We consider the system Applicative_first_order_05__#3.16.

  Alphabet:

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

  Rules:

    times(x, 0) => 0 
    times(x, s(y)) => plus(times(x, y), x) 
    plus(x, 0) => x 
    plus(0, x) => x 
    plus(x, s(y)) => s(plus(x, y)) 
    plus(s(x), y) => s(plus(x, 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] times#(X, s(Y)) =#> plus#(times(X, Y), X)   
    1] times#(X, s(Y)) =#> times#(X, Y)   
    2] plus#(X, s(Y)) =#> plus#(X, Y)   
    3] plus#(s(X), Y) =#> plus#(X, Y)   
    4] map#(F, cons(X, Y)) =#> map#(F, Y)   
    5] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    6] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    7] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    times(X, 0) => 0 
    times(X, s(Y)) => plus(times(X, Y), X) 
    plus(X, 0) => X 
    plus(0, X) => X 
    plus(X, s(Y)) => s(plus(X, Y)) 
    plus(s(X), Y) => s(plus(X, 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 ::=

  times(X, 0) => 0 
  times(X, s(Y)) => plus(times(X, Y), X) 
  plus(X, 0) => X 
  plus(0, X) => X 
  plus(X, s(Y)) => s(plus(X, Y)) 
  plus(s(X), Y) => s(plus(X, 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:

  times#(X, s(Y)) >? plus#(times(X, Y), X) 
  times#(X, s(Y)) >? times#(X, Y) 
  plus#(X, s(Y)) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, 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) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_3, x_1) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_2, x_4) 

We choose Lex = {filter, filter2} and Mul = {@_{o -> o}, cons, false, filter2#, filter#, map, map#, plus, plus#, s, times, times#, true}, and the following precedence: times# > plus# > filter2# = filter# > map > filter = filter2 > cons > times > false > map# > @_{o -> o} > true > plus > s

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

  times#(X, s(Y)) >= plus#(times(X, Y), X) 
  times#(X, s(Y)) >= times#(X, Y) 
  plus#(X, s(Y)) > plus#(X, Y) 
  plus#(s(X), Y) >= plus#(X, Y) 
  map#(F, cons(X, Y)) >= map#(F, Y) 
  filter#(F, cons(X, Y)) >= filter2#(F, Y) 
  filter2#(F, X) >= filter#(F, X) 
  filter2#(F, X) >= filter#(F, X) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

With these choices, we have:

  1] times#(X, s(Y)) >= plus#(times(X, Y), X)  because [2], by (Star) 
  2] times#*(X, s(Y)) >= plus#(times(X, Y), X)  because times# > plus#, [3] and [4], by (Copy) 
  3] times#*(X, s(Y)) >= times(X, Y)  because times# > times, [4] and [6], by (Copy) 
  4] times#*(X, s(Y)) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 
  6] times#*(X, s(Y)) >= Y  because [7], by (Select) 
  7] s(Y) >= Y  because [8], by (Star) 
  8] s*(Y) >= Y  because [9], by (Select) 
  9] Y >= Y  by (Meta) 

  10] times#(X, s(Y)) >= times#(X, Y)  because times# in Mul, [11] and [12], by (Fun) 
  11] X >= X  by (Meta) 
  12] s(Y) >= Y  because [8], by (Star) 

  13] plus#(X, s(Y)) > plus#(X, Y)  because [14], by definition 
  14] plus#*(X, s(Y)) >= plus#(X, Y)  because plus# in Mul, [15] and [16], by (Stat) 
  15] X >= X  by (Meta) 
  16] s(Y) > Y  because [17], by definition 
  17] s*(Y) >= Y  because [18], by (Select) 
  18] Y >= Y  by (Meta) 

  19] plus#(s(X), Y) >= plus#(X, Y)  because plus# in Mul, [20] and [23], by (Fun) 
  20] s(X) >= X  because [21], by (Star) 
  21] s*(X) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 
  23] Y >= Y  by (Meta) 

