We consider the system Applicative_first_order_05__#3.57.

  Alphabet:

    0 : [] --> b 
    app : [c * c] --> c 
    cons : [b * c] --> c 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    map : [b -> b * c] --> c 
    minus : [b * b] --> b 
    nil : [] --> c 
    plus : [b * b] --> b 
    quot : [b * b] --> b 
    s : [b] --> b 
    sum : [c] --> c 
    true : [] --> a 

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    minus(minus(x, y), z) => minus(x, plus(y, z)) 
    quot(0, s(x)) => 0 
    quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) 
    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    app(nil, x) => x 
    app(x, nil) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    sum(cons(x, nil)) => cons(x, nil) 
    sum(cons(x, cons(y, z))) => sum(cons(plus(x, y), z)) 
    sum(app(x, cons(y, cons(z, u)))) => sum(app(x, sum(cons(y, cons(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) 

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] minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z))   
    2] minus#(minus(X, Y), Z) =#> plus#(Y, Z)   
    3] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y))   
    4] quot#(s(X), s(Y)) =#> minus#(X, Y)   
    5] plus#(s(X), Y) =#> plus#(X, Y)   
    6] app#(cons(X, Y), Z) =#> app#(Y, Z)   
    7] sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
    8] sum#(cons(X, cons(Y, Z))) =#> plus#(X, Y)   
    9] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
    10] sum#(app(X, cons(Y, cons(Z, U)))) =#> app#(X, sum(cons(Y, cons(Z, U))))   
    11] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U)))   
    12] map#(F, cons(X, Y)) =#> map#(F, Y)   
    13] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    14] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    15] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    app(nil, X) => X 
    app(X, nil) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    sum(cons(X, nil)) => cons(X, nil) 
    sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) 
    sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(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 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#(minus(X, Y), Z) >? minus#(X, plus(Y, Z)) 
  minus#(minus(X, Y), Z) >? plus#(Y, Z) 
  quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) 
  quot#(s(X), s(Y)) >? minus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, Y) 
  app#(cons(X, Y), Z) >? app#(Y, Z) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(cons(X, cons(Y, Z))) >? plus#(X, Y) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? app#(X, sum(cons(Y, cons(Z, U)))) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(cons(Y, cons(Z, U))) 
  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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.2y1 + 3y0 
  app# = \y0y1.0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.2G0(0) 
  filter2 = \y0G1y2y3.2G1(0) 
  filter2# = \y0G1y2y3.1 
  filter# = \G0y1.1 
  map = \G0y1.2G0(y1) 
  map# = \G0y1.0 
  minus = \y0y1.y0 
  minus# = \y0y1.0 
  nil = 0 
  plus = \y0y1.y0 + y1 
  plus# = \y0y1.0 
  quot = \y0y1.2y0 
  quot# = \y0y1.3 + y0 + 3y1 
  s = \y0.1 + y0 
  sum = \y0.0 
  sum# = \y0.3 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[minus#(minus(_x0, _x1), _x2)]] = 0 >= 0 = [[minus#(_x0, plus(_x1, _x2))]] 
  [[minus#(minus(_x0, _x1), _x2)]] = 0 >= 0 = [[plus#(_x1, _x2)]] 
  [[quot#(s(_x0), s(_x1))]] = 7 + x0 + 3x1 > 6 + x0 + 3x1 = [[quot#(minus(_x0, _x1), s(_x1))]] 
  [[quot#(s(_x0), s(_x1))]] = 7 + x0 + 3x1 > 0 = [[minus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 3 >= 3 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 3 > 0 = [[plus#(_x0, _x1)]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 3 >= 3 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 3 > 0 = [[app#(_x0, sum(cons(_x1, cons(_x2, _x3))))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 3 >= 3 = [[sum#(cons(_x1, cons(_x2, _x3)))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 1 >= 1 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 1 >= 1 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 1 >= 1 = [[filter#(_F0, _x2)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 2 + 2x0 >= 1 + 2x0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[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)]] 

