We consider the system AotoYamada_05__007.

  Alphabet:

    0 : [] --> b 
    cons : [b * a] --> a 
    inc : [] --> a -> a 
    map : [b -> b] --> a -> a 
    nil : [] --> a 
    plus : [b] --> b -> b 
    s : [b] --> b 

  Rules:

    plus(0) x => x 
    plus(s(x)) y => s(plus(x) y) 
    map(f) nil => nil 
    map(f) cons(x, y) => cons(f x, map(f) y) 
    inc => map(plus(s(0))) 

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]).

In order to do so, we start by eta-expanding the system, which gives:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  inc(X) => map(/\x.plus(s(0), x), X) 

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

  Dependency Pairs P_0:

    0] plus#(s(X), Y) =#> plus#(X, Y)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] inc#(X) =#> map#(/\x.plus(s(0), x), X)   
    3] inc#(X) =#> plus#(s(0), Y)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    inc(X) => map(/\x.plus(s(0), 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 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:

  plus#(s(X), Y) >? plus#(X, Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  inc#(X) >? map#(/\x.plus(s(0), x), X) 
  inc#(X) >? plus#(s(0), Y) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  inc(X) >= map(/\x.plus(s(0), x), X) 

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

  pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-inc# = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.3 
  inc = \y0.0 
  inc# = \y0.3 + 2y0 
  map = \G0y1.3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) 
  map# = \G0y1.0 
  nil = 0 
  plus = \y0y1.y1 
  plus# = \y0y1.0 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[inc#(_x0)]] = 3 + 2x0 > 0 = [[map#(/\x.plus(s(0), x), _x0)]] 
  [[inc#(_x0)]] = 3 + 2x0 > 0 = [[plus#(s(0), _x1)]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, nil)]] = 4F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 9 + 2F0(0) + 11F0(3) >= 3 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 3 + 3x0x0 + 5x0 >= 3x0x0 + 5x0 = [[map(/\x.plus(s(0), x), _x0)]] 

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:

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

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:

  plus#(s(X), Y) >? plus#(X, Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  inc(X) >= map(/\x.plus(s(0), x), X) 

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

  pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-inc# = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.3 + y0 + y1 
  inc = \y0.0 
  map = \G0y1.2y1 + 2G0(y1) + 3y1G0(y1) 
  map# = \G0y1.0 
  nil = 2 
  plus = \y0y1.y0 + y1 
  plus# = \y0y1.y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(_x0, _x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, nil)]] = 4 + 8F0(2) >= 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) >= 3 + 2x2 + F0(x1) + 2F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 5 + 3x0x0 + 7x0 >= 2 + 3x0x0 + 7x0 = [[map(/\x.plus(s(0), x), _x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

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

Thus, the original system is terminating if (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:

  map#(F, cons(X, Y)) >? map#(F, Y) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  inc(X) >= map(/\x.plus(s(0), x), X) 

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

  pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-inc# = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.2 + y0 + 2y1 
  inc = \y0.0 
  map = \G0y1.y1 + y1G0(y1) 
  map# = \G0y1.y1 
  nil = 0 
  plus = \y0y1.3 + y1 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 > x2 = [[map#(_F0, _x2)]] 
  [[plus(0, _x0)]] = 3 + x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 3 + x1 >= 0 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 + 2x2F0(2 + x1 + 2x2) + 2F0(2 + x1 + 2x2) + x1F0(2 + x1 + 2x2) >= 2 + 2x2 + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 3 + 4x0 + x0x0 >= 4x0 + x0x0 = [[map(/\x.plus(s(0), x), _x0)]] 

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.

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.