  24] map#(F, cons(X, Y)) >= map#(F, Y)  because map# in Mul, [25] and [26], by (Fun) 
  25] F >= F  by (Meta) 
  26] cons(X, Y) >= Y  because [27], by (Star) 
  27] cons*(X, Y) >= Y  because [28], by (Select) 
  28] Y >= Y  by (Meta) 

  29] filter#(F, cons(X, Y)) >= filter2#(F, Y)  because filter# = filter2#, filter# in Mul, [30] and [31], by (Fun) 
  30] F >= F  by (Meta) 
  31] cons(X, Y) >= Y  because [32], by (Star) 
  32] cons*(X, Y) >= Y  because [33], by (Select) 
  33] Y >= Y  by (Meta) 

  34] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [35] and [36], by (Fun) 
  35] F >= F  by (Meta) 
  36] X >= X  by (Meta) 

  37] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [38] and [39], by (Fun) 
  38] F >= F  by (Meta) 
  39] X >= X  by (Meta) 

  40] times(X, _|_) >= _|_  by (Bot) 

  41] times(X, s(Y)) >= plus(times(X, Y), X)  because [42], by (Star) 
  42] times*(X, s(Y)) >= plus(times(X, Y), X)  because times > plus, [43] and [46], by (Copy) 
  43] times*(X, s(Y)) >= times(X, Y)  because times in Mul, [11] and [44], by (Stat) 
  44] s(Y) > Y  because [45], by definition 
  45] s*(Y) >= Y  because [9], by (Select) 
  46] times*(X, s(Y)) >= X  because [11], by (Select) 

  47] plus(X, _|_) >= X  because [48], by (Star) 
  48] plus*(X, _|_) >= X  because [49], by (Select) 
  49] X >= X  by (Meta) 

  50] plus(_|_, X) >= X  because [51], by (Star) 
  51] plus*(_|_, X) >= X  because [52], by (Select) 
  52] X >= X  by (Meta) 

  53] plus(X, s(Y)) >= s(plus(X, Y))  because [54], by (Star) 
  54] plus*(X, s(Y)) >= s(plus(X, Y))  because plus > s and [55], by (Copy) 
  55] plus*(X, s(Y)) >= plus(X, Y)  because plus in Mul, [15] and [16], by (Stat) 

  56] plus(s(X), Y) >= s(plus(X, Y))  because [57], by (Star) 
  57] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [58], by (Copy) 
  58] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [59] and [23], by (Stat) 
  59] s(X) > X  because [60], by definition 
  60] s*(X) >= X  because [22], by (Select) 

  61] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [62], by (Star) 
  62] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [63] and [69], by (Copy) 
  63] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [64] and [65], by (Copy) 
  64] map*(F, cons(X, Y)) >= F  because [25], by (Select) 
  65] map*(F, cons(X, Y)) >= X  because [66], by (Select) 
  66] cons(X, Y) >= X  because [67], by (Star) 
  67] cons*(X, Y) >= X  because [68], by (Select) 
  68] X >= X  by (Meta) 
  69] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [25] and [70], by (Stat) 
  70] cons(X, Y) > Y  because [71], by definition 
  71] cons*(X, Y) >= Y  because [28], by (Select) 

  72] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [73], by (Star) 
  73] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [30], [74], [76], [77], [78] and [82], by (Stat) 
  74] cons(X, Y) > Y  because [75], by definition 
  75] cons*(X, Y) >= Y  because [33], by (Select) 
  76] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [77] and [78], by (Copy) 
  77] filter*(F, cons(X, Y)) >= F  because [30], by (Select) 
  78] filter*(F, cons(X, Y)) >= X  because [79], by (Select) 
  79] cons(X, Y) >= X  because [80], by (Star) 
  80] cons*(X, Y) >= X  because [81], by (Select) 
  81] X >= X  by (Meta) 
  82] filter*(F, cons(X, Y)) >= Y  because [31], by (Select) 

