We consider the system merge.

  Alphabet:

    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    merge : [list * list * list] --> list 
    nil : [] --> list 

  Rules:

    merge(nil, nil, x) => x 
    merge(nil, cons(x, y), z) => merge(y, nil, cons(x, z)) 
    merge(cons(x, y), z, u) => merge(z, y, cons(x, u)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, 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] merge#(nil, cons(X, Y), Z) =#> merge#(Y, nil, cons(X, Z))   
    1] merge#(cons(X, Y), Z, U) =#> merge#(Z, Y, cons(X, U))   
    2] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    merge(nil, nil, X) => X 
    merge(nil, cons(X, Y), Z) => merge(Y, nil, cons(X, Z)) 
    merge(cons(X, Y), Z, U) => merge(Z, Y, cons(X, U)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, 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 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:

  merge#(nil, cons(X, Y), Z) >? merge#(Y, nil, cons(X, Z)) 
  merge#(cons(X, Y), Z, U) >? merge#(Z, Y, cons(X, U)) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  merge(nil, nil, X) >= X 
  merge(nil, cons(X, Y), Z) >= merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) >= merge(Z, Y, cons(X, U)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  merge = \y0y1y2.y0 + y1 + y2 
  merge# = \y0y1y2.0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[merge#(nil, cons(_x0, _x1), _x2)]] = 0 >= 0 = [[merge#(_x1, nil, cons(_x0, _x2))]] 
  [[merge#(cons(_x0, _x1), _x2, _x3)]] = 0 >= 0 = [[merge#(_x2, _x1, cons(_x0, _x3))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[merge(nil, nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 1 + x1 + x2 + x3 >= 1 + x1 + x2 + x3 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 

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:

  merge#(nil, cons(X, Y), Z) =#> merge#(Y, nil, cons(X, Z))   
  merge#(cons(X, Y), Z, U) =#> merge#(Z, Y, cons(X, U))   

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:

  merge#(nil, cons(X, Y), Z) >? merge#(Y, nil, cons(X, Z)) 
  merge#(cons(X, Y), Z, U) >? merge#(Z, Y, cons(X, U)) 
  merge(nil, nil, X) >= X 
  merge(nil, cons(X, Y), Z) >= merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) >= merge(Z, Y, cons(X, U)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  merge = \y0y1y2.y0 + y1 + y2 
  merge# = \y0y1y2.2y0 + 2y1 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[merge#(nil, cons(_x0, _x1), _x2)]] = 2 + 2x1 > 2x1 = [[merge#(_x1, nil, cons(_x0, _x2))]] 
  [[merge#(cons(_x0, _x1), _x2, _x3)]] = 2 + 2x1 + 2x2 > 2x1 + 2x2 = [[merge#(_x2, _x1, cons(_x0, _x3))]] 
  [[merge(nil, nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 1 + x1 + x2 + x3 >= 1 + x1 + x2 + x3 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 

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.