We consider the system Applicative_first_order_05__31.

  Alphabet:

    !faccolon : [] --> b -> b -> b 
    !facplus : [] --> b -> b -> b 
    a : [] --> b 
    cons : [] --> d -> e -> e 
    false : [] --> c 
    filter : [] --> (d -> c) -> e -> e 
    filter2 : [] --> c -> (d -> c) -> d -> e -> e 
    g : [] --> b -> a -> b 
    map : [] --> (d -> d) -> e -> e 
    nil : [] --> e 
    true : [] --> c 

  Rules:

    !faccolon (!faccolon x y) z => !faccolon x (!faccolon y z) 
    !faccolon (!facplus x y) z => !facplus (!faccolon x z) (!faccolon y z) 
    !faccolon x (!facplus y (f z)) => !faccolon (g x z) (!facplus y a) 
    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 

Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.

We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:

  Alphabet:

    !faccolon : [b * b] --> b 
    !facplus : [b * b] --> b 
    a : [] --> b 
    cons : [d * e] --> e 
    false : [] --> c 
    filter : [d -> c * e] --> e 
    filter2 : [c * d -> c * d * e] --> e 
    g : [b * a] --> b 
    map : [d -> d * e] --> e 
    nil : [] --> e 
    true : [] --> c 
    ~AP1 : [a -> b * a] --> b 

  Rules:

    !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
    !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
    !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(g(X, Z), !facplus(Y, a)) 
    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) 
    !faccolon(X, !facplus(Y, g(Z, U))) => !faccolon(g(X, U), !facplus(Y, a)) 
    ~AP1(F, X) => F X 

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] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
    1] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
    2] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
    3] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
    4] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(g(X, Z), !facplus(Y, a))   
    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)   
    9] !faccolon#(X, !facplus(Y, g(Z, U))) =#> !faccolon#(g(X, U), !facplus(Y, a))   

  Rules R_0:

    !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
    !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
    !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(g(X, Z), !facplus(Y, a)) 
    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) 
    !faccolon(X, !facplus(Y, g(Z, U))) => !faccolon(g(X, U), !facplus(Y, a)) 
    ~AP1(F, X) => F 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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? !faccolon#(g(X, Z), !facplus(Y, a)) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  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) 
  !faccolon#(X, !facplus(Y, g(Z, U))) >? !faccolon#(g(X, U), !facplus(Y, a)) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.0 
  !faccolon# = \y0y1.y1 
  !facplus = \y0y1.y1 
  a = 0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0y1.0 
  map = \G0y1.0 
  map# = \G0y1.0 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + 2G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x0, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 3 + 2F2(x3) > 0 = [[!faccolon#(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[!faccolon#(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon#(g(_x0, _x3), !facplus(_x1, a))]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + 2F0(x1) >= F0(x1) = [[_F0 _x1]] 

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:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   
  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)   
  !faccolon#(X, !facplus(Y, g(Z, U))) =#> !faccolon#(g(X, U), !facplus(Y, a))   

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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  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) 
  !faccolon#(X, !facplus(Y, g(Z, U))) >? !faccolon#(g(X, U), !facplus(Y, a)) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.0 
  !faccolon# = \y0y1.y1 
  !facplus = \y0y1.y1 
  a = 0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0y1.1 
  map = \G0y1.0 
  map# = \G0y1.0 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x0, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[!faccolon#(_x0, !facplus(_x1, g(_x2, _x3)))]] = 1 > 0 = [[!faccolon#(g(_x0, _x3), !facplus(_x1, a))]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

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:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   
  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)   

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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  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) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.0 
  !faccolon# = \y0y1.0 
  !facplus = \y0y1.0 
  a = 0 
  cons = \y0y1.1 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0y1.0 
  map = \G0y1.2y1 
  map# = \G0y1.y1 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x0, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x1, _x2)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 >= 1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + x2 >= x2 = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

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:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
  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)   

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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  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) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.0 
  !faccolon# = \y0y1.2y1 
  !facplus = \y0y1.0 
  a = 0 
  cons = \y0y1.2 + y0 + y1 
  false = 3 
  filter = \G0y1.2y1 + 2y1G0(y1) 
  filter2 = \y0G1y2y3.2 + y2 + 2y3 + 2y3G1(y3) 
  filter2# = \y0G1y2y3.2y3 
  filter# = \G0y1.2y1 
  g = \y0y1.0 
  map = \G0y1.y1 + y1G0(y1) 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 2x2 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 2x2 >= 2x2 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 2x2 >= 2x2 = [[!faccolon#(_x0, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 2x2 >= 2x2 = [[!faccolon#(_x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 > 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2x2 >= 2x2 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 2x2 >= 2x2 = [[filter#(_F0, _x2)]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 4F0(2 + x1 + x2) >= 2 + x1 + 2x2 + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + 2x2 + 2x2F0(x2) >= 2 + x1 + 2x2 + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + 2x2 + 2x2F0(x2) >= 2x2 + 2x2F0(x2) = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