  83] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [84], by (Star) 
  84] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [85] and [87], by (Copy) 
  85] filter2*(true, F, X, Y) >= X  because [86], by (Select) 
  86] X >= X  by (Meta) 
  87] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [35], [36], [88] and [89], by (Stat) 
  88] filter2*(true, F, X, Y) >= F  because [35], by (Select) 
  89] filter2*(true, F, X, Y) >= Y  because [36], by (Select) 

  90] filter2(false, F, X, Y) >= filter(F, Y)  because [91], by (Star) 
  91] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [38], [39], [92] and [93], by (Stat) 
  92] filter2*(false, F, X, Y) >= F  because [38], by (Select) 
  93] filter2*(false, F, X, Y) >= Y  because [39], by (Select) 

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:

  times#(X, s(Y)) =#> plus#(times(X, Y), X)   
  times#(X, s(Y)) =#> times#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, 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:

  times#(X, s(Y)) >? plus#(times(X, Y), X) 
  times#(X, s(Y)) >? times#(X, Y) 
  plus#(s(X), Y) >? plus#(X, 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) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[cons(x_1, x_2)]] = x_2 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_2, x_4) 
  [[map(x_1, x_2)]] = _|_ 
  [[plus#(x_1, x_2)]] = plus# 
  [[times#(x_1, x_2)]] = x_2 

We choose Lex = {} and Mul = {@_{o -> o}, false, filter, filter2, filter2#, filter#, map#, plus, plus#, s, times, true}, and the following precedence: false > filter2# = filter# > map# > @_{o -> o} > filter = filter2 > times > plus > s > plus# > true

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

  s(X) > plus# 
  s(X) >= X 
  plus# >= plus# 
  map#(F, X) >= map#(F, X) 
  filter#(F, X) >= filter2#(F, X) 
  filter2#(F, X) >= filter#(F, X) 
  filter2#(F, X) >= filter#(F, X) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  _|_ >= _|_ 
  filter(F, X) >= filter2(F, X) 
  filter2(F, X) >= filter(F, X) 
  filter2(F, X) >= filter(F, X) 

With these choices, we have:

  1] s(X) > plus#  because [2], by definition 
  2] s*(X) >= plus#  because s > plus#, by (Copy) 

  3] s(X) >= X  because [4], by (Star) 
  4] s*(X) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 

  6] plus# >= plus#  because plus# in Mul, by (Fun) 

  7] map#(F, X) >= map#(F, X)  because map# in Mul, [8] and [9], by (Fun) 
  8] F >= F  by (Meta) 
  9] X >= X  by (Meta) 

  10] filter#(F, X) >= filter2#(F, X)  because filter# = filter2#, filter# in Mul, [11] and [12], by (Fun) 
  11] F >= F  by (Meta) 
  12] X >= X  by (Meta) 

  13] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [14] and [15], by (Fun) 
  14] F >= F  by (Meta) 
  15] X >= X  by (Meta) 

  16] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [17] and [18], by (Fun) 
  17] F >= F  by (Meta) 
  18] X >= X  by (Meta) 

  19] times(X, _|_) >= _|_  by (Bot) 

  20] times(X, s(Y)) >= plus(times(X, Y), X)  because [21], by (Star) 
  21] times*(X, s(Y)) >= plus(times(X, Y), X)  because times > plus, [22] and [26], by (Copy) 
  22] times*(X, s(Y)) >= times(X, Y)  because times in Mul, [23] and [24], by (Stat) 
  23] X >= X  by (Meta) 
  24] s(Y) > Y  because [25], by definition 
  25] s*(Y) >= Y  because [5], by (Select) 
  26] times*(X, s(Y)) >= X  because [23], by (Select) 

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

  30] plus(_|_, X) >= X  because [31], by (Star) 
  31] plus*(_|_, X) >= X  because [32], by (Select) 
  32] X >= X  by (Meta) 

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

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

  47] _|_ >= _|_  by (Bot) 

