We consider the system Applicative_first_order_05__17.

  Alphabet:

    !facdot : [a * a] --> a 
    1 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    i : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    !facdot(1, x) => x 
    !facdot(x, 1) => x 
    !facdot(i(x), x) => 1 
    !facdot(x, i(x)) => 1 
    !facdot(i(x), !facdot(x, y)) => y 
    !facdot(x, !facdot(i(x), y)) => y 
    !facdot(!facdot(x, y), z) => !facdot(x, !facdot(y, z)) 
    i(1) => 1 
    i(i(x)) => x 
    i(!facdot(x, y)) => !facdot(i(y), i(x)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(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, all):

  Dependency Pairs P_0:

    0] !facdot#(!facdot(X, Y), Z) =#> !facdot#(X, !facdot(Y, Z))   
    1] !facdot#(!facdot(X, Y), Z) =#> !facdot#(Y, Z)   
    2] i#(!facdot(X, Y)) =#> !facdot#(i(Y), i(X))   
    3] i#(!facdot(X, Y)) =#> i#(Y)   
    4] i#(!facdot(X, Y)) =#> i#(X)   
    5] map#(F, cons(X, Y)) =#> map#(F, Y)   
    6] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    7] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    8] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    !facdot(1, X) => X 
    !facdot(X, 1) => X 
    !facdot(i(X), X) => 1 
    !facdot(X, i(X)) => 1 
    !facdot(i(X), !facdot(X, Y)) => Y 
    !facdot(X, !facdot(i(X), Y)) => Y 
    !facdot(!facdot(X, Y), Z) => !facdot(X, !facdot(Y, Z)) 
    i(1) => 1 
    i(i(X)) => X 
    i(!facdot(X, Y)) => !facdot(i(Y), i(X)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

Thus, the original system is terminating if (P_0, R_0, static, all) is finite.

We consider the dependency pair problem (P_0, R_0, static, all).

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 : 0, 1 
    * 2 : 0, 1 
    * 3 : 2, 3, 4 
    * 4 : 2, 3, 4 
    * 5 : 5 
    * 6 : 7, 8 
    * 7 : 6 
    * 8 : 6 

This graph has the following strongly connected components:

  P_1:

    !facdot#(!facdot(X, Y), Z) =#> !facdot#(X, !facdot(Y, Z))   
    !facdot#(!facdot(X, Y), Z) =#> !facdot#(Y, Z)   

  P_2:

    i#(!facdot(X, Y)) =#> i#(Y)   
    i#(!facdot(X, Y)) =#> i#(X)   

  P_3:

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

  P_4:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, 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), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, all), (P_2, R_0, static, all), (P_3, R_0, static, all) and (P_4, R_0, static, all) is finite.

We consider the dependency pair problem (P_4, R_0, static, all).

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:

  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  !facdot(1, X) >= X 
  !facdot(X, 1) >= X 
  !facdot(i(X), X) >= 1 
  !facdot(X, i(X)) >= 1 
  !facdot(i(X), !facdot(X, Y)) >= Y 
  !facdot(X, !facdot(i(X), Y)) >= Y 
  !facdot(!facdot(X, Y), Z) >= !facdot(X, !facdot(Y, Z)) 
  i(1) >= 1 
  i(i(X)) >= X 
  i(!facdot(X, Y)) >= !facdot(i(Y), i(X)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

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

The following interpretation satisfies the requirements:

