We consider the system Applicative_first_order_05__#3.18.

  Alphabet:

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

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    double(0) => 0 
    double(s(x)) => s(s(double(x))) 
    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    plus(s(x), y) => plus(x, s(y)) 
    plus(s(x), y) => s(plus(minus(x, y), double(y))) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] minus#(s(X), s(Y)) =#> minus#(X, Y)   
    1] double#(s(X)) =#> double#(X)   
    2] plus#(s(X), Y) =#> plus#(X, Y)   
    3] plus#(s(X), Y) =#> plus#(X, s(Y))   
    4] plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y))   
    5] plus#(s(X), Y) =#> minus#(X, Y)   
    6] plus#(s(X), Y) =#> double#(Y)   
    7] map#(F, cons(X, Y)) =#> map#(F, Y)   
    8] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    9] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    10] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    double(0) => 0 
    double(s(X)) => s(s(double(X))) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    plus(s(X), Y) => plus(X, s(Y)) 
    plus(s(X), Y) => s(plus(minus(X, Y), double(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).

The formative rules of (P_0, R_0) are R_1 ::=

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  double(0) => 0 
  double(s(X)) => s(s(double(X))) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  plus(s(X), Y) => plus(X, s(Y)) 
  plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  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) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, static, formative) by (P_0, R_1, static, formative).

Thus, the original system is terminating if (P_0, R_1, static, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) 
  plus#(s(X), Y) >? minus#(X, Y) 
  plus#(s(X), Y) >? double#(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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  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 
  double = \y0.0 
  double# = \y0.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  map = \G0y1.2G0(y1) + 3G0(0) 
  map# = \G0y1.0 
  minus = \y0y1.y0 
  minus# = \y0y1.0 
  plus = \y0y1.y1 
  plus# = \y0y1.3 
  s = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] 
  [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(_x0, s(_x1))]] 
  [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(minus(_x0, _x1), double(_x1))]] 
  [[plus#(s(_x0), _x1)]] = 3 > 0 = [[minus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 3 > 0 = [[double#(_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)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] 
  [[double(0)]] = 0 >= 0 = [[0]] 
  [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[plus(_x0, s(_x1))]] 
  [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_x1)))]] 
  [[map(_F0, cons(_x1, _x2))]] = 5F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[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_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  double#(s(X)) =#> double#(X)   
  plus#(s(X), Y) =#> plus#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, s(Y))   
  plus#(s(X), Y) =#> plus#(minus(X, Y), double(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_1, static, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(s(X), Y) >? plus#(minus(X, Y), double(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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  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 + y0 + 2y1 
  double = \y0.0 
  double# = \y0.0 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + y2 + 2y3 
  filter2# = \y0G1y2y3.2 + 2y3 
  filter# = \G0y1.2y1 
  map = \G0y1.y1 + G0(y1) + 2y1G0(y1) 
  map# = \G0y1.0 
  minus = \y0y1.y0 
  minus# = \y0y1.0 
  plus = \y0y1.y1 
  plus# = \y0y1.0 
  s = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(minus(_x0, _x1), double(_x1))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 4x2 > 2 + 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 2 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] 
  [[double(0)]] = 0 >= 0 = [[0]] 
  [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = x1 >= x1 = [[plus(_x0, s(_x1))]] 
  [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_x1)))]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 + 2x1F0(2 + x1 + 2x2) + 4x2F0(2 + x1 + 2x2) + 5F0(2 + x1 + 2x2) >= 2 + 2x2 + F0(x1) + 2F0(x2) + 4x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 >= 2 + x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 2 + x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of:

  minus#(s(X), s(Y)) =#> minus#(X, Y)   
  double#(s(X)) =#> double#(X)   
  plus#(s(X), Y) =#> plus#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, s(Y))   
  plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y))   
  map#(F, cons(X, Y)) =#> map#(F, Y)   

Thus, the original system is terminating if (P_2, R_1, static, formative) is finite.

