We consider the system Applicative_first_order_05__#3.45.

  Alphabet:

    0 : [] --> d 
    cons : [d * d] --> d 
    f : [d] --> a 
    false : [] --> c 
    filter : [d -> c * d] --> d 
    filter2 : [c * d -> c * d * d] --> d 
    g : [d] --> d 
    h : [d] --> b 
    map : [d -> d * d] --> d 
    nil : [] --> d 
    s : [d] --> d 
    true : [] --> c 

  Rules:

    f(s(x)) => f(x) 
    g(cons(0, x)) => g(x) 
    g(cons(s(x), y)) => s(x) 
    h(cons(x, y)) => h(g(cons(x, y))) 
    map(i, nil) => nil 
    map(i, cons(x, y)) => cons(i x, map(i, y)) 
    filter(i, nil) => nil 
    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
    filter2(true, i, x, y) => cons(x, filter(i, y)) 
    filter2(false, i, x, y) => filter(i, 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] f#(s(X)) =#> f#(X)   
    1] g#(cons(0, X)) =#> g#(X)   
    2] h#(cons(X, Y)) =#> h#(g(cons(X, Y)))   
    3] h#(cons(X, Y)) =#> g#(cons(X, Y))   
    4] map#(F, cons(X, Y)) =#> map#(F, Y)   
    5] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    6] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    7] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    f(s(X)) => f(X) 
    g(cons(0, X)) => g(X) 
    g(cons(s(X), Y)) => s(X) 
    h(cons(X, Y)) => h(g(cons(X, 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 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:

  f#(s(X)) >? f#(X) 
  g#(cons(0, X)) >? g#(X) 
  h#(cons(X, Y)) >? h#(g(cons(X, Y))) 
  h#(cons(X, Y)) >? g#(cons(X, Y)) 
  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) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 3 
  cons = \y0y1.0 
  f = \y0.0 
  f# = \y0.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0.2 
  g# = \y0.0 
  h = \y0.0 
  h# = \y0.3 
  map = \G0y1.2 + 2G0(0) 
  map# = \G0y1.0 
  nil = 0 
  s = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] 
  [[g#(cons(0, _x0))]] = 0 >= 0 = [[g#(_x0)]] 
  [[h#(cons(_x0, _x1))]] = 3 >= 3 = [[h#(g(cons(_x0, _x1)))]] 
  [[h#(cons(_x0, _x1))]] = 3 > 0 = [[g#(cons(_x0, _x1))]] 
  [[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)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 2 >= 2 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 2 >= 0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _x1)))]] 
  [[map(_F0, nil)]] = 2 + 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2F0(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)]] 

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:

  f#(s(X)) =#> f#(X)   
  g#(cons(0, X)) =#> g#(X)   
  h#(cons(X, Y)) =#> h#(g(cons(X, Y)))   
  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_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:

  f#(s(X)) >? f#(X) 
  g#(cons(0, X)) >? g#(X) 
  h#(cons(X, Y)) >? h#(g(cons(X, Y))) 
  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) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 3 
  cons = \y0y1.1 + y1 
  f = \y0.0 
  f# = \y0.0 
  false = 3 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.2 + 2y3 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0.2 
  g# = \y0.0 
  h = \y0.0 
  h# = \y0.0 
  map = \G0y1.2 + y1 
  map# = \G0y1.y1 
  nil = 0 
  s = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] 
  [[g#(cons(0, _x0))]] = 0 >= 0 = [[g#(_x0)]] 
  [[h#(cons(_x0, _x1))]] = 0 >= 0 = [[h#(g(cons(_x0, _x1)))]] 
  [[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)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 2 >= 2 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 2 >= 0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _x1)))]] 
  [[map(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[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 >= 1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 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_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  f#(s(X)) =#> f#(X)   
  g#(cons(0, X)) =#> g#(X)   
  h#(cons(X, Y)) =#> h#(g(cons(X, 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:

