We consider the system qsort.

  Alphabet:

    0 : [] --> nat 
    app : [list * list] --> list 
    cons : [nat * list] --> list 
    false : [] --> bool 
    filter : [nat -> bool * list] --> list 
    gr : [nat * nat] --> bool 
    if : [bool * list * list] --> list 
    le : [nat * nat] --> bool 
    nil : [] --> list 
    qsort : [list] --> list 
    s : [nat] --> nat 
    true : [] --> bool 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    gr(0, x) => false 
    gr(s(x), 0) => true 
    gr(s(x), s(y)) => gr(x, y) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(f, y)) 
    qsort(nil) => nil 
    qsort(cons(x, y)) => app(qsort(filter(/\z.le(z, x), y)), app(cons(x, nil), qsort(filter(/\u.gr(u, x), 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] app#(cons(X, Y), Z) =#> app#(Y, Z)   
    1] le#(s(X), s(Y)) =#> le#(X, Y)   
    2] gr#(s(X), s(Y)) =#> gr#(X, Y)   
    3] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    4] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    5] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    6] qsort#(cons(X, Y)) =#> app#(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y))))   
    7] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    8] qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
    9] qsort#(cons(X, Y)) =#> le#(Z, X)   
    10] qsort#(cons(X, Y)) =#> app#(cons(X, nil), qsort(filter(/\x.gr(x, X), Y)))   
    11] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
    12] qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), Y)   
    13] qsort#(cons(X, Y)) =#> gr#(Z, X)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    gr(0, X) => false 
    gr(s(X), 0) => true 
    gr(s(X), s(Y)) => gr(X, Y) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
    qsort(nil) => nil 
    qsort(cons(X, Y)) => app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), 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 : 2 
    * 3 :  
    * 4 : 3, 4, 5 
    * 5 : 3, 4, 5 
    * 6 : 0 
    * 7 : 6, 7, 8, 9, 10, 11, 12, 13 
    * 8 : 3, 4, 5 
    * 9 : 1 
    * 10 : 0 
    * 11 : 6, 7, 8, 9, 10, 11, 12, 13 
    * 12 : 3, 4, 5 
    * 13 : 2 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

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

  P_4:

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

  P_5:

    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), 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:

  qsort#(cons(X, Y)) >? qsort#(filter(/\x.le(x, X), Y)) 
  qsort#(cons(X, Y)) >? qsort#(filter(/\x.gr(x, X), Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0, X) >= false 
  gr(s(X), 0) >= true 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
  qsort(nil) >= nil 
  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), 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:

  [[false]] = _|_ 
  [[filter(x_1, x_2)]] = filter(x_2) 
  [[gr(x_1, x_2)]] = gr(x_1) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {0, @_{o -> o}, app, cons, filter, gr, if, le, qsort, qsort#, s}, and the following precedence: 0 > qsort > app > le > cons = filter > if > qsort# > gr > s > @_{o -> o}

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

  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
  qsort#(cons(X, Y)) >= qsort#(filter(Y)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  app(_|_, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= _|_ 
  le(s(X), 0) >= _|_ 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0) >= _|_ 
  gr(s(X)) >= _|_ 
  gr(s(X)) >= gr(X) 
  filter(_|_) >= _|_ 
  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
  qsort(_|_) >= _|_ 
  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 

With these choices, we have:

  1] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [2], by definition 
  2] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [3], by (Stat) 
  3] cons(X, Y) > filter(Y)  because [4], by definition 
  4] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 
  5] Y >= Y  by (Meta) 

  6] qsort#(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [7], by (Fun) 
  7] cons(X, Y) >= filter(Y)  because [8], by (Star) 
  8] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 

  9] if(X, Y) >= X  because [10], by (Star) 
  10] if*(X, Y) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] if(X, Y) >= Y  because [13], by (Star) 
  13] if*(X, Y) >= Y  because [14], by (Select) 
  14] Y >= Y  by (Meta) 

  15] app(_|_, X) >= X  because [16], by (Star) 
  16] app*(_|_, X) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 