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:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.0 
  !faccolon# = \y0y1.0 
  !facplus = \y0y1.0 
  a = 0 
  cons = \y0y1.2 + y0 + y1 
  false = 3 
  filter = \G0y1.2y1 + 2y1G0(y1) + 2G0(0) 
  filter2 = \y0G1y2y3.2 + y2 + 2y3 + G1(0) + G1(y2) + 2y3G1(y3) 
  filter2# = \y0G1y2y3.3 
  filter# = \G0y1.0 
  g = \y0y1.0 
  map = \G0y1.2y1 + y1G0(y1) 
  nil = 2 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x0, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon#(_x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 4 + 2F0(2) >= 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + 2x2 + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 4 + 2F0(0) + 4F0(2) >= 2 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 2F0(0) + 4F0(2 + x1 + x2) >= 2 + x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) >= 2 + x1 + 2x2 + 2x2F0(x2) + 2F0(0) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) >= 2x2 + 2x2F0(x2) + 2F0(0) = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

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:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   

Thus, the original system is terminating if (P_5, R_0, static, formative) is finite.

We consider the dependency pair problem (P_5, 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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[!faccolon#(x_1, x_2)]] = !faccolon#(x_1) 
  [[a]] = _|_ 
  [[cons(x_1, x_2)]] = _|_ 
  [[filter(x_1, x_2)]] = _|_ 
  [[filter2(x_1, x_2, x_3, x_4)]] = _|_ 
  [[g(x_1, x_2)]] = x_1 
  [[map(x_1, x_2)]] = map 
  [[nil]] = _|_ 

We choose Lex = {!faccolon} and Mul = {!faccolon#, !facplus, @_{o -> o}, false, map, true, ~AP1}, and the following precedence: !faccolon > !faccolon# > false > map > !facplus > true > ~AP1 > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  !faccolon#(!faccolon(X, Y)) >= !faccolon#(X) 
  !faccolon#(!faccolon(X, Y)) >= !faccolon#(Y) 
  !faccolon#(!facplus(X, Y)) > !faccolon#(X) 
  !faccolon#(!facplus(X, Y)) >= !faccolon#(Y) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(X, !facplus(Y, _|_)) 
  map >= _|_ 
  map >= _|_ 
  _|_ >= _|_ 
  _|_ >= _|_ 
  _|_ >= _|_ 
  _|_ >= _|_ 
  !faccolon(X, !facplus(Y, Z)) >= !faccolon(X, !facplus(Y, _|_)) 
  ~AP1(F, X) >= @_{o -> o}(F, X) 

With these choices, we have:

  1] !faccolon#(!faccolon(X, Y)) >= !faccolon#(X)  because !faccolon# in Mul and [2], by (Fun) 
  2] !faccolon(X, Y) >= X  because [3], by (Star) 
  3] !faccolon*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] !faccolon#(!faccolon(X, Y)) >= !faccolon#(Y)  because !faccolon# in Mul and [6], by (Fun) 
  6] !faccolon(X, Y) >= Y  because [7], by (Star) 
  7] !faccolon*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] !faccolon#(!facplus(X, Y)) > !faccolon#(X)  because [10], by definition 
  10] !faccolon#*(!facplus(X, Y)) >= !faccolon#(X)  because !faccolon# in Mul and [11], by (Stat) 
  11] !facplus(X, Y) > X  because [12], by definition 
  12] !facplus*(X, Y) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] !faccolon#(!facplus(X, Y)) >= !faccolon#(Y)  because !faccolon# in Mul and [15], by (Fun) 
  15] !facplus(X, Y) >= Y  because [16], by (Star) 
  16] !facplus*(X, Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 

  18] !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z))  because [19], by (Star) 
  19] !faccolon*(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z))  because [20], [22] and [23], by (Stat) 
  20] !faccolon(X, Y) > X  because [21], by definition 
  21] !faccolon*(X, Y) >= X  because [4], by (Select) 
  22] !faccolon*(!faccolon(X, Y), Z) >= X  because [2], by (Select) 
  23] !faccolon*(!faccolon(X, Y), Z) >= !faccolon(Y, Z)  because [24], [26] and [27], by (Stat) 
  24] !faccolon(X, Y) > Y  because [25], by definition 
  25] !faccolon*(X, Y) >= Y  because [8], by (Select) 
  26] !faccolon*(!faccolon(X, Y), Z) >= Y  because [6], by (Select) 
  27] !faccolon*(!faccolon(X, Y), Z) >= Z  because [28], by (Select) 
  28] Z >= Z  by (Meta) 