We consider the dependency pair problem (P_2, R_1, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  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 + y0 + y1 
  double = \y0.0 
  double# = \y0.0 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y2 + y3 
  map = \G0y1.y1 + 2y1G0(y1) 
  map# = \G0y1.y1 
  minus = \y0y1.y0 
  minus# = \y0y1.0 
  plus = \y0y1.2y1 
  plus# = \y0y1.0 
  s = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] 
  [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] 
  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(minus(_x0, _x1), double(_x1))]] 
  [[map#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] 
  [[double(0)]] = 0 >= 0 = [[0]] 
  [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] 
  [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[plus(_x0, s(_x1))]] 
  [[plus(s(_x0), _x1)]] = 2x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_x1)))]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) >= 1 + x2 + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + x1 + 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_2, R_1, static, formative) by (P_3, R_1, static, formative), where P_3 consists of:

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

Thus, the original system is terminating if (P_3, R_1, static, formative) is finite.

We consider the dependency pair problem (P_3, R_1, static, formative).

The formative rules of (P_3, R_1) are R_2 ::=

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  double(0) => 0 
  double(s(X)) => s(s(double(X))) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  plus(s(X), Y) => plus(X, s(Y)) 
  plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_1, static, formative) by (P_3, R_2, static, formative).

Thus, the original system is terminating if (P_3, R_2, static, formative) is finite.

We consider the dependency pair problem (P_3, R_2, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

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:

  [[0]] = _|_ 
  [[minus(x_1, x_2)]] = minus(x_1) 
  [[minus#(x_1, x_2)]] = x_1 
  [[plus#(x_1, x_2)]] = plus#(x_1) 

We choose Lex = {minus, plus} and Mul = {double, double#, plus#, s}, and the following precedence: double# > plus > double > s > minus > plus#

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

  s(X) >= X 
  double#(s(X)) >= double#(X) 
  plus#(s(X)) >= plus#(X) 
  plus#(s(X)) >= plus#(X) 
  plus#(s(X)) > plus#(minus(X, Y)) 
  minus(X, _|_) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(_|_) >= _|_ 
  double(s(X)) >= s(s(double(X))) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

With these choices, we have:

  1] s(X) >= X  because [2], by (Star) 
  2] s*(X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] double#(s(X)) >= double#(X)  because double# in Mul and [5], by (Fun) 
  5] s(X) >= X  because [6], by (Star) 
  6] s*(X) >= X  because [7], by (Select) 
  7] X >= X  by (Meta) 

  8] plus#(s(X)) >= plus#(X)  because plus# in Mul and [9], by (Fun) 
  9] s(X) >= X  because [10], by (Star) 
  10] s*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] plus#(s(X)) >= plus#(X)  because plus# in Mul and [13], by (Fun) 
  13] s(X) >= X  because [14], by (Star) 
  14] s*(X) >= X  because [15], by (Select) 
  15] X >= X  by (Meta) 

  16] plus#(s(X)) > plus#(minus(X, Y))  because [17], by definition 
  17] plus#*(s(X)) >= plus#(minus(X, Y))  because [18], by (Select) 
  18] s(X) >= plus#(minus(X, Y))  because [19], by (Star) 
  19] s*(X) >= plus#(minus(X, Y))  because s > plus# and [20], by (Copy) 
  20] s*(X) >= minus(X, Y)  because s > minus and [21], by (Copy) 
  21] s*(X) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 

  23] minus(X, _|_) >= X  because [24], by (Star) 
  24] minus*(X, _|_) >= X  because [25], by (Select) 
  25] X >= X  by (Meta) 

  26] minus(s(X), s(Y)) >= minus(X, Y)  because [27], by (Fun) 
  27] s(X) >= X  because [2], by (Star) 

  28] double(_|_) >= _|_  by (Bot) 

  29] double(s(X)) >= s(s(double(X)))  because [30], by (Star) 
  30] double*(s(X)) >= s(s(double(X)))  because double > s and [31], by (Copy) 
  31] double*(s(X)) >= s(double(X))  because double > s and [32], by (Copy) 
  32] double*(s(X)) >= double(X)  because double in Mul and [33], by (Stat) 
  33] s(X) > X  because [34], by definition 
  34] s*(X) >= X  because [7], by (Select) 

  35] plus(_|_, X) >= X  because [36], by (Star) 
  36] plus*(_|_, X) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 

