We consider the system prefixsum.

  Alphabet:

    !plus : [nat * nat] --> nat 
    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    nil : [] --> list 
    ps : [list] --> list 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    ps(nil) => nil 
    ps(cons(x, y)) => cons(x, ps(map(/\z.!plus(x, z), y))) 

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] map#(F, cons(X, Y)) =#> map#(F, Y)   
    1] ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y))   
    2] ps#(cons(X, Y)) =#> map#(/\x.!plus(X, x), Y)   

  Rules R_0:

    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    ps(nil) => nil 
    ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 

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 : 1, 2 
    * 2 : 0 

This graph has the following strongly connected components:

  P_1:

    map#(F, cons(X, Y)) =#> map#(F, Y)   

  P_2:

    ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, 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) and (P_2, R_0, m, f).

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

  ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  ps(nil) >= nil 
  ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.0 
  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  nil = 0 
  ps = \y0.y0 
  ps# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[ps#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[ps(nil)]] = 0 >= 0 = [[nil]] 
  [[ps(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] 

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

  map#(F, cons(X, Y)) >? map#(F, Y) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  ps(nil) >= nil 
  ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.0 
  cons = \y0y1.2 + y1 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  nil = 0 
  ps = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[ps(nil)]] = 0 >= 0 = [[nil]] 
  [[ps(cons(_x0, _x1))]] = 4 + 2x1 >= 2 + 2x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _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.