  48] filter(F, X) >= filter2(F, X)  because filter = filter2, filter in Mul, [11] and [12], by (Fun) 

  49] filter2(F, X) >= filter(F, X)  because filter2 = filter, filter2 in Mul, [14] and [15], by (Fun) 

  50] filter2(F, X) >= filter(F, X)  because filter2 = filter, filter2 in Mul, [17] and [18], by (Fun) 

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:

  times#(X, s(Y)) =#> times#(X, Y)   
  plus#(s(X), Y) =#> plus#(X, 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_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:

  times#(X, s(Y)) >? times#(X, Y) 
  plus#(s(X), Y) >? plus#(X, 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) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) 
  [[filter(x_1, x_2)]] = filter(x_2, x_1) 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_1, x_3) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_4, x_2, x_1, x_3) 
  [[filter#(x_1, x_2)]] = filter#(x_2, x_1) 
  [[map#(x_1, x_2)]] = _|_ 

We choose Lex = {@_{o -> o}, filter, filter2, filter2#, filter#} and Mul = {cons, false, map, plus, plus#, s, times, times#, true}, and the following precedence: false > filter = filter2 > map > cons > @_{o -> o} = filter2# = filter# > plus# > times > plus > s > times# > true

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

  times#(X, s(Y)) > times#(X, Y) 
  plus#(s(X), Y) >= plus#(X, Y) 
  _|_ >= _|_ 
  filter#(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y) 
  filter2#(true, F, X, Y) >= filter#(F, Y) 
  filter2#(false, F, X, Y) >= filter#(F, Y) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

With these choices, we have:

  1] times#(X, s(Y)) > times#(X, Y)  because [2], by definition 
  2] times#*(X, s(Y)) >= times#(X, Y)  because times# in Mul, [3] and [4], by (Stat) 
  3] X >= X  by (Meta) 
  4] s(Y) > Y  because [5], by definition 
  5] s*(Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 

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

  12] _|_ >= _|_  by (Bot) 

  13] filter#(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y)  because [14], by (Star) 
  14] filter#*(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y)  because filter# = filter2#, [15], [18], [22], [24] and [26], by (Stat) 
  15] cons(X, Y) > Y  because [16], by definition 
  16] cons*(X, Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 
  18] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter# = @_{o -> o}, [19], [22] and [24], by (Stat) 
  19] cons(X, Y) > X  because [20], by definition 
  20] cons*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] filter#*(F, cons(X, Y)) >= F  because [23], by (Select) 
  23] F >= F  by (Meta) 
  24] filter#*(F, cons(X, Y)) >= X  because [25], by (Select) 
  25] cons(X, Y) >= X  because [20], by (Star) 
  26] filter#*(F, cons(X, Y)) >= Y  because [27], by (Select) 
  27] cons(X, Y) >= Y  because [16], by (Star) 

  28] filter2#(true, F, X, Y) >= filter#(F, Y)  because [29], by (Star) 
  29] filter2#*(true, F, X, Y) >= filter#(F, Y)  because filter2# = filter#, [30], [31], [32] and [33], by (Stat) 
  30] F >= F  by (Meta) 
  31] Y >= Y  by (Meta) 
  32] filter2#*(true, F, X, Y) >= F  because [30], by (Select) 
  33] filter2#*(true, F, X, Y) >= Y  because [31], by (Select) 

  34] filter2#(false, F, X, Y) >= filter#(F, Y)  because [35], by (Star) 
  35] filter2#*(false, F, X, Y) >= filter#(F, Y)  because filter2# = filter#, [36], [37], [38] and [39], by (Stat) 
  36] F >= F  by (Meta) 
  37] Y >= Y  by (Meta) 
  38] filter2#*(false, F, X, Y) >= F  because [36], by (Select) 
  39] filter2#*(false, F, X, Y) >= Y  because [37], by (Select) 

  40] times(X, _|_) >= _|_  by (Bot) 