  18] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [19], by (Star) 
  19] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [20] and [24], by (Copy) 
  20] app*(cons(X, Y), Z) >= X  because [21], by (Select) 
  21] cons(X, Y) >= X  because [22], by (Star) 
  22] cons*(X, Y) >= X  because [23], by (Select) 
  23] X >= X  by (Meta) 
  24] app*(cons(X, Y), Z) >= app(Y, Z)  because app in Mul, [25] and [28], by (Stat) 
  25] cons(X, Y) > Y  because [26], by definition 
  26] cons*(X, Y) >= Y  because [27], by (Select) 
  27] Y >= Y  by (Meta) 
  28] Z >= Z  by (Meta) 

  29] le(0, X) >= _|_  by (Bot) 

  30] le(s(X), 0) >= _|_  by (Bot) 

  31] le(s(X), s(Y)) >= le(X, Y)  because [32], by (Star) 
  32] le*(s(X), s(Y)) >= le(X, Y)  because le in Mul, [33] and [36], by (Stat) 
  33] s(X) > X  because [34], by definition 
  34] s*(X) >= X  because [35], by (Select) 
  35] X >= X  by (Meta) 
  36] s(Y) >= Y  because [37], by (Star) 
  37] s*(Y) >= Y  because [38], by (Select) 
  38] Y >= Y  by (Meta) 

  39] gr(0) >= _|_  by (Bot) 

  40] gr(s(X)) >= _|_  by (Bot) 

  41] gr(s(X)) >= gr(X)  because [42], by (Star) 
  42] gr*(s(X)) >= gr(X)  because gr in Mul and [43], by (Stat) 
  43] s(X) > X  because [44], by definition 
  44] s*(X) >= X  because [45], by (Select) 
  45] X >= X  by (Meta) 

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

  47] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [48], by (Star) 
  48] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [49] and [56], by (Copy) 
  49] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [50] and [53], by (Stat) 
  50] cons(X, Y) > X  because [51], by definition 
  51] cons*(X, Y) >= X  because [52], by (Select) 
  52] X >= X  by (Meta) 
  53] cons(X, Y) > filter(Y)  because [54], by definition 
  54] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [55], by (Stat) 
  55] Y >= Y  by (Meta) 
  56] filter*(cons(X, Y)) >= filter(Y)  because filter in Mul and [57], by (Stat) 
  57] cons(X, Y) > Y  because [58], by definition 
  58] cons*(X, Y) >= Y  because [55], by (Select) 

  59] qsort(_|_) >= _|_  by (Bot) 

  60] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [61], by (Star) 
  61] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [62] and [63], by (Copy) 
  62] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [3], by (Stat) 
  63] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [64] and [68], by (Copy) 
  64] qsort*(cons(X, Y)) >= cons(X, _|_)  because [65], by (Select) 
  65] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [66] and [67], by (Fun) 
  66] X >= X  by (Meta) 
  67] Y >= _|_  by (Bot) 
  68] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [69], by (Stat) 
  69] cons(X, Y) > filter(Y)  because [70], by definition 
  70] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 

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:

  qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, 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 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:

  qsort#(cons(X, Y)) >? qsort#(filter(/\x.gr(x, X), Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0, X) >= false 
  gr(s(X), 0) >= true 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
  qsort(nil) >= nil 
  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), 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:

  [[false]] = _|_ 
  [[filter(x_1, x_2)]] = filter(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[le(x_1, x_2)]] = le(x_1) 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {0, @_{o -> o}, app, cons, filter, gr, if, le, qsort, qsort#, s}, and the following precedence: 0 > @_{o -> o} > qsort > app > cons = filter > qsort# > gr > if > le = s

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

  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  app(_|_, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0) >= _|_ 
  le(s(X)) >= _|_ 
  le(s(X)) >= le(X) 
  gr(0, X) >= _|_ 
  gr(s(X), 0) >= _|_ 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(_|_) >= _|_ 
  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
  qsort(_|_) >= _|_ 
  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 

With these choices, we have:

  1] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [2], by definition 
  2] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [3], by (Stat) 
  3] cons(X, Y) > filter(Y)  because [4], by definition 
  4] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 
  5] Y >= Y  by (Meta) 

