We consider the system Applicative_first_order_05__12.

  Alphabet:

    and : [a * a] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    not : [a] --> a 
    or : [a * a] --> a 
    true : [] --> b 

  Rules:

    not(not(x)) => x 
    not(or(x, y)) => and(not(x), not(y)) 
    not(and(x, y)) => or(not(x), not(y)) 
    and(x, or(y, z)) => or(and(x, y), and(x, z)) 
    and(or(x, y), z) => or(and(z, x), and(z, y)) 
    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, formative):

  Dependency Pairs P_0:

    0] not#(or(X, Y)) =#> and#(not(X), not(Y))   
    1] not#(or(X, Y)) =#> not#(X)   
    2] not#(or(X, Y)) =#> not#(Y)   
    3] not#(and(X, Y)) =#> not#(X)   
    4] not#(and(X, Y)) =#> not#(Y)   
    5] and#(X, or(Y, Z)) =#> and#(X, Y)   
    6] and#(X, or(Y, Z)) =#> and#(X, Z)   
    7] and#(or(X, Y), Z) =#> and#(Z, X)   
    8] and#(or(X, Y), Z) =#> and#(Z, Y)   
    9] map#(F, cons(X, Y)) =#> map#(F, Y)   
    10] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    11] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    12] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    not(not(X)) => X 
    not(or(X, Y)) => and(not(X), not(Y)) 
    not(and(X, Y)) => or(not(X), not(Y)) 
    and(X, or(Y, Z)) => or(and(X, Y), and(X, Z)) 
    and(or(X, Y), Z) => or(and(Z, X), and(Z, Y)) 
    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, 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 : 5, 6, 7, 8 
    * 1 : 0, 1, 2, 3, 4 
    * 2 : 0, 1, 2, 3, 4 
    * 3 : 0, 1, 2, 3, 4 
    * 4 : 0, 1, 2, 3, 4 
    * 5 : 5, 6, 7, 8 
    * 6 : 5, 6, 7, 8 
    * 7 : 5, 6, 7, 8 
    * 8 : 5, 6, 7, 8 
    * 9 : 9 
    * 10 : 11, 12 
    * 11 : 10 
    * 12 : 10 

This graph has the following strongly connected components:

  P_1:

    not#(or(X, Y)) =#> not#(X)   
    not#(or(X, Y)) =#> not#(Y)   
    not#(and(X, Y)) =#> not#(X)   
    not#(and(X, Y)) =#> not#(Y)   

  P_2:

    and#(X, or(Y, Z)) =#> and#(X, Y)   
    and#(X, or(Y, Z)) =#> and#(X, Z)   
    and#(or(X, Y), Z) =#> and#(Z, X)   
    and#(or(X, Y), Z) =#> and#(Z, Y)   

  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, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (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:

  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) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 
  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:

  and = \y0y1.0 
  cons = \y0y1.1 + y1 
  false = 3 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.1 + 2y3 
  filter2# = \y0G1y2y3.y3 
  filter# = \G0y1.y1 
  map = \G0y1.y1 
  nil = 0 
  not = \y0.2 + y0 
  or = \y0y1.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] 
  [[not(not(_x0))]] = 4 + x0 >= x0 = [[_x0]] 
  [[not(or(_x0, _x1))]] = 2 >= 0 = [[and(not(_x0), not(_x1))]] 
  [[not(and(_x0, _x1))]] = 2 >= 0 = [[or(not(_x0), not(_x1))]] 
  [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] 
  [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (P_5, R_0, static, formative) is finite.

We consider the dependency pair problem (P_5, 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 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (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 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, 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 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:

  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  and#(or(X, Y), Z) >? and#(Z, Y) 

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

The following interpretation satisfies the requirements:

  and# = \y0y1.y0 + 2y1 
  or = \y0y1.3 + 2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[and#(_x0, or(_x1, _x2))]] = 6 + x0 + 4x1 + 4x2 > x0 + 2x1 = [[and#(_x0, _x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 6 + x0 + 4x1 + 4x2 > x0 + 2x2 = [[and#(_x0, _x2)]] 
  [[and#(or(_x0, _x1), _x2)]] = 3 + 2x0 + 2x1 + 2x2 > x2 + 2x0 = [[and#(_x2, _x0)]] 
  [[and#(or(_x0, _x1), _x2)]] = 3 + 2x0 + 2x1 + 2x2 > x2 + 2x1 = [[and#(_x2, _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 with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

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

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.3 + y0 + y1 
  not# = \y0.y0 
  or = \y0y1.3 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[not#(or(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[not#(_x0)]] 
  [[not#(or(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[not#(_x1)]] 
  [[not#(and(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[not#(_x0)]] 
  [[not#(and(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[not#(_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.