  29] !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z))  because [30], by (Star) 
  30] !faccolon*(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z))  because !faccolon > !facplus, [31] and [36], by (Copy) 
  31] !faccolon*(!facplus(X, Y), Z) >= !faccolon(X, Z)  because [11], [32] and [34], by (Stat) 
  32] !faccolon*(!facplus(X, Y), Z) >= X  because [33], by (Select) 
  33] !facplus(X, Y) >= X  because [12], by (Star) 
  34] !faccolon*(!facplus(X, Y), Z) >= Z  because [35], by (Select) 
  35] Z >= Z  by (Meta) 
  36] !faccolon*(!facplus(X, Y), Z) >= !faccolon(Y, Z)  because [37], [39] and [34], by (Stat) 
  37] !facplus(X, Y) > Y  because [38], by definition 
  38] !facplus*(X, Y) >= Y  because [17], by (Select) 
  39] !faccolon*(!facplus(X, Y), Z) >= Y  because [15], by (Select) 

  40] !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(X, !facplus(Y, _|_))  because [41] and [42], by (Fun) 
  41] X >= X  by (Meta) 
  42] !facplus(Y, ~AP1(F, Z)) >= !facplus(Y, _|_)  because !facplus in Mul, [43] and [44], by (Fun) 
  43] Y >= Y  by (Meta) 
  44] ~AP1(F, Z) >= _|_  by (Bot) 

  45] map >= _|_  by (Bot) 

  46] map >= _|_  by (Bot) 

  47] _|_ >= _|_  by (Bot) 

  48] _|_ >= _|_  by (Bot) 

  49] _|_ >= _|_  by (Bot) 

  50] _|_ >= _|_  by (Bot) 

  51] !faccolon(X, !facplus(Y, Z)) >= !faccolon(X, !facplus(Y, _|_))  because [52] and [53], by (Fun) 
  52] X >= X  by (Meta) 
  53] !facplus(Y, Z) >= !facplus(Y, _|_)  because !facplus in Mul, [54] and [55], by (Fun) 
  54] Y >= Y  by (Meta) 
  55] Z >= _|_  by (Bot) 

  56] ~AP1(F, X) >= @_{o -> o}(F, X)  because [57], by (Star) 
  57] ~AP1*(F, X) >= @_{o -> o}(F, X)  because ~AP1 > @_{o -> o}, [58] and [60], by (Copy) 
  58] ~AP1*(F, X) >= F  because [59], by (Select) 
  59] F >= F  by (Meta) 
  60] ~AP1*(F, X) >= X  because [61], by (Select) 
  61] X >= X  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of:

  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   

Thus, the original system is terminating if (P_6, R_0, static, formative) is finite.

We consider the dependency pair problem (P_6, 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:

  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.2 + y1 + 2y0 
  !faccolon# = \y0y1.y0 
  !facplus = \y0y1.y1 
  a = 0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.2G0(0) 
  filter2 = \y0G1y2y3.2G1(0) 
  g = \y0y1.0 
  map = \G0y1.2G0(y1) 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 2 + x1 + 2x0 > x0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 2 + x1 + 2x0 > x1 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x1 >= x1 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 >= 4 + x2 + 2x0 + 2x1 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 2 + x2 + 2x1 >= 2 + x2 + 2x1 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 5 + 2x0 + F2(x3) >= 2 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2F0(0) >= 2F0(0) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2F0(0) >= 0 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2F0(0) >= 2F0(0) = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = 2 + 2x0 >= 2 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, static, formative) by (P_7, R_0, static, formative), where P_7 consists of:

  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   

Thus, the original system is terminating if (P_7, R_0, static, formative) is finite.

We consider the dependency pair problem (P_7, 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:

  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(g(X, Z), !facplus(Y, a)) 
  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) 
  !faccolon(X, !facplus(Y, g(Z, U))) >= !faccolon(g(X, U), !facplus(Y, a)) 
  ~AP1(F, X) >= F X 

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

The following interpretation satisfies the requirements:

  !faccolon = \y0y1.y0 
  !faccolon# = \y0y1.y0 
  !facplus = \y0y1.1 + 2y1 
  a = 0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  g = \y0y1.0 
  map = \G0y1.2 
  nil = 0 
  true = 3 
  ~AP1 = \G0y1.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 1 + 2x1 > x1 = [[!faccolon#(_x1, _x2)]] 
  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = x0 >= x0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 1 + 2x1 >= 1 + 2x1 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = x0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[map(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] 
  [[!faccolon(_x0, !facplus(_x1, g(_x2, _x3)))]] = x0 >= 0 = [[!faccolon(g(_x0, _x3), !facplus(_x1, a))]] 
  [[~AP1(_F0, _x1)]] = 3 + F0(x1) >= F0(x1) = [[_F0 _x1]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_7, 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 +++

[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
[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.