  41] times(X, s(Y)) >= plus(times(X, Y), X)  because [42], by (Star) 
  42] times*(X, s(Y)) >= plus(times(X, Y), X)  because times > plus, [43] and [44], by (Copy) 
  43] times*(X, s(Y)) >= times(X, Y)  because times in Mul, [3] and [4], by (Stat) 
  44] times*(X, s(Y)) >= X  because [3], by (Select) 

  45] plus(X, _|_) >= X  because [46], by (Star) 
  46] plus*(X, _|_) >= X  because [47], by (Select) 
  47] X >= X  by (Meta) 

  48] plus(_|_, X) >= X  because [49], by (Star) 
  49] plus*(_|_, X) >= X  because [50], by (Select) 
  50] X >= X  by (Meta) 

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

  58] plus(s(X), Y) >= s(plus(X, Y))  because [59], by (Star) 
  59] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [60], by (Copy) 
  60] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [61] and [11], by (Stat) 
  61] s(X) > X  because [62], by definition 
  62] s*(X) >= X  because [10], by (Select) 

  63] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [64], by (Star) 
  64] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [65] and [72], by (Copy) 
  65] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [66] and [68], by (Copy) 
  66] map*(F, cons(X, Y)) >= F  because [67], by (Select) 
  67] F >= F  by (Meta) 
  68] map*(F, cons(X, Y)) >= X  because [69], by (Select) 
  69] cons(X, Y) >= X  because [70], by (Star) 
  70] cons*(X, Y) >= X  because [71], by (Select) 
  71] X >= X  by (Meta) 
  72] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [73] and [74], by (Stat) 
  73] F >= F  by (Meta) 
  74] cons(X, Y) > Y  because [75], by definition 
  75] cons*(X, Y) >= Y  because [76], by (Select) 
  76] Y >= Y  by (Meta) 

  77] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [78], by (Star) 
  78] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [15], [79], [80], [81] and [82], by (Stat) 
  79] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [80] and [81], by (Copy) 
  80] filter*(F, cons(X, Y)) >= F  because [23], by (Select) 
  81] filter*(F, cons(X, Y)) >= X  because [25], by (Select) 
  82] filter*(F, cons(X, Y)) >= Y  because [27], by (Select) 

  83] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [84], by (Star) 
  84] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [85] and [87], by (Copy) 
  85] filter2*(true, F, X, Y) >= X  because [86], by (Select) 
  86] X >= X  by (Meta) 
  87] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [30], [31], [88] and [89], by (Stat) 
  88] filter2*(true, F, X, Y) >= F  because [30], by (Select) 
  89] filter2*(true, F, X, Y) >= Y  because [31], by (Select) 

  90] filter2(false, F, X, Y) >= filter(F, Y)  because [91], by (Star) 
  91] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [36], [37], [92] and [93], by (Stat) 
  92] filter2*(false, F, X, Y) >= F  because [36], by (Select) 
  93] filter2*(false, F, X, Y) >= Y  because [37], by (Select) 

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:

  plus#(s(X), Y) =#> plus#(X, 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_3, R_1, static, formative) is finite.

We consider the dependency pair problem (P_3, 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:

  plus#(s(X), Y) >? plus#(X, 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) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[filter(x_1, x_2)]] = filter(x_2, x_1) 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_3, x_1) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_2, x_4) 

We choose Lex = {filter, filter2} and Mul = {@_{o -> o}, cons, false, filter2#, filter#, map, map#, plus, plus#, s, times, true}, and the following precedence: false > filter = filter2 > filter2# = filter# > map# > plus# > map > times > plus > s > true > cons > @_{o -> o}

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

  plus#(s(X), Y) >= plus#(X, Y) 
  map#(F, cons(X, Y)) > map#(F, Y) 
  filter#(F, cons(X, Y)) >= filter2#(F, Y) 
  filter2#(F, X) >= filter#(F, X) 
  filter2#(F, X) >= filter#(F, X) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