  38] plus(s(X), Y) >= s(plus(X, Y))  because [39], by (Star) 
  39] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [40], by (Copy) 
  40] plus*(s(X), Y) >= plus(X, Y)  because [41], [43] and [44], by (Stat) 
  41] s(X) > X  because [42], by definition 
  42] s*(X) >= X  because [11], by (Select) 
  43] plus*(s(X), Y) >= X  because [9], by (Select) 
  44] plus*(s(X), Y) >= Y  because [45], by (Select) 
  45] Y >= Y  by (Meta) 

  46] plus(s(X), Y) >= plus(X, s(Y))  because [47], by (Star) 
  47] plus*(s(X), Y) >= plus(X, s(Y))  because [48], [50] and [51], by (Stat) 
  48] s(X) > X  because [49], by definition 
  49] s*(X) >= X  because [15], by (Select) 
  50] plus*(s(X), Y) >= X  because [13], by (Select) 
  51] plus*(s(X), Y) >= s(Y)  because plus > s and [52], by (Copy) 
  52] plus*(s(X), Y) >= Y  because [53], by (Select) 
  53] Y >= Y  by (Meta) 

  54] plus(s(X), Y) >= s(plus(minus(X, Y), double(Y)))  because [55], by (Star) 
  55] plus*(s(X), Y) >= s(plus(minus(X, Y), double(Y)))  because plus > s and [56], by (Copy) 
  56] plus*(s(X), Y) >= plus(minus(X, Y), double(Y))  because [57], [58] and [61], by (Stat) 
  57] s(X) > minus(X, Y)  because [20], by definition 
  58] plus*(s(X), Y) >= minus(X, Y)  because plus > minus and [59], by (Copy) 
  59] plus*(s(X), Y) >= X  because [60], by (Select) 
  60] s(X) >= X  because [21], by (Star) 
  61] plus*(s(X), Y) >= double(Y)  because plus > double and [62], by (Copy) 
  62] plus*(s(X), Y) >= Y  because [63], by (Select) 
  63] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_2, static, formative) by (P_4, R_2, static, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_2, static, formative) is finite.

We consider the dependency pair problem (P_4, R_2, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

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:

  [[0]] = _|_ 
  [[minus(x_1, x_2)]] = x_1 

We choose Lex = {plus, plus#} and Mul = {double, double#, minus#, s}, and the following precedence: double# > minus# > plus > double > plus# > s

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

  minus#(s(X), s(Y)) >= minus#(X, Y) 
  double#(s(X)) >= double#(X) 
  plus#(s(X), Y) >= plus#(X, Y) 
  plus#(s(X), Y) > plus#(X, s(Y)) 
  X >= X 
  s(X) >= X 
  double(_|_) >= _|_ 
  double(s(X)) >= s(s(double(X))) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(X, double(Y))) 

With these choices, we have:

  1] minus#(s(X), s(Y)) >= minus#(X, Y)  because minus# in Mul, [2] and [5], by (Fun) 
  2] s(X) >= X  because [3], by (Star) 
  3] s*(X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 
  5] s(Y) >= Y  because [6], by (Star) 
  6] s*(Y) >= Y  because [7], by (Select) 
  7] Y >= Y  by (Meta) 

  8] double#(s(X)) >= double#(X)  because double# in Mul and [9], by (Fun) 
  9] s(X) >= X  because [10], by (Star) 
  10] s*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] plus#(s(X), Y) >= plus#(X, Y)  because [13] and [16], by (Fun) 
  13] s(X) >= X  because [14], by (Star) 
  14] s*(X) >= X  because [15], by (Select) 
  15] X >= X  by (Meta) 
  16] Y >= Y  by (Meta) 

  17] plus#(s(X), Y) > plus#(X, s(Y))  because [18], by definition 
  18] plus#*(s(X), Y) >= plus#(X, s(Y))  because [19], [22] and [24], by (Stat) 
  19] s(X) > X  because [20], by definition 
  20] s*(X) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] plus#*(s(X), Y) >= X  because [23], by (Select) 
  23] s(X) >= X  because [20], by (Star) 
  24] plus#*(s(X), Y) >= s(Y)  because plus# > s and [25], by (Copy) 
  25] plus#*(s(X), Y) >= Y  because [26], by (Select) 
  26] Y >= Y  by (Meta) 

