We consider the system from.

  Alphabet:

    chain : [N -> N * list] --> list 
    cons : [N * list] --> list 
    false : [] --> B 
    from : [N * list] --> list 
    if : [B * list * list] --> list 
    incch : [list] --> list 
    lteq : [N * N] --> B 
    nil : [] --> list 
    o : [] --> N 
    s : [N] --> N 
    true : [] --> B 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    lteq(s(x), o) => false 
    lteq(o, x) => true 
    lteq(s(x), s(y)) => lteq(x, y) 
    from(x, nil) => nil 
    from(x, cons(y, z)) => if(lteq(x, y), cons(y, z), from(x, z)) 
    chain(f, nil) => nil 
    chain(f, cons(x, y)) => cons(f x, chain(f, from(f x, y))) 
    incch(x) => chain(/\y.s(y), x) 

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] lteq#(s(X), s(Y)) =#> lteq#(X, Y)   
    1] from#(X, cons(Y, Z)) =#> if#(lteq(X, Y), cons(Y, Z), from(X, Z))   
    2] from#(X, cons(Y, Z)) =#> lteq#(X, Y)   
    3] from#(X, cons(Y, Z)) =#> from#(X, Z)   
    4] chain#(F, cons(X, Y)) =#> chain#(F, from(F X, Y))   
    5] chain#(F, cons(X, Y)) =#> from#(F X, Y)   
    6] incch#(X) =#> chain#(/\x.s(x), X)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    lteq(s(X), o) => false 
    lteq(o, X) => true 
    lteq(s(X), s(Y)) => lteq(X, Y) 
    from(X, nil) => nil 
    from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
    chain(F, nil) => nil 
    chain(F, cons(X, Y)) => cons(F X, chain(F, from(F X, Y))) 
    incch(X) => chain(/\x.s(x), X) 

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:

  lteq#(s(X), s(Y)) >? lteq#(X, Y) 
  from#(X, cons(Y, Z)) >? if#(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  from#(X, cons(Y, Z)) >? lteq#(X, Y) 
  from#(X, cons(Y, Z)) >? from#(X, Z) 
  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  chain#(F, cons(X, Y)) >? from#(F X, Y) 
  incch#(X) >? chain#(/\x.s(x), X) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 
  pi( incch#(X) ) = #argfun-incch##(chain#(/\x.s(x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-incch# = \y0.3 + y0 
  #argfun-incch## = \y0.3 + y0 
  chain = \G0y1.2y1 + 2G0(0) + 2G0(y1) 
  chain# = \G0y1.3 
  cons = \y0y1.0 
  false = 0 
  from = \y0y1.2 
  from# = \y0y1.0 
  if = \y0y1y2.y1 + y2 
  if# = \y0y1y2.0 
  incch = \y0.0 
  incch# = \y0.0 
  lteq = \y0y1.y1 + 2y0 
  lteq# = \y0y1.0 
  nil = 2 
  o = 3 
  s = \y0.3 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[lteq#(s(_x0), s(_x1))]] = 0 >= 0 = [[lteq#(_x0, _x1)]] 
  [[from#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[if#(lteq(_x0, _x1), cons(_x1, _x2), from(_x0, _x2))]] 
  [[from#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[lteq#(_x0, _x1)]] 
  [[from#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[from#(_x0, _x2)]] 
  [[chain#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[chain#(_F0, from(_F0 _x1, _x2))]] 
  [[chain#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[from#(_F0 _x1, _x2)]] 
  [[#argfun-incch##(chain#(/\x.s(x), _x0))]] = 6 > 3 = [[chain#(/\x.s(x), _x0)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[lteq(s(_x0), o)]] = 9 + 4x0 >= 0 = [[false]] 
  [[lteq(o, _x0)]] = 6 + x0 >= 0 = [[true]] 
  [[lteq(s(_x0), s(_x1))]] = 9 + 2x1 + 4x0 >= x1 + 2x0 = [[lteq(_x0, _x1)]] 
  [[from(_x0, nil)]] = 2 >= 2 = [[nil]] 
  [[from(_x0, cons(_x1, _x2))]] = 2 >= 2 = [[if(lteq(_x0, _x1), cons(_x1, _x2), from(_x0, _x2))]] 
  [[chain(_F0, nil)]] = 4 + 2F0(0) + 2F0(2) >= 2 = [[nil]] 
  [[chain(_F0, cons(_x1, _x2))]] = 4F0(0) >= 0 = [[cons(_F0 _x1, chain(_F0, from(_F0 _x1, _x2)))]] 
  [[#argfun-incch#(chain(/\x.s(x), _x0))]] = 15 + 6x0 >= 12 + 6x0 = [[chain(/\x.s(x), _x0)]] 