  !facdot = \y0y1.y0 + y1 
  1 = 0 
  cons = \y0y1.2 + 2y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + 2y3 
  filter2# = \y0G1y2y3.1 + 2y3 + 2y3G1(y3) 
  filter# = \G0y1.2y1 + 2y1G0(y1) 
  i = \y0.2y0 
  map = \G0y1.y1 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 4x2 + 4x2F0(2 + 2x2) + 4F0(2 + 2x2) > 1 + 2x2 + 2x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 + 2x2F0(x2) > 2x2 + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 + 2x2F0(x2) > 2x2 + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[!facdot(1, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!facdot(_x0, 1)]] = x0 >= x0 = [[_x0]] 
  [[!facdot(i(_x0), _x0)]] = 3x0 >= 0 = [[1]] 
  [[!facdot(_x0, i(_x0))]] = 3x0 >= 0 = [[1]] 
  [[!facdot(i(_x0), !facdot(_x0, _x1))]] = x1 + 3x0 >= x1 = [[_x1]] 
  [[!facdot(_x0, !facdot(i(_x0), _x1))]] = x1 + 3x0 >= x1 = [[_x1]] 
  [[!facdot(!facdot(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!facdot(_x0, !facdot(_x1, _x2))]] 
  [[i(1)]] = 0 >= 0 = [[1]] 
  [[i(i(_x0))]] = 4x0 >= x0 = [[_x0]] 
  [[i(!facdot(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!facdot(i(_x1), i(_x0))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x2 >= x2 = [[filter(_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.

Thus, the original system is terminating if each of (P_1, R_0, static, all), (P_2, R_0, static, all) and (P_3, R_0, static, all) is finite.

We consider the dependency pair problem (P_3, R_0, static, all).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_3, R_0) has no usable rules.

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

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

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y1 
  map# = \G0y1.y1 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[map#(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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 each of (P_1, R_0, static, all) and (P_2, R_0, static, all) is finite.

We consider the dependency pair problem (P_2, R_0, static, all).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

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

  i#(!facdot(X, Y)) >? i#(Y) 
  i#(!facdot(X, Y)) >? i#(X) 

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

The following interpretation satisfies the requirements:

  !facdot = \y0y1.3 + y0 + y1 
  i# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[i#(!facdot(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[i#(_x1)]] 
  [[i#(!facdot(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[i#(_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.

Thus, the original system is terminating if (P_1, R_0, static, all) is finite.

We consider the dependency pair problem (P_1, R_0, static, all).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  !facdot(1, X) => X 
  !facdot(X, 1) => X 
  !facdot(i(X), X) => 1 
  !facdot(X, i(X)) => 1 
  !facdot(i(X), !facdot(X, Y)) => Y 
  !facdot(X, !facdot(i(X), Y)) => Y 
  !facdot(!facdot(X, Y), Z) => !facdot(X, !facdot(Y, Z)) 

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

  !facdot#(!facdot(X, Y), Z) >? !facdot#(X, !facdot(Y, Z)) 
  !facdot#(!facdot(X, Y), Z) >? !facdot#(Y, Z) 
  !facdot(1, X) >= X 
  !facdot(X, 1) >= X 
  !facdot(i(X), X) >= 1 
  !facdot(X, i(X)) >= 1 
  !facdot(i(X), !facdot(X, Y)) >= Y 
  !facdot(X, !facdot(i(X), Y)) >= Y 
  !facdot(!facdot(X, Y), Z) >= !facdot(X, !facdot(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !facdot = \y0y1.1 + y0 + y1 
  !facdot# = \y0y1.2y0 
  1 = 0 
  i = \y0.3 

Using this interpretation, the requirements translate to:

  [[!facdot#(!facdot(_x0, _x1), _x2)]] = 2 + 2x0 + 2x1 > 2x0 = [[!facdot#(_x0, !facdot(_x1, _x2))]] 
  [[!facdot#(!facdot(_x0, _x1), _x2)]] = 2 + 2x0 + 2x1 > 2x1 = [[!facdot#(_x1, _x2)]] 
  [[!facdot(1, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[!facdot(_x0, 1)]] = 1 + x0 >= x0 = [[_x0]] 
  [[!facdot(i(_x0), _x0)]] = 4 + x0 >= 0 = [[1]] 
  [[!facdot(_x0, i(_x0))]] = 4 + x0 >= 0 = [[1]] 
  [[!facdot(i(_x0), !facdot(_x0, _x1))]] = 5 + x0 + x1 >= x1 = [[_x1]] 
  [[!facdot(_x0, !facdot(i(_x0), _x1))]] = 5 + x0 + x1 >= x1 = [[_x1]] 
  [[!facdot(!facdot(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[!facdot(_x0, !facdot(_x1, _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.