  27] X >= X  by (Meta) 

  28] s(X) >= X  because [3], by (Star) 

  29] double(_|_) >= _|_  by (Bot) 

  30] double(s(X)) >= s(s(double(X)))  because [31], by (Star) 
  31] double*(s(X)) >= s(s(double(X)))  because double > s and [32], by (Copy) 
  32] double*(s(X)) >= s(double(X))  because double > s and [33], by (Copy) 
  33] double*(s(X)) >= double(X)  because double in Mul and [34], by (Stat) 
  34] s(X) > X  because [35], by definition 
  35] s*(X) >= X  because [11], by (Select) 

  36] plus(_|_, X) >= X  because [37], by (Star) 
  37] plus*(_|_, X) >= X  because [38], by (Select) 
  38] X >= X  by (Meta) 

  39] plus(s(X), Y) >= s(plus(X, Y))  because [40], by (Star) 
  40] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [41], by (Copy) 
  41] plus*(s(X), Y) >= plus(X, Y)  because [42], [44] and [45], by (Stat) 
  42] s(X) > X  because [43], by definition 
  43] s*(X) >= X  because [15], by (Select) 
  44] plus*(s(X), Y) >= X  because [13], by (Select) 
  45] plus*(s(X), Y) >= Y  because [16], by (Select) 

  46] plus(s(X), Y) >= plus(X, s(Y))  because [47], by (Star) 
  47] plus*(s(X), Y) >= plus(X, s(Y))  because [19], [48] and [49], by (Stat) 
  48] plus*(s(X), Y) >= X  because [23], by (Select) 
  49] plus*(s(X), Y) >= s(Y)  because plus > s and [50], by (Copy) 
  50] plus*(s(X), Y) >= Y  because [26], by (Select) 

  51] plus(s(X), Y) >= s(plus(X, double(Y)))  because [52], by (Star) 
  52] plus*(s(X), Y) >= s(plus(X, double(Y)))  because plus > s and [53], by (Copy) 
  53] plus*(s(X), Y) >= plus(X, double(Y))  because [54], [57] and [59], by (Stat) 
  54] s(X) > X  because [55], by definition 
  55] s*(X) >= X  because [56], by (Select) 
  56] X >= X  by (Meta) 
  57] plus*(s(X), Y) >= X  because [58], by (Select) 
  58] s(X) >= X  because [55], by (Star) 
  59] plus*(s(X), Y) >= double(Y)  because plus > double and [60], by (Copy) 
  60] plus*(s(X), Y) >= Y  because [61], by (Select) 
  61] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_2, static, formative) by (P_5, R_2, static, formative), where P_5 consists of:

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

Thus, the original system is terminating if (P_5, R_2, static, formative) is finite.

We consider the dependency pair problem (P_5, R_2, 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) 
  double#(s(X)) >? double#(X) 
  plus#(s(X), Y) >? plus#(X, Y) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

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:

  [[0]] = _|_ 
  [[minus(x_1, x_2)]] = minus(x_1) 

We choose Lex = {plus} and Mul = {double, double#, minus, minus#, plus#, s}, and the following precedence: minus# > plus# > plus > double > double# = s > minus

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

  minus#(s(X), s(Y)) >= minus#(X, Y) 
  double#(s(X)) > double#(X) 
  plus#(s(X), Y) >= plus#(X, Y) 
  minus(X) >= X 
  minus(s(X)) >= minus(X) 
  double(_|_) >= _|_ 
  double(s(X)) >= s(s(double(X))) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X), double(Y))) 

With these choices, we have:

  1] minus#(s(X), s(Y)) >= minus#(X, Y)  because minus# in Mul, [2] and [5], by (Fun) 
  2] s(X) >= X  because [3], by (Star) 
  3] s*(X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 
  5] s(Y) >= Y  because [6], by (Star) 
  6] s*(Y) >= Y  because [7], by (Select) 
  7] Y >= Y  by (Meta) 