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:

  lteq#(s(X), s(Y)) =#> lteq#(X, Y)   
  from#(X, cons(Y, Z)) =#> if#(lteq(X, Y), cons(Y, Z), from(X, Z))   
  from#(X, cons(Y, Z)) =#> lteq#(X, Y)   
  from#(X, cons(Y, Z)) =#> from#(X, Z)   
  chain#(F, cons(X, Y)) =#> chain#(F, from(F X, 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:

  lteq#(s(X), s(Y)) >? lteq#(X, Y) 
  from#(X, cons(Y, Z)) >? if#(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  from#(X, cons(Y, Z)) >? lteq#(X, Y) 
  from#(X, cons(Y, Z)) >? from#(X, Z) 
  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-incch# = \y0.3 + y0 
  chain = \G0y1.2 + 2G0(0) 
  chain# = \G0y1.0 
  cons = \y0y1.0 
  false = 0 
  from = \y0y1.0 
  from# = \y0y1.3 + 2y0 
  if = \y0y1y2.y1 + y2 
  if# = \y0y1y2.0 
  incch = \y0.0 
  lteq = \y0y1.0 
  lteq# = \y0y1.0 
  nil = 0 
  o = 3 
  s = \y0.3 
  true = 0 

Using this interpretation, the requirements translate to:

  [[lteq#(s(_x0), s(_x1))]] = 0 >= 0 = [[lteq#(_x0, _x1)]] 
  [[from#(_x0, cons(_x1, _x2))]] = 3 + 2x0 > 0 = [[if#(lteq(_x0, _x1), cons(_x1, _x2), from(_x0, _x2))]] 
  [[from#(_x0, cons(_x1, _x2))]] = 3 + 2x0 > 0 = [[lteq#(_x0, _x1)]] 
  [[from#(_x0, cons(_x1, _x2))]] = 3 + 2x0 >= 3 + 2x0 = [[from#(_x0, _x2)]] 
  [[chain#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[chain#(_F0, from(_F0 _x1, _x2))]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[lteq(s(_x0), o)]] = 0 >= 0 = [[false]] 
  [[lteq(o, _x0)]] = 0 >= 0 = [[true]] 
  [[lteq(s(_x0), s(_x1))]] = 0 >= 0 = [[lteq(_x0, _x1)]] 
  [[from(_x0, nil)]] = 0 >= 0 = [[nil]] 
  [[from(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[if(lteq(_x0, _x1), cons(_x1, _x2), from(_x0, _x2))]] 
  [[chain(_F0, nil)]] = 2 + 2F0(0) >= 0 = [[nil]] 
  [[chain(_F0, cons(_x1, _x2))]] = 2 + 2F0(0) >= 0 = [[cons(_F0 _x1, chain(_F0, from(_F0 _x1, _x2)))]] 
  [[#argfun-incch#(chain(/\x.s(x), _x0))]] = 11 >= 8 = [[chain(/\x.s(x), _x0)]] 

