We consider the system Applicative_first_order_05__perfect2.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    f : [a * a * a * a] --> b 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    if : [b * b * b] --> b 
    le : [a * a] --> b 
    map : [c -> c * d] --> d 
    minus : [a * a] --> a 
    nil : [] --> d 
    perfectp : [a] --> b 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    minus(0, x) => 0 
    minus(s(x), 0) => s(x) 
    minus(s(x), s(y)) => minus(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    if(true, x, y) => x 
    if(false, x, y) => y 
    perfectp(0) => false 
    perfectp(s(x)) => f(x, s(0), s(x), s(x)) 
    f(0, x, 0, y) => true 
    f(0, x, s(y), z) => false 
    f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) 
    f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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] minus#(s(X), s(Y)) =#> minus#(X, Y)   
    1] le#(s(X), s(Y)) =#> le#(X, Y)   
    2] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X))   
    3] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z)   
    4] f#(s(X), 0, Y, Z) =#> minus#(Y, s(X))   
    5] f#(s(X), s(Y), Z, U) =#> if#(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U))   
    6] f#(s(X), s(Y), Z, U) =#> le#(X, Y)   
    7] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U)   
    8] f#(s(X), s(Y), Z, U) =#> minus#(Y, X)   
    9] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U)   
    10] map#(F, cons(X, Y)) =#> map#(F, Y)   
    11] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    12] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    13] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(0, X) => 0 
    minus(s(X), 0) => s(X) 
    minus(s(X), s(Y)) => minus(X, Y) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    perfectp(0) => false 
    perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
    f(0, X, 0, Y) => true 
    f(0, X, s(Y), Z) => false 
    f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
    f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
    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 : 0 
    * 1 : 1 
    * 2 : 5, 6, 7, 8, 9 
    * 3 : 3, 4, 5, 6, 7, 8, 9 
    * 4 : 0 
    * 5 :  
    * 6 : 1 
    * 7 : 3, 4, 5, 6, 7, 8, 9 
    * 8 : 0 
    * 9 : 3, 4, 5, 6, 7, 8, 9 
    * 10 : 10 
    * 11 : 12, 13 
    * 12 : 11 
    * 13 : 11 

This graph has the following strongly connected components:

  P_1:

    minus#(s(X), s(Y)) =#> minus#(X, Y)   

  P_2:

    le#(s(X), s(Y)) =#> le#(X, Y)   

  P_3:

    f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z)   
    f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U)   
    f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U)   

  P_4:

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

  P_5:

    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), (P_4, R_0, m, f) and (P_5, 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), (P_4, 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 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) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 + y1 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  filter2# = \y0G1y2y3.1 + y3 
  filter# = \G0y1.y1 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.2y1 
  minus = \y0y1.3 
  nil = 0 
  perfectp = \y0.3 
  s = \y0.0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 1 + 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)]] 
  [[minus(0, _x0)]] = 3 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 3 >= 0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 3 >= 3 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[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)]] 

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:

  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, 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), (P_4, R_0, static, formative) and (P_6, R_0, static, formative) is finite.

We consider the dependency pair problem (P_6, 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 :  

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), (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:

  map#(F, cons(X, Y)) >? map#(F, Y) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 + y1 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.2 + 2y3 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  minus = \y0y1.3 
  nil = 0 
  perfectp = \y0.3 
  s = \y0.0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[minus(0, _x0)]] = 3 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 3 >= 0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 3 >= 3 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[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 >= 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 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, 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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(s(X), 0, Y, Z) >? f#(X, Z, minus(Y, s(X)), Z) 
  f#(s(X), s(Y), Z, U) >? f#(s(X), minus(Y, X), Z, U) 
  f#(s(X), s(Y), Z, U) >? f#(X, U, Z, U) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 = \y0y1.2 
  f = \y0y1y2y3.0 
  f# = \y0y1y2y3.2y0 
  false = 0 
  filter = \G0y1.2 
  filter2 = \y0G1y2y3.2 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.2 + 2y1G0(y1) 
  minus = \y0y1.2y0 
  nil = 0 
  perfectp = \y0.3 
  s = \y0.1 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0), 0, _x1, _x2)]] = 2 + 2x0 > 2x0 = [[f#(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f#(s(_x0), s(_x1), _x2, _x3)]] = 2 + 2x0 >= 2 + 2x0 = [[f#(s(_x0), minus(_x1, _x0), _x2, _x3)]] 
  [[f#(s(_x0), s(_x1), _x2, _x3)]] = 2 + 2x0 > 2x0 = [[f#(_x0, _x3, _x2, _x3)]] 
  [[minus(0, _x0)]] = 0 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 2 + 2x0 >= 1 + x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 2 + 2x0 >= 2x0 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[map(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 4F0(2) >= 2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 >= 2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 >= 2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 >= 2 = [[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_7, R_0, static, formative), where P_7 consists of:

  f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U)   

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

  f#(s(X), s(Y), Z, U) >? f#(s(X), minus(Y, X), Z, U) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 = \y0y1.0 
  f = \y0y1y2y3.0 
  f# = \y0y1y2y3.2y1 
  false = 0 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.0 
  minus = \y0y1.y0 
  nil = 0 
  perfectp = \y0.3 
  s = \y0.2 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0), s(_x1), _x2, _x3)]] = 4 + 2x1 > 2x1 = [[f#(s(_x0), minus(_x1, _x0), _x2, _x3)]] 
  [[minus(0, _x0)]] = 0 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 2 + x0 >= 2 + x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 2 + x0 >= x0 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[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)]] 

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.

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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  le#(s(X), s(Y)) >? le#(X, Y) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 = \y0y1.0 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  le# = \y0y1.y1 
  map = \G0y1.0 
  minus = \y0y1.3 + 2y0 
  nil = 0 
  perfectp = \y0.3 + 2y0 
  s = \y0.2 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[le#(s(_x0), s(_x1))]] = 2 + x1 > x1 = [[le#(_x0, _x1)]] 
  [[minus(0, _x0)]] = 3 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 7 + 2x0 >= 2 + x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 7 + 2x0 >= 3 + 2x0 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 7 + 2x0 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[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)]] 

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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  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 = \y0y1.0 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.0 
  minus = \y0y1.3 + 2y0 
  minus# = \y0y1.y1 
  nil = 0 
  perfectp = \y0.3 + 2y0 
  s = \y0.2 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 2 + x1 > x1 = [[minus#(_x0, _x1)]] 
  [[minus(0, _x0)]] = 3 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 7 + 2x0 >= 2 + x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 7 + 2x0 >= 3 + 2x0 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 7 + 2x0 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[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)]] 

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.