With these choices, we have:

  1] plus#(s(X), Y) >= plus#(X, Y)  because plus# 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] Y >= Y  by (Meta) 

  6] map#(F, cons(X, Y)) > map#(F, Y)  because [7], by definition 
  7] map#*(F, cons(X, Y)) >= map#(F, Y)  because map# in Mul, [8] and [9], by (Stat) 
  8] F >= F  by (Meta) 
  9] cons(X, Y) > Y  because [10], by definition 
  10] cons*(X, Y) >= Y  because [11], by (Select) 
  11] Y >= Y  by (Meta) 

  12] filter#(F, cons(X, Y)) >= filter2#(F, Y)  because filter# = filter2#, filter# in Mul, [13] and [14], by (Fun) 
  13] F >= F  by (Meta) 
  14] cons(X, Y) >= Y  because [15], by (Star) 
  15] cons*(X, Y) >= Y  because [16], by (Select) 
  16] Y >= Y  by (Meta) 

  17] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [18] and [19], by (Fun) 
  18] F >= F  by (Meta) 
  19] X >= X  by (Meta) 

  20] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [21] and [22], by (Fun) 
  21] F >= F  by (Meta) 
  22] X >= X  by (Meta) 

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

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

  32] plus(X, _|_) >= X  because [33], by (Star) 
  33] plus*(X, _|_) >= X  because [34], by (Select) 
  34] X >= X  by (Meta) 

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

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

  45] plus(s(X), Y) >= s(plus(X, Y))  because [46], by (Star) 
  46] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [47], by (Copy) 
  47] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [48] and [5], by (Stat) 
  48] s(X) > X  because [49], by definition 
  49] s*(X) >= X  because [4], by (Select) 

  50] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [51], by (Star) 
  51] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [52] and [58], by (Copy) 
  52] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [53] and [54], by (Copy) 
  53] map*(F, cons(X, Y)) >= F  because [8], by (Select) 
  54] map*(F, cons(X, Y)) >= X  because [55], by (Select) 
  55] cons(X, Y) >= X  because [56], by (Star) 
  56] cons*(X, Y) >= X  because [57], by (Select) 
  57] X >= X  by (Meta) 
  58] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [8] and [9], by (Stat) 

  59] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [60], by (Star) 
  60] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [61], [63], [64], [65] and [69], by (Stat) 
  61] cons(X, Y) > Y  because [62], by definition 
  62] cons*(X, Y) >= Y  because [16], by (Select) 
  63] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [64] and [65], by (Copy) 
  64] filter*(F, cons(X, Y)) >= F  because [13], by (Select) 
  65] filter*(F, cons(X, Y)) >= X  because [66], by (Select) 
  66] cons(X, Y) >= X  because [67], by (Star) 
  67] cons*(X, Y) >= X  because [68], by (Select) 
  68] X >= X  by (Meta) 
  69] filter*(F, cons(X, Y)) >= Y  because [14], by (Select) 

  70] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [71], by (Star) 
  71] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [72] and [74], by (Copy) 
  72] filter2*(true, F, X, Y) >= X  because [73], by (Select) 
  73] X >= X  by (Meta) 
  74] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [18], [19], [75] and [76], by (Stat) 
  75] filter2*(true, F, X, Y) >= F  because [18], by (Select) 
  76] filter2*(true, F, X, Y) >= Y  because [19], by (Select) 

  77] filter2(false, F, X, Y) >= filter(F, Y)  because [78], by (Star) 
  78] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [21], [22], [79] and [80], by (Stat) 
  79] filter2*(false, F, X, Y) >= F  because [21], by (Select) 
  80] filter2*(false, F, X, Y) >= Y  because [22], by (Select) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, static, formative) by (P_4, R_1, static, formative), where P_4 consists of:

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

We consider the dependency pair problem (P_4, 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:

  plus#(s(X), Y) >? plus#(X, 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) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_3, x_4) 
  [[filter#(x_1, x_2)]] = filter#(x_2) 