  8] double#(s(X)) > double#(X)  because [9], by definition 
  9] double#*(s(X)) >= double#(X)  because [10], by (Select) 
  10] s(X) >= double#(X)  because s = double#, s in Mul and [11], by (Fun) 
  11] X >= X  by (Meta) 

  12] plus#(s(X), Y) >= plus#(X, Y)  because plus# in Mul, [13] and [16], by (Fun) 
  13] s(X) >= X  because [14], by (Star) 
  14] s*(X) >= X  because [15], by (Select) 
  15] X >= X  by (Meta) 
  16] Y >= Y  by (Meta) 

  17] minus(X) >= X  because [18], by (Star) 
  18] minus*(X) >= X  because [19], by (Select) 
  19] X >= X  by (Meta) 

  20] minus(s(X)) >= minus(X)  because minus in Mul and [2], by (Fun) 

  21] double(_|_) >= _|_  by (Bot) 

  22] double(s(X)) >= s(s(double(X)))  because [23], by (Star) 
  23] double*(s(X)) >= s(s(double(X)))  because double > s and [24], by (Copy) 
  24] double*(s(X)) >= s(double(X))  because double > s and [25], by (Copy) 
  25] double*(s(X)) >= double(X)  because double in Mul and [26], by (Stat) 
  26] s(X) > X  because [27], by definition 
  27] s*(X) >= X  because [11], by (Select) 

  28] plus(_|_, X) >= X  because [29], by (Star) 
  29] plus*(_|_, X) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 

  31] plus(s(X), Y) >= s(plus(X, Y))  because [32], by (Star) 
  32] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [33], by (Copy) 
  33] plus*(s(X), Y) >= plus(X, Y)  because [34], [36] and [37], by (Stat) 
  34] s(X) > X  because [35], by definition 
  35] s*(X) >= X  because [15], by (Select) 
  36] plus*(s(X), Y) >= X  because [13], by (Select) 
  37] plus*(s(X), Y) >= Y  because [16], by (Select) 

  38] plus(s(X), Y) >= plus(X, s(Y))  because [39], by (Star) 
  39] plus*(s(X), Y) >= plus(X, s(Y))  because [40], [43] and [45], by (Stat) 
  40] s(X) > X  because [41], by definition 
  41] s*(X) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] plus*(s(X), Y) >= X  because [44], by (Select) 
  44] s(X) >= X  because [41], by (Star) 
  45] plus*(s(X), Y) >= s(Y)  because plus > s and [46], by (Copy) 
  46] plus*(s(X), Y) >= Y  because [47], by (Select) 
  47] Y >= Y  by (Meta) 

  48] plus(s(X), Y) >= s(plus(minus(X), double(Y)))  because [49], by (Star) 
  49] plus*(s(X), Y) >= s(plus(minus(X), double(Y)))  because plus > s and [50], by (Copy) 
  50] plus*(s(X), Y) >= plus(minus(X), double(Y))  because [51], [55] and [58], by (Stat) 
  51] s(X) > minus(X)  because [52], by definition 
  52] s*(X) >= minus(X)  because s > minus and [53], by (Copy) 
  53] s*(X) >= X  because [54], by (Select) 
  54] X >= X  by (Meta) 
  55] plus*(s(X), Y) >= minus(X)  because plus > minus and [56], by (Copy) 
  56] plus*(s(X), Y) >= X  because [57], by (Select) 
  57] s(X) >= X  because [53], by (Star) 
  58] plus*(s(X), Y) >= double(Y)  because plus > double and [59], by (Copy) 
  59] plus*(s(X), Y) >= Y  because [60], by (Select) 
  60] Y >= Y  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_2, static, formative) by (P_6, R_2, static, formative), where P_6 consists of:

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

Thus, the original system is terminating if (P_6, R_2, static, formative) is finite.