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:

  lteq#(s(X), s(Y)) =#> lteq#(X, Y)   
  from#(X, cons(Y, Z)) =#> from#(X, Z)   
  chain#(F, cons(X, Y)) =#> chain#(F, from(F X, 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:

  lteq#(s(X), s(Y)) >? lteq#(X, Y) 
  from#(X, cons(Y, Z)) >? from#(X, Z) 
  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-incch# = \y0.3 + y0 
  chain = \G0y1.2 + 2G0(y1) + 3G0(0) 
  chain# = \G0y1.0 
  cons = \y0y1.0 
  false = 0 
  from = \y0y1.2 
  from# = \y0y1.0 
  if = \y0y1y2.y1 + y2 
  incch = \y0.0 
  lteq = \y0y1.y0 + y1 
  lteq# = \y0y1.y0 + y1 
  nil = 0 
  o = 3 
  s = \y0.3 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[lteq#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[lteq#(_x0, _x1)]] 
  [[from#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[from#(_x0, _x2)]] 
  [[chain#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[chain#(_F0, from(_F0 _x1, _x2))]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[lteq(s(_x0), o)]] = 6 + 2x0 >= 0 = [[false]] 
  [[lteq(o, _x0)]] = 3 + x0 >= 0 = [[true]] 
  [[lteq(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 >= x0 + x1 = [[lteq(_x0, _x1)]] 
  [[from(_x0, nil)]] = 2 >= 0 = [[nil]] 
  [[from(_x0, cons(_x1, _x2))]] = 2 >= 2 = [[if(lteq(_x0, _x1), cons(_x1, _x2), from(_x0, _x2))]] 
  [[chain(_F0, nil)]] = 2 + 5F0(0) >= 0 = [[nil]] 
  [[chain(_F0, cons(_x1, _x2))]] = 2 + 5F0(0) >= 0 = [[cons(_F0 _x1, chain(_F0, from(_F0 _x1, _x2)))]] 
  [[#argfun-incch#(chain(/\x.s(x), _x0))]] = 20 + 4x0 >= 17 + 4x0 = [[chain(/\x.s(x), _x0)]] 

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:

  from#(X, cons(Y, Z)) =#> from#(X, Z)   
  chain#(F, cons(X, Y)) =#> chain#(F, from(F X, 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:

  from#(X, cons(Y, Z)) >? from#(X, Z) 
  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

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:

  [[chain#(x_1, x_2)]] = chain#(x_2, x_1) 
  [[false]] = _|_ 
  [[from(x_1, x_2)]] = from(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[incch(x_1)]] = x_1 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {chain#} and Mul = {#argfun-incch#, @_{o -> o}, chain, cons, from, from#, if, lteq, o, s}, and the following precedence: chain# > #argfun-incch# > chain > @_{o -> o} > cons > from > from# > if > lteq > s > o

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

  from#(X, cons(Y, Z)) >= from#(X, Z) 
  chain#(F, cons(X, Y)) > chain#(F, from(Y)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  lteq(s(X), o) >= _|_ 
  lteq(o, X) >= _|_ 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(_|_) >= _|_ 
  from(cons(X, Y)) >= if(cons(X, Y), from(Y)) 
  chain(F, _|_) >= _|_ 
  chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y))) 
  #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X) 

With these choices, we have:

  1] from#(X, cons(Y, Z)) >= from#(X, Z)  because from# in Mul, [2] and [3], by (Fun) 
  2] X >= X  by (Meta) 
  3] cons(Y, Z) >= Z  because [4], by (Star) 
  4] cons*(Y, Z) >= Z  because [5], by (Select) 
  5] Z >= Z  by (Meta) 

  6] chain#(F, cons(X, Y)) > chain#(F, from(Y))  because [7], by definition 
  7] chain#*(F, cons(X, Y)) >= chain#(F, from(Y))  because [8], [12] and [14], by (Stat) 
  8] cons(X, Y) > from(Y)  because [9], by definition 
  9] cons*(X, Y) >= from(Y)  because cons > from and [10], by (Copy) 
  10] cons*(X, Y) >= Y  because [11], by (Select) 
  11] Y >= Y  by (Meta) 
  12] chain#*(F, cons(X, Y)) >= F  because [13], by (Select) 
  13] F >= F  by (Meta) 
  14] chain#*(F, cons(X, Y)) >= from(Y)  because [15], by (Select) 
  15] cons(X, Y) >= from(Y)  because [9], by (Star) 