We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {cons, false, filter2#, filter#, map, plus, plus#, s, times, true}, and the following precedence: false > map > plus# > @_{o -> o} = filter = filter2 > cons > filter2# > filter# > times > plus > s > true

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

  plus#(s(X), Y) >= plus#(X, Y) 
  filter#(cons(X, Y)) >= filter2#(X, Y) 
  filter2#(X, Y) >= filter#(Y) 
  filter2#(X, Y) > filter#(Y) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

With these choices, we have:

  1] plus#(s(X), Y) >= plus#(X, Y)  because plus# 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] Y >= Y  by (Meta) 

  6] filter#(cons(X, Y)) >= filter2#(X, Y)  because [7], by (Star) 
  7] filter#*(cons(X, Y)) >= filter2#(X, Y)  because [8], by (Select) 
  8] cons(X, Y) >= filter2#(X, Y)  because [9], by (Star) 
  9] cons*(X, Y) >= filter2#(X, Y)  because cons > filter2#, [10] and [12], by (Copy) 
  10] cons*(X, Y) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 

  14] filter2#(X, Y) >= filter#(Y)  because [15], by (Star) 
  15] filter2#*(X, Y) >= filter#(Y)  because filter2# > filter# and [16], by (Copy) 
  16] filter2#*(X, Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 

  18] filter2#(X, Y) > filter#(Y)  because [19], by definition 
  19] filter2#*(X, Y) >= filter#(Y)  because filter2# > filter# and [20], by (Copy) 
  20] filter2#*(X, Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] times(X, _|_) >= _|_  by (Bot) 

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

  31] plus(X, _|_) >= X  because [32], by (Star) 
  32] plus*(X, _|_) >= X  because [33], by (Select) 
  33] X >= X  by (Meta) 

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

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

  44] plus(s(X), Y) >= s(plus(X, Y))  because [45], by (Star) 
  45] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [46], by (Copy) 
  46] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [47] and [5], by (Stat) 
  47] s(X) > X  because [48], by definition 
  48] s*(X) >= X  because [4], by (Select) 

  49] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [50], by (Star) 
  50] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [51] and [58], by (Copy) 
  51] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [52] and [54], by (Copy) 
  52] map*(F, cons(X, Y)) >= F  because [53], by (Select) 
  53] F >= F  by (Meta) 
  54] map*(F, cons(X, Y)) >= X  because [55], by (Select) 
  55] cons(X, Y) >= X  because [56], by (Star) 
  56] cons*(X, Y) >= X  because [57], by (Select) 
  57] X >= X  by (Meta) 
  58] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [59] and [60], by (Stat) 
  59] F >= F  by (Meta) 
  60] cons(X, Y) > Y  because [61], by definition 
  61] cons*(X, Y) >= Y  because [62], by (Select) 
  62] Y >= Y  by (Meta) 

  63] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [64], by (Star) 
  64] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [65], [66], [67], [70], [71] and [73], by (Stat) 
  65] F >= F  by (Meta) 
  66] cons(X, Y) > Y  because [12], by definition 
  67] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter = @_{o -> o}, [65], [68], [70] and [71], by (Stat) 
  68] cons(X, Y) > X  because [69], by definition 
  69] cons*(X, Y) >= X  because [11], by (Select) 
  70] filter*(F, cons(X, Y)) >= F  because [65], by (Select) 
  71] filter*(F, cons(X, Y)) >= X  because [72], by (Select) 
  72] cons(X, Y) >= X  because [69], by (Star) 
  73] filter*(F, cons(X, Y)) >= Y  because [74], by (Select) 
  74] cons(X, Y) >= Y  because [12], by (Star) 