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:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z))   
  minus#(minus(X, Y), Z) =#> plus#(Y, Z)   
  plus#(s(X), Y) =#> plus#(X, Y)   
  app#(cons(X, Y), Z) =#> app#(Y, Z)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U)))   
  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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  minus#(minus(X, Y), Z) >? minus#(X, plus(Y, Z)) 
  minus#(minus(X, Y), Z) >? plus#(Y, Z) 
  plus#(s(X), Y) >? plus#(X, Y) 
  app#(cons(X, Y), Z) >? app#(Y, Z) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(cons(Y, cons(Z, U))) 
  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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.2y1 + 3y0 
  app# = \y0y1.0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  map = \G0y1.0 
  map# = \G0y1.0 
  minus = \y0y1.1 + y0 + y1 
  minus# = \y0y1.3 + 2y0 
  nil = 0 
  plus = \y0y1.y1 
  plus# = \y0y1.0 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.0 
  sum# = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 3 + 2x0 >= 3 + 2x0 = [[minus#(_x0, _x1)]] 
  [[minus#(minus(_x0, _x1), _x2)]] = 5 + 2x0 + 2x1 > 3 + 2x0 = [[minus#(_x0, plus(_x1, _x2))]] 
  [[minus#(minus(_x0, _x1), _x2)]] = 5 + 2x0 + 2x1 > 0 = [[plus#(_x1, _x2)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(cons(_x1, cons(_x2, _x3)))]] 
  [[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)]] 
  [[minus(_x0, 0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 1 + x0 + x2 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_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 the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, Y)   
  app#(cons(X, Y), Z) =#> app#(Y, Z)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U)))   
  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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, Y) 
  app#(cons(X, Y), Z) >? app#(Y, Z) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(cons(Y, cons(Z, U))) 
  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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.2 + y1 + 3y0 
  app# = \y0y1.0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  map = \G0y1.0 
  map# = \G0y1.0 
  minus = \y0y1.2y0 
  minus# = \y0y1.0 
  nil = 0 
  plus = \y0y1.y1 
  plus# = \y0y1.0 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.0 
  sum# = \y0.2y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 4 + 6x0 >= 4 + 6x0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 4 + 6x0 > 0 = [[sum#(cons(_x1, cons(_x2, _x3)))]] 
  [[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)]] 
  [[minus(_x0, 0)]] = 2x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 2x0 >= 2x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = 4x0 >= 2x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 2 + x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 2 + 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2 + x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_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 the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, Y)   
  app#(cons(X, Y), Z) =#> app#(Y, Z)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
  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_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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, Y) 
  app#(cons(X, Y), Z) >? app#(Y, Z) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.y1 + 3y0 
  app# = \y0y1.0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  map = \G0y1.0 
  map# = \G0y1.0 
  minus = \y0y1.y0 
  minus# = \y0y1.0 
  nil = 0 
  plus = \y0y1.y0 + 2y1 
  plus# = \y0y1.y0 
  quot = \y0y1.y0 
  s = \y0.1 + y0 
  sum = \y0.0 
  sum# = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(_x0, _x1)]] 
  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[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)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_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 the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  app#(cons(X, Y), Z) =#> app#(Y, Z)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
  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_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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  app#(cons(X, Y), Z) >? app#(Y, Z) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.2y1 + 3y0 
  app# = \y0y1.y0 
  cons = \y0y1.2 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + y3 
  filter2# = \y0G1y2y3.y3 + G1(0) + 2y3G1(y3) + 2G1(y3) 
  filter# = \G0y1.y1 + G0(0) + 2y1G0(y1) 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  minus = \y0y1.2y0 
  minus# = \y0y1.0 
  nil = 0 
  plus = \y0y1.2y1 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.2 
  sum# = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[app#(cons(_x0, _x1), _x2)]] = 2 + x1 > x1 = [[app#(_x1, _x2)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 2 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + F0(0) + 2x2F0(2 + x2) + 4F0(2 + x2) > x2 + F0(0) + 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = x2 + F0(0) + 2x2F0(x2) + 2F0(x2) >= x2 + F0(0) + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = x2 + F0(0) + 2x2F0(x2) + 2F0(x2) >= x2 + F0(0) + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[minus(_x0, 0)]] = 2x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 2x0 >= 2x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = 4x0 >= 2x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 6 + 2x2 + 3x1 >= 2 + 2x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 2 >= 2 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 2 >= 2 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 2 >= 2 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 >= 2 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 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:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.y1 + 3y0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.3 
  filter# = \G0y1.0 
  map = \G0y1.0 
  minus = \y0y1.3 + 3y0 
  minus# = \y0y1.0 
  nil = 0 
  plus = \y0y1.y1 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.0 
  sum# = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] 
  [[minus(_x0, 0)]] = 3 + 3x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 3 + 3x0 >= 3 + 3x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = 12 + 9x0 >= 3 + 3x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_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 the dependency pair problem (P_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   

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:

  minus#(s(X), s(Y)) >? minus#(X, Y) 
  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.y1 + 3y0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  map = \G0y1.0 
  minus = \y0y1.y0 
  minus# = \y0y1.y0 
  nil = 0 
  plus = \y0y1.y0 + y1 
  quot = \y0y1.2y0 + 2y1 
  s = \y0.2 + y0 
  sum = \y0.0 
  sum# = \y0.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 2 + x0 > x0 = [[minus#(_x0, _x1)]] 
  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 2 + x0 >= x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 4 + 2x0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 8 + 2x0 + 2x1 >= 6 + 2x0 + 2x1 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 0 >= 0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 0 >= 0 = [[sum(app(_x0, sum(cons(_x1, cons(_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 the dependency pair problem (P_6, R_0, static, formative) by (P_7, R_0, static, formative), where P_7 consists of:

  sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z))   
  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   

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:

  sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) 
  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.y1 + 3y0 
  cons = \y0y1.1 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  map = \G0y1.2y1 
  minus = \y0y1.y0 
  nil = 1 
  plus = \y0y1.2y1 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.2 
  sum# = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[sum#(cons(_x0, cons(_x1, _x2)))]] = 2 + x2 > 1 + x2 = [[sum#(cons(plus(_x0, _x1), _x2))]] 
  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 2 + x3 + 3x0 >= 2 + 3x0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 3 + x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 1 + 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 3 + x2 + 3x1 >= 1 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 2 >= 2 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 2 >= 2 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 2 >= 2 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[map(_F0, nil)]] = 2 >= 1 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 1 >= 1 = [[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_7, R_0, static, formative) by (P_8, R_0, static, formative), where P_8 consists of:

  sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U)))))   

Thus, the original system is terminating if (P_8, R_0, static, formative) is finite.

We consider the dependency pair problem (P_8, 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:

  sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(Y))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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 
  app = \y0y1.2y1 + 3y0 
  cons = \y0y1.2 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + y3 
  map = \G0y1.2y1 
  minus = \y0y1.y0 
  nil = 0 
  plus = \y0y1.2y1 
  quot = \y0y1.0 
  s = \y0.y0 
  sum = \y0.2 
  sum# = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 8 + 2x3 + 3x0 > 4 + 3x0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] 
  [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] 
  [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[s(plus(_x0, _x1))]] 
  [[app(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 6 + 2x2 + 3x1 >= 2 + 2x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 2 >= 2 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 2 >= 2 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 2 >= 2 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 >= 2 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 >= 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_8, 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.