We consider the dependency pair problem (P_6, R_2, 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) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

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:

  [[0]] = _|_ 
  [[minus(x_1, x_2)]] = minus(x_1) 
  [[minus#(x_1, x_2)]] = minus#(x_1) 

We choose Lex = {plus} and Mul = {double, minus, minus#, plus#, s}, and the following precedence: plus > double > plus# > minus# = s > minus

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

  minus#(s(X)) >= minus#(X) 
  plus#(s(X), Y) > plus#(X, Y) 
  minus(X) >= X 
  minus(s(X)) >= minus(X) 
  double(_|_) >= _|_ 
  double(s(X)) >= s(s(double(X))) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X), double(Y))) 

With these choices, we have:

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

  5] plus#(s(X), Y) > plus#(X, Y)  because [6], by definition 
  6] plus#*(s(X), Y) >= plus#(X, Y)  because plus# in Mul, [7] and [10], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] Y >= Y  by (Meta) 

  11] minus(X) >= X  because [12], by (Star) 
  12] minus*(X) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] minus(s(X)) >= minus(X)  because [15], by (Star) 
  15] minus*(s(X)) >= minus(X)  because [16], by (Select) 
  16] s(X) >= minus(X)  because [17], by (Star) 
  17] s*(X) >= minus(X)  because s > minus and [18], by (Copy) 
  18] s*(X) >= X  because [4], by (Select) 

  19] double(_|_) >= _|_  by (Bot) 

  20] double(s(X)) >= s(s(double(X)))  because [21], by (Star) 
  21] double*(s(X)) >= s(s(double(X)))  because double > s and [22], by (Copy) 
  22] double*(s(X)) >= s(double(X))  because double > s and [23], by (Copy) 
  23] double*(s(X)) >= double(X)  because double in Mul and [24], by (Stat) 
  24] s(X) > X  because [25], by definition 
  25] s*(X) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 

  27] plus(_|_, X) >= X  because [28], by (Star) 
  28] plus*(_|_, X) >= X  because [29], by (Select) 
  29] X >= X  by (Meta) 

  30] plus(s(X), Y) >= s(plus(X, Y))  because [31], by (Star) 
  31] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [32], by (Copy) 
  32] plus*(s(X), Y) >= plus(X, Y)  because [7], [33] and [35], by (Stat) 
  33] plus*(s(X), Y) >= X  because [34], by (Select) 
  34] s(X) >= X  because [8], by (Star) 
  35] plus*(s(X), Y) >= Y  because [10], by (Select) 

  36] plus(s(X), Y) >= plus(X, s(Y))  because [37], by (Star) 
  37] plus*(s(X), Y) >= plus(X, s(Y))  because [38], [41] and [43], by (Stat) 
  38] s(X) > X  because [39], by definition 
  39] s*(X) >= X  because [40], by (Select) 
  40] X >= X  by (Meta) 
  41] plus*(s(X), Y) >= X  because [42], by (Select) 
  42] s(X) >= X  because [39], by (Star) 
  43] plus*(s(X), Y) >= s(Y)  because plus > s and [44], by (Copy) 
  44] plus*(s(X), Y) >= Y  because [45], by (Select) 
  45] Y >= Y  by (Meta) 

  46] plus(s(X), Y) >= s(plus(minus(X), double(Y)))  because [47], by (Star) 
  47] plus*(s(X), Y) >= s(plus(minus(X), double(Y)))  because plus > s and [48], by (Copy) 
  48] plus*(s(X), Y) >= plus(minus(X), double(Y))  because [49], [53] and [56], by (Stat) 
  49] s(X) > minus(X)  because [50], by definition 
  50] s*(X) >= minus(X)  because s > minus and [51], by (Copy) 
  51] s*(X) >= X  because [52], by (Select) 
  52] X >= X  by (Meta) 
  53] plus*(s(X), Y) >= minus(X)  because plus > minus and [54], by (Copy) 
  54] plus*(s(X), Y) >= X  because [55], by (Select) 
  55] s(X) >= X  because [51], by (Star) 
  56] plus*(s(X), Y) >= double(Y)  because plus > double and [57], by (Copy) 
  57] plus*(s(X), Y) >= Y  because [58], by (Select) 
  58] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_2, static, formative) by (P_7, R_2, static, formative), where P_7 consists of:

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

Thus, the original system is terminating if (P_7, R_2, static, formative) is finite.