  16] if(X, Y) >= X  because [17], by (Star) 
  17] if*(X, Y) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 

  19] if(X, Y) >= Y  because [20], by (Star) 
  20] if*(X, Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] lteq(s(X), o) >= _|_  by (Bot) 

  23] lteq(o, X) >= _|_  by (Bot) 

  24] lteq(s(X), s(Y)) >= lteq(X, Y)  because [25], by (Star) 
  25] lteq*(s(X), s(Y)) >= lteq(X, Y)  because lteq in Mul, [26] and [29], by (Stat) 
  26] s(X) >= X  because [27], by (Star) 
  27] s*(X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 
  29] s(Y) > Y  because [30], by definition 
  30] s*(Y) >= Y  because [31], by (Select) 
  31] Y >= Y  by (Meta) 

  32] from(_|_) >= _|_  by (Bot) 

  33] from(cons(X, Y)) >= if(cons(X, Y), from(Y))  because [34], by (Star) 
  34] from*(cons(X, Y)) >= if(cons(X, Y), from(Y))  because from > if, [35] and [39], by (Copy) 
  35] from*(cons(X, Y)) >= cons(X, Y)  because [36], by (Select) 
  36] cons(X, Y) >= cons(X, Y)  because cons in Mul, [37] and [38], by (Fun) 
  37] X >= X  by (Meta) 
  38] Y >= Y  by (Meta) 
  39] from*(cons(X, Y)) >= from(Y)  because [40], by (Select) 
  40] cons(X, Y) >= from(Y)  because [41], by (Star) 
  41] cons*(X, Y) >= from(Y)  because cons > from and [42], by (Copy) 
  42] cons*(X, Y) >= Y  because [38], by (Select) 

  43] chain(F, _|_) >= _|_  by (Bot) 

  44] chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because [45], by (Star) 
  45] chain*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because chain > cons, [46] and [52], by (Copy) 
  46] chain*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because chain > @_{o -> o}, [47] and [48], by (Copy) 
  47] chain*(F, cons(X, Y)) >= F  because [13], by (Select) 
  48] chain*(F, cons(X, Y)) >= X  because [49], by (Select) 
  49] cons(X, Y) >= X  because [50], by (Star) 
  50] cons*(X, Y) >= X  because [51], by (Select) 
  51] X >= X  by (Meta) 
  52] chain*(F, cons(X, Y)) >= chain(F, from(Y))  because chain in Mul, [53] and [8], by (Stat) 
  53] F >= F  by (Meta) 

  54] #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because [55], by (Star) 
  55] #argfun-incch#*(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because #argfun-incch# > chain, [56] and [60], by (Copy) 
  56] #argfun-incch#*(chain(/\x.s(x), X)) >= /\x.s(x)  because [57], by (F-Abs) 
  57] #argfun-incch#*(chain(/\x.s(x), X), y) >= s(y)  because #argfun-incch# > s and [58], by (Copy) 
  58] #argfun-incch#*(chain(/\x.s(x), X), y) >= y  because [59], by (Select) 
  59] y >= y  by (Var) 
  60] #argfun-incch#*(chain(/\x.s(x), X)) >= X  because [61], by (Select) 
  61] chain(/\x.s(x), X) >= X  because [62], by (Star) 
  62] chain*(/\x.s(x), X) >= X  because [63], by (Select) 
  63] X >= X  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_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

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

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:

  from#(X, cons(Y, Z)) >? from#(X, Z) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, nil) >= nil 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, nil) >= nil 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

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]] = _|_ 
  [[from(x_1, x_2)]] = from(x_2) 
  [[from#(x_1, x_2)]] = from#(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[incch(x_1)]] = x_1 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {#argfun-incch#, @_{o -> o}, chain, cons, from, from#, if, lteq, o, s}, and the following precedence: #argfun-incch# > o > chain > @_{o -> o} > s > cons = from > from# > lteq > if