  6] if(X, Y) >= X  because [7], by (Star) 
  7] if*(X, Y) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] if(X, Y) >= Y  because [10], by (Star) 
  10] if*(X, Y) >= Y  because [11], by (Select) 
  11] Y >= Y  by (Meta) 

  12] app(_|_, X) >= X  because [13], by (Star) 
  13] app*(_|_, X) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 

  15] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [16], by (Star) 
  16] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [17] and [21], by (Copy) 
  17] app*(cons(X, Y), Z) >= X  because [18], by (Select) 
  18] cons(X, Y) >= X  because [19], by (Star) 
  19] cons*(X, Y) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 
  21] app*(cons(X, Y), Z) >= app(Y, Z)  because app in Mul, [22] and [25], by (Stat) 
  22] cons(X, Y) > Y  because [23], by definition 
  23] cons*(X, Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 
  25] Z >= Z  by (Meta) 

  26] le(0) >= _|_  by (Bot) 

  27] le(s(X)) >= _|_  by (Bot) 

  28] le(s(X)) >= le(X)  because [29], by (Star) 
  29] le*(s(X)) >= le(X)  because [30], by (Select) 
  30] s(X) >= le(X)  because s = le, s in Mul and [31], by (Fun) 
  31] X >= X  by (Meta) 

  32] gr(0, X) >= _|_  by (Bot) 

  33] gr(s(X), 0) >= _|_  by (Bot) 

  34] gr(s(X), s(Y)) >= gr(X, Y)  because gr in Mul, [35] and [38], by (Fun) 
  35] s(X) >= X  because [36], by (Star) 
  36] s*(X) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] s(Y) >= Y  because [39], by (Star) 
  39] s*(Y) >= Y  because [40], by (Select) 
  40] Y >= Y  by (Meta) 

  41] filter(_|_) >= _|_  by (Bot) 

  42] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [43], by (Star) 
  43] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [44] and [51], by (Copy) 
  44] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [45] and [48], by (Stat) 
  45] cons(X, Y) > X  because [46], by definition 
  46] cons*(X, Y) >= X  because [47], by (Select) 
  47] X >= X  by (Meta) 
  48] cons(X, Y) > filter(Y)  because [49], by definition 
  49] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [50], by (Stat) 
  50] Y >= Y  by (Meta) 
  51] filter*(cons(X, Y)) >= filter(Y)  because filter in Mul and [52], by (Stat) 
  52] cons(X, Y) > Y  because [53], by definition 
  53] cons*(X, Y) >= Y  because [50], by (Select) 

  54] qsort(_|_) >= _|_  by (Bot) 

  55] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [56], by (Star) 
  56] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [57] and [60], by (Copy) 
  57] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [58], by (Stat) 
  58] cons(X, Y) > filter(Y)  because [59], by definition 
  59] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 
  60] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [61] and [65], by (Copy) 
  61] qsort*(cons(X, Y)) >= cons(X, _|_)  because [62], by (Select) 
  62] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [63] and [64], by (Fun) 
  63] X >= X  by (Meta) 
  64] Y >= _|_  by (Bot) 
  65] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [3], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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), (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 with usable rules [Kop12, Thm. 7.44].  (P_4, R_0) has no usable rules.

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

  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y1 
  filter# = \G0y1.y1 + y1G0(y1) 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 3 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > x2 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > x2 + x2F0(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_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 with usable rules [Kop12, Thm. 7.44].  (P_3, R_0) has no usable rules.

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

  gr#(s(X), s(Y)) >? gr#(X, Y) 

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

The following interpretation satisfies the requirements:

  gr# = \y0y1.y0 + y1 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[gr#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[gr#(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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 with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

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

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

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

The following interpretation satisfies the requirements:

  le# = \y0y1.y0 + y1 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[le#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[le#(_x0, _x1)]] 

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 with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

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

  app#(cons(X, Y), Z) >? app#(Y, Z) 

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

The following interpretation satisfies the requirements:

  app# = \y0y1.y0 
  cons = \y0y1.1 + 2y1 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 1 + 2x1 > x1 = [[app#(_x1, _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.