We consider the system Applicative_05__Ex3Lists.

  Alphabet:

    append : [b * b] --> b 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    append(append(x, y), z) => append(x, append(y, z)) 
    map(f, append(x, y)) => append(map(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] append#(cons(X, Y), Z) =#> append#(Y, Z)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] append#(append(X, Y), Z) =#> append#(X, append(Y, Z))   
    3] append#(append(X, Y), Z) =#> append#(Y, Z)   
    4] map#(F, append(X, Y)) =#> append#(map(F, X), map(F, Y))   
    5] map#(F, append(X, Y)) =#> map#(F, X)   
    6] map#(F, append(X, Y)) =#> map#(F, Y)   

  Rules R_0:

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

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

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

The following interpretation satisfies the requirements:

  append = \y0y1.y1 
  append# = \y0y1.0 
  cons = \y0y1.0 
  map = \G0y1.2y1 
  map# = \G0y1.3 + 2G0(0) 
  nil = 1 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(0) >= 3 + 2F0(0) = [[map#(_F0, _x2)]] 
  [[append#(append(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x0, append(_x1, _x2))]] 
  [[append#(append(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[map#(_F0, append(_x1, _x2))]] = 3 + 2F0(0) > 0 = [[append#(map(_F0, _x1), map(_F0, _x2))]] 
  [[map#(_F0, append(_x1, _x2))]] = 3 + 2F0(0) >= 3 + 2F0(0) = [[map#(_F0, _x1)]] 
  [[map#(_F0, append(_x1, _x2))]] = 3 + 2F0(0) >= 3 + 2F0(0) = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 2 >= 1 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = x2 >= x2 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 2x2 >= 2x2 = [[append(map(_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:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   
  append#(append(X, Y), Z) =#> append#(X, append(Y, Z))   
  append#(append(X, Y), Z) =#> append#(Y, Z)   
  map#(F, append(X, Y)) =#> map#(F, X)   
  map#(F, append(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:

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

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

The following interpretation satisfies the requirements:

  append = \y0y1.3 + y1 + 2y0 
  append# = \y0y1.y1 + 2y0 
  cons = \y0y1.y1 
  map = \G0y1.2y1 
  map# = \G0y1.0 
  nil = 2 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = x2 + 2x1 >= x2 + 2x1 = [[append#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append#(append(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 > 3 + x2 + 2x0 + 2x1 = [[append#(_x0, append(_x1, _x2))]] 
  [[append#(append(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 > x2 + 2x1 = [[append#(_x1, _x2)]] 
  [[map#(_F0, append(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x1)]] 
  [[map#(_F0, append(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 7 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 3 + x2 + 2x1 >= 3 + x2 + 2x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 4 >= 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2x2 >= 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = 9 + x2 + 2x1 + 4x0 >= 6 + x2 + 2x0 + 2x1 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 6 + 2x2 + 4x1 >= 3 + 2x2 + 4x1 = [[append(map(_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_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   
  map#(F, append(X, Y)) =#> map#(F, X)   
  map#(F, append(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:

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

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

The following interpretation satisfies the requirements:

  append = \y0y1.1 + y0 + y1 
  append# = \y0y1.0 
  cons = \y0y1.y1 
  map = \G0y1.2y1 
  map# = \G0y1.2y1 
  nil = 2 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 2x2 >= 2x2 = [[map#(_F0, _x2)]] 
  [[map#(_F0, append(_x1, _x2))]] = 2 + 2x1 + 2x2 > 2x1 = [[map#(_F0, _x1)]] 
  [[map#(_F0, append(_x1, _x2))]] = 2 + 2x1 + 2x2 > 2x2 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 3 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 4 >= 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2x2 >= 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 2 + 2x1 + 2x2 >= 1 + 2x1 + 2x2 = [[append(map(_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_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

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

Thus, the original system is terminating if (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:

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

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

The following interpretation satisfies the requirements:

  append = \y0y1.y1 + 2y0 
  append# = \y0y1.y0 
  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  map# = \G0y1.0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 1 + x1 > x1 = [[append#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x2 + 2x1 >= 1 + x2 + 2x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = x2 + 2x1 + 4x0 >= x2 + 2x0 + 2x1 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = x2 + 2x1 >= x2 + 2x1 = [[append(map(_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_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_0, static, formative) is finite.

We consider the dependency pair problem (P_4, 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) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  append(append(X, Y), Z) >= append(X, append(Y, Z)) 
  map(F, append(X, Y)) >= append(map(F, X), map(F, Y)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  cons = \y0y1.2 + y1 
  map = \G0y1.2y1 
  map# = \G0y1.y1 
  nil = 2 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 2 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 4 >= 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 2x1 + 2x2 >= 2x1 + 2x2 = [[append(map(_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_4, 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.