  75] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [76], by (Star) 
  76] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [77] and [79], by (Copy) 
  77] filter2*(true, F, X, Y) >= X  because [78], by (Select) 
  78] X >= X  by (Meta) 
  79] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [80], [81], [82] and [83], by (Stat) 
  80] F >= F  by (Meta) 
  81] Y >= Y  by (Meta) 
  82] filter2*(true, F, X, Y) >= F  because [80], by (Select) 
  83] filter2*(true, F, X, Y) >= Y  because [81], by (Select) 

  84] filter2(false, F, X, Y) >= filter(F, Y)  because [85], by (Star) 
  85] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [86], [87], [88] and [89], by (Stat) 
  86] F >= F  by (Meta) 
  87] Y >= Y  by (Meta) 
  88] filter2*(false, F, X, Y) >= F  because [86], by (Select) 
  89] filter2*(false, F, X, Y) >= Y  because [87], by (Select) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_1, static, formative) by (P_5, R_1, static, formative), where P_5 consists of:

  plus#(s(X), Y) =#> plus#(X, Y)   
  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if (P_5, R_1, static, formative) is finite.

We consider the dependency pair problem (P_5, 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:

  plus#(s(X), Y) >? plus#(X, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, 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) 

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]] = _|_ 
  [[cons(x_1, x_2)]] = x_2 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4) 
  [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_2, x_4) 
  [[map(x_1, x_2)]] = map(x_2) 

We choose Lex = {} and Mul = {@_{o -> o}, false, filter, filter2, filter2#, filter#, map, plus, plus#, s, times, true}, and the following precedence: @_{o -> o} > filter = filter2 > filter2# = filter# > map > plus# > false > times > plus > s > true

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

  plus#(s(X), Y) > plus#(X, Y) 
  filter#(F, X) >= filter2#(F, X) 
  filter2#(F, X) >= filter#(F, X) 
  times(X, _|_) >= _|_ 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, _|_) >= X 
  plus(_|_, X) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(X) >= map(X) 
  filter(F, X) >= filter2(F, X) 
  filter2(F, X) >= filter(F, X) 
  filter2(F, X) >= filter(F, X) 

With these choices, we have:

  1] plus#(s(X), Y) > plus#(X, Y)  because [2], by definition 
  2] plus#*(s(X), Y) >= plus#(X, Y)  because plus# in Mul, [3] and [6], by (Stat) 
  3] s(X) > X  because [4], by definition 
  4] s*(X) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 
  6] Y >= Y  by (Meta) 

  7] filter#(F, X) >= filter2#(F, X)  because filter# = filter2#, filter# in Mul, [8] and [9], by (Fun) 
  8] F >= F  by (Meta) 
  9] X >= X  by (Meta) 

  10] filter2#(F, X) >= filter#(F, X)  because filter2# = filter#, filter2# in Mul, [11] and [12], by (Fun) 
  11] F >= F  by (Meta) 
  12] X >= X  by (Meta) 

  13] times(X, _|_) >= _|_  by (Bot) 

  14] times(X, s(Y)) >= plus(times(X, Y), X)  because [15], by (Star) 
  15] times*(X, s(Y)) >= plus(times(X, Y), X)  because times > plus, [16] and [21], by (Copy) 
  16] times*(X, s(Y)) >= times(X, Y)  because times in Mul, [17] and [18], by (Stat) 
  17] X >= X  by (Meta) 
  18] s(Y) > Y  because [19], by definition 
  19] s*(Y) >= Y  because [20], by (Select) 
  20] Y >= Y  by (Meta) 
  21] times*(X, s(Y)) >= X  because [17], by (Select) 

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

  25] plus(_|_, X) >= X  because [26], by (Star) 
  26] plus*(_|_, X) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 

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

  35] plus(s(X), Y) >= s(plus(X, Y))  because [36], by (Star) 
  36] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [37], by (Copy) 
  37] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [3] and [6], by (Stat) 

  38] map(X) >= map(X)  because map in Mul and [39], by (Fun) 
  39] X >= X  by (Meta) 

  40] filter(F, X) >= filter2(F, X)  because filter = filter2, filter in Mul, [8] and [9], by (