun) 
  6] /\x./\y.s(y) >= /\x./\y.s(y)  because [7], by (Abs) 
  7] /\y.s(y) >= /\y.s(y)  because [8], by (Abs) 
  8] s(x) >= s(x)  because s in Mul and [9], by (Fun) 
  9] x >= x  by (Var) 

  10] foldr(F, X, nil) >= X  because [11], by (Star) 
  11] foldr*(F, X, nil) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 

  13] foldr(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because [14], by (Star) 
  14] foldr*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because foldr > @_{o -> o}, [15] and [21], by (Copy) 
  15] foldr*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, Y)  because foldr > @_{o -> o -> o}, [16] and [17], by (Copy) 
  16] foldr*(F, X, cons(Y, Z)) >= F  because [2], by (Select) 
  17] foldr*(F, X, cons(Y, Z)) >= Y  because [18], by (Select) 
  18] cons(Y, Z) >= Y  because [19], by (Star) 
  19] cons*(Y, Z) >= Y  because [20], by (Select) 
  20] Y >= Y  by (Meta) 
  21] foldr*(F, X, cons(Y, Z)) >= foldr(F, X, Z)  because foldr in Mul, [2], [22] and [23], by (Stat) 
  22] X >= X  by (Meta) 
  23] cons(Y, Z) > Z  because [24], by definition 
  24] cons*(Y, Z) >= Z  because [25], by (Select) 
  25] Z >= Z  by (Meta) 

  26] foldr(/\x./\y.s(y), _|_, X) >= foldr(/\x./\y.s(y), _|_, X)  because foldr in Mul, [6], [27] and [28], by (Fun) 
  27] _|_ >= _|_  by (Bot) 
  28] X >= X  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

  foldr#(F, X, cons(Y, Z)) =#> foldr#(F, X, Z)   

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

  foldr#(F, X, cons(Y, Z)) >? foldr#(F, X, Z) 
  foldr(F, X, nil) >= X 
  foldr(F, X, cons(Y, Z)) >= F Y foldr(F, X, Z) 
  length(X) >= foldr(/\x./\y.s(y), 0, X) 

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

  pi( length(X) ) = #argfun-length#(foldr(/\x./\y.s(y), 0, 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:

  [[length(x_1)]] = x_1 

We choose Lex = {} and Mul = {#argfun-length#, 0, @_{o -> o -> o}, @_{o -> o}, cons, foldr, foldr#, nil, s}, and the following precedence: #argfun-length# > 0 > cons > foldr# > nil > foldr > s > @_{o -> o -> o} > @_{o -> o}

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

  foldr#(F, X, cons(Y, Z)) > foldr#(F, X, Z) 
  foldr(F, X, nil) >= X 
  foldr(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) 
  #argfun-length#(foldr(/\x./\y.s(y), 0, X)) >= foldr(/\x./\y.s(y), 0, X) 

With these choices, we have:

  1] foldr#(F, X, cons(Y, Z)) > foldr#(F, X, Z)  because [2], by definition 
  2] foldr#*(F, X, cons(Y, Z)) >= foldr#(F, X, Z)  because foldr# in Mul, [3], [4] and [5], by (Stat) 
  3] F >= F  by (Meta) 
  4] X >= X  by (Meta) 
  5] cons(Y, Z) > Z  because [6], by definition 
  6] cons*(Y, Z) >= Z  because [7], by (Select) 
  7] Z >= Z  by (Meta) 

  8] foldr(F, X, nil) >= X  because [9], by (Star) 
  9] foldr*(F, X, nil) >= X  because [10], by (Select) 
  10] X >= X  by (Meta) 

  11] foldr(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because [12], by (Star) 
  12] foldr*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because foldr > @_{o -> o}, [13] and [19], by (Copy) 
  13] foldr*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, Y)  because foldr > @_{o -> o -> o}, [14] and [15], by (Copy) 
  14] foldr*(F, X, cons(Y, Z)) >= F  because [3], by (Select) 
  15] foldr*(F, X, cons(Y, Z)) >= Y  because [16], by (Select) 
  16] cons(Y, Z) >= Y  because [17], by (Star) 
  17] cons*(Y, Z) >= Y  because [18], by (Select) 
  18] Y >= Y  by (Meta) 
  19] foldr*(F, X, cons(Y, Z)) >= foldr(F, X, Z)  because foldr in Mul, [3], [4] and [5], by (Stat) 

  20] #argfun-length#(foldr(/\x./\y.s(y), 0, X)) >= foldr(/\x./\y.s(y), 0, X)  because [21], by (Star) 
  21] #argfun-length#*(foldr(/\x./\y.s(y), 0, X)) >= foldr(/\x./\y.s(y), 0, X)  because #argfun-length# > foldr, [22], [27] and [28], by (Copy) 
  22] #argfun-length#*(foldr(/\x./\y.s(y), 0, X)) >= /\x./\y.s(y)  because [23], by (F-Abs) 
  23] #argfun-length#*(foldr(/\x./\y.s(y), 0, X), z) >= /\x.s(x)  because [24], by (F-Abs) 
  24] #argfun-length#*(foldr(/\x./\y.s(y), 0, X), z, u) >= s(u)  because #argfun-length# > s and [25], by (Copy) 
  25] #argfun-length#*(foldr(/\x./\y.s(y), 0, X), z, u) >= u  because [26], by (Select) 
  26] u >= u  by (Var) 
  27] #argfun-length#*(foldr(/\x./\y.s(y), 0, X)) >= 0  because #argfun-length# > 0, by (Copy) 
  28] #argfun-length#*(foldr(/\x./\y.s(y), 0, X)) >= X  because [29], by (Select) 
  29] foldr(/\x./\y.s(y), 0, X) >= X  because [30], by (Star) 
  30] foldr*(/\x./\y.s(y), 0, X) >= X  because [31], by (Select) 
  31] 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_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.