  f#(s(X)) >? f#(X) 
  g#(cons(0, X)) >? g#(X) 
  h#(cons(X, Y)) >? h#(g(cons(X, 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) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 3 
  cons = \y0y1.1 
  f = \y0.0 
  f# = \y0.0 
  false = 3 
  filter = \G0y1.1 
  filter2 = \y0G1y2y3.1 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  g = \y0.0 
  g# = \y0.0 
  h = \y0.0 
  h# = \y0.2y0 
  map = \G0y1.y1 + 2G0(0) + 3G0(y1) 
  nil = 0 
  s = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] 
  [[g#(cons(0, _x0))]] = 0 >= 0 = [[g#(_x0)]] 
  [[h#(cons(_x0, _x1))]] = 2 > 0 = [[h#(g(cons(_x0, _x1)))]] 
  [[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)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 0 >= 0 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 0 >= 0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _x1)))]] 
  [[map(_F0, nil)]] = 5F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + 2F0(0) + 3F0(1) >= 1 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 1 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 >= 1 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 >= 1 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 >= 1 = [[filter(_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:

  f#(s(X)) =#> f#(X)   
  g#(cons(0, X)) =#> g#(X)   
  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:

  f#(s(X)) >? f#(X) 
  g#(cons(0, X)) >? g#(X) 
  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) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 3 
  cons = \y0y1.2 + y1 
  f = \y0.0 
  f# = \y0.0 
  false = 3 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.2 + 2y3 
  filter2# = \y0G1y2y3.1 + y3 
  filter# = \G0y1.y1 
  g = \y0.2 
  g# = \y0.0 
  h = \y0.0 
  map = \G0y1.y1 + 2G0(0) + 3y1G0(y1) 
  nil = 0 
  s = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] 
  [[g#(cons(0, _x0))]] = 0 >= 0 = [[g#(_x0)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 > 1 + x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 2 >= 2 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 2 >= 0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _x1)))]] 
  [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x2 + 2F0(0) + 3x2F0(2 + x2) + 6F0(2 + x2) >= 2 + x2 + 2F0(0) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 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 >= 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_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

  f#(s(X)) =#> f#(X)   
  g#(cons(0, X)) =#> g#(X)   

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:

  f#(s(X)) >? f#(X) 
  g#(cons(0, X)) >? g#(X) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 3 
  cons = \y0y1.1 + y1 
  f = \y0.0 
  f# = \y0.0 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  g = \y0.0 
  g# = \y0.y0 
  h = \y0.0 
  map = \G0y1.y1 
  nil = 0 
  s = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] 
  [[g#(cons(0, _x0))]] = 1 + x0 > x0 = [[g#(_x0)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 0 >= 0 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 0 >= 0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _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))]] = 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)]] 

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:

  f#(s(X)) =#> f#(X)   

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:

  f#(s(X)) >? f#(X) 
  f(s(X)) >= f(X) 
  g(cons(0, X)) >= g(X) 
  g(cons(s(X), Y)) >= s(X) 
  h(cons(X, Y)) >= h(g(cons(X, 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:

  0 = 0 
  cons = \y0y1.1 + 2y0 + 2y1 
  f = \y0.0 
  f# = \y0.y0 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + 2y2 + 2y3 
  g = \y0.y0 
  h = \y0.0 
  map = \G0y1.y1 + 2y1G0(y1) 
  nil = 0 
  s = \y0.1 + y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 1 + x0 > x0 = [[f#(_x0)]] 
  [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] 
  [[g(cons(0, _x0))]] = 1 + 2x0 >= x0 = [[g(_x0)]] 
  [[g(cons(s(_x0), _x1))]] = 3 + 2x0 + 2x1 >= 1 + x0 = [[s(_x0)]] 
  [[h(cons(_x0, _x1))]] = 0 >= 0 = [[h(g(cons(_x0, _x1)))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + 2x1 + 2x2 + 2F0(1 + 2x1 + 2x2) + 4x1F0(1 + 2x1 + 2x2) + 4x2F0(1 + 2x1 + 2x2) >= 1 + 2x2 + 2F0(x1) + 4x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + 2x1 + 2x2 >= 1 + 2x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x1 + 2x2 >= 1 + 2x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x1 + 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_5, 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.