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

  from#(cons(X, Y)) > from#(Y) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  lteq(s(X), o) >= _|_ 
  lteq(o, X) >= _|_ 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(_|_) >= _|_ 
  from(cons(X, Y)) >= if(cons(X, Y), from(Y)) 
  chain(F, _|_) >= _|_ 
  chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y))) 
  #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X) 

With these choices, we have:

  1] from#(cons(X, Y)) > from#(Y)  because [2], by definition 
  2] from#*(cons(X, Y)) >= from#(Y)  because [3], by (Select) 
  3] cons(X, Y) >= from#(Y)  because [4], by (Star) 
  4] cons*(X, Y) >= from#(Y)  because cons > from# and [5], by (Copy) 
  5] cons*(X, Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 

  7] if(X, Y) >= X  because [8], by (Star) 
  8] if*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 

  10] if(X, Y) >= Y  because [11], by (Star) 
  11] if*(X, Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] lteq(s(X), o) >= _|_  by (Bot) 

  14] lteq(o, X) >= _|_  by (Bot) 

  15] lteq(s(X), s(Y)) >= lteq(X, Y)  because lteq in Mul, [16] and [19], by (Fun) 
  16] s(X) >= X  because [17], by (Star) 
  17] s*(X) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 
  19] s(Y) >= Y  because [20], by (Star) 
  20] s*(Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] from(_|_) >= _|_  by (Bot) 

  23] from(cons(X, Y)) >= if(cons(X, Y), from(Y))  because [24], by (Star) 
  24] from*(cons(X, Y)) >= if(cons(X, Y), from(Y))  because from > if, [25] and [29], by (Copy) 
  25] from*(cons(X, Y)) >= cons(X, Y)  because [26], by (Select) 
  26] cons(X, Y) >= cons(X, Y)  because cons in Mul, [27] and [28], by (Fun) 
  27] X >= X  by (Meta) 
  28] Y >= Y  by (Meta) 
  29] from*(cons(X, Y)) >= from(Y)  because [30], by (Select) 
  30] cons(X, Y) >= from(Y)  because [31], by (Star) 
  31] cons*(X, Y) >= from(Y)  because cons = from, cons in Mul and [28], by (Stat) 

  32] chain(F, _|_) >= _|_  by (Bot) 

  33] chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because [34], by (Star) 
  34] chain*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because chain > cons, [35] and [42], by (Copy) 
  35] chain*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because chain > @_{o -> o}, [36] and [38], by (Copy) 
  36] chain*(F, cons(X, Y)) >= F  because [37], by (Select) 
  37] F >= F  by (Meta) 
  38] chain*(F, cons(X, Y)) >= X  because [39], by (Select) 
  39] cons(X, Y) >= X  because [40], by (Star) 
  40] cons*(X, Y) >= X  because [41], by (Select) 
  41] X >= X  by (Meta) 
  42] chain*(F, cons(X, Y)) >= chain(F, from(Y))  because chain in Mul, [43] and [44], by (Stat) 
  43] F >= F  by (Meta) 
  44] cons(X, Y) > from(Y)  because [45], by definition 
  45] cons*(X, Y) >= from(Y)  because cons = from, cons in Mul and [46], by (Stat) 
  46] Y >= Y  by (Meta) 

  47] #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because [48], by (Star) 
  48] #argfun-incch#*(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because #argfun-incch# > chain, [49] and [53], by (Copy) 
  49] #argfun-incch#*(chain(/\x.s(x), X)) >= /\x.s(x)  because [50], by (F-Abs) 
  50] #argfun-incch#*(chain(/\x.s(x), X), y) >= s(y)  because #argfun-incch# > s and [51], by (Copy) 
  51] #argfun-incch#*(chain(/\x.s(x), X), y) >= y  because [52], by (Select) 
  52] y >= y  by (Var) 
  53] #argfun-incch#*(chain(/\x.s(x), X)) >= X  because [54], by (Select) 
  54] chain(/\x.s(x), X) >= X  because [55], by (Star) 
  55] chain*(/\x.s(x), X) >= X  because [56], by (Select) 
  56] X >= X  by (Meta) 

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.

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.