We consider the dependency pair problem (P_7, R_2, 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(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X, Y), double(Y))) 

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:

  [[0]] = _|_ 
  [[minus(x_1, x_2)]] = minus(x_1) 
  [[minus#(x_1, x_2)]] = minus#(x_1) 

We choose Lex = {plus} and Mul = {double, minus, minus#, s}, and the following precedence: plus > double > minus# = s > minus

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

  minus#(s(X)) > minus#(X) 
  minus(X) >= X 
  minus(s(X)) >= minus(X) 
  double(_|_) >= _|_ 
  double(s(X)) >= s(s(double(X))) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(s(X), Y) >= s(plus(minus(X), double(Y))) 

With these choices, we have:

  1] minus#(s(X)) > minus#(X)  because [2], by definition 
  2] minus#*(s(X)) >= minus#(X)  because [3], by (Select) 
  3] s(X) >= minus#(X)  because s = minus#, s in Mul and [4], by (Fun) 
  4] X >= X  by (Meta) 

  5] minus(X) >= X  because [6], by (Star) 
  6] minus*(X) >= X  because [7], by (Select) 
  7] X >= X  by (Meta) 

  8] minus(s(X)) >= minus(X)  because minus in Mul and [9], by (Fun) 
  9] s(X) >= X  because [10], by (Star) 
  10] s*(X) >= X  because [4], by (Select) 

  11] double(_|_) >= _|_  by (Bot) 

  12] double(s(X)) >= s(s(double(X)))  because [13], by (Star) 
  13] double*(s(X)) >= s(s(double(X)))  because double > s and [14], by (Copy) 
  14] double*(s(X)) >= s(double(X))  because double > s and [15], by (Copy) 
  15] double*(s(X)) >= double(X)  because double in Mul and [16], by (Stat) 
  16] s(X) > X  because [17], by definition 
  17] s*(X) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 

  19] plus(_|_, X) >= X  because [20], by (Star) 
  20] plus*(_|_, X) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 

  22] plus(s(X), Y) >= s(plus(X, Y))  because [23], by (Star) 
  23] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [24], by (Copy) 
  24] plus*(s(X), Y) >= plus(X, Y)  because [25], [28] and [30], by (Stat) 
  25] s(X) > X  because [26], by definition 
  26] s*(X) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 
  28] plus*(s(X), Y) >= X  because [29], by (Select) 
  29] s(X) >= X  because [26], by (Star) 
  30] plus*(s(X), Y) >= Y  because [31], by (Select) 
  31] Y >= Y  by (Meta) 

  32] plus(s(X), Y) >= plus(X, s(Y))  because [33], by (Star) 
  33] plus*(s(X), Y) >= plus(X, s(Y))  because [34], [37] and [39], by (Stat) 
  34] s(X) > X  because [35], by definition 
  35] s*(X) >= X  because [36], by (Select) 
  36] X >= X  by (Meta) 
  37] plus*(s(X), Y) >= X  because [38], by (Select) 
  38] s(X) >= X  because [35], by (Star) 
  39] plus*(s(X), Y) >= s(Y)  because plus > s and [40], by (Copy) 
  40] plus*(s(X), Y) >= Y  because [41], by (Select) 
  41] Y >= Y  by (Meta) 

  42] plus(s(X), Y) >= s(plus(minus(X), double(Y)))  because [43], by (Star) 
  43] plus*(s(X), Y) >= s(plus(minus(X), double(Y)))  because plus > s and [44], by (Copy) 
  44] plus*(s(X), Y) >= plus(minus(X), double(Y))  because [45], [49] and [51], by (Stat) 
  45] s(X) > minus(X)  because [46], by definition 
  46] s*(X) >= minus(X)  because s > minus and [47], by (Copy) 
  47] s*(X) >= X  because [48], by (Select) 
  48] X >= X  by (Meta) 
  49] plus*(s(X), Y) >= minus(X)  because [50], by (Select) 
  50] s(X) >= minus(X)  because [46], by (Star) 
  51] plus*(s(X), Y) >= double(Y)  because plus > double and [52], by (Copy) 
  52] plus*(s(X), Y) >= Y  because [53], by (Select) 
  53] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_7, R_2) by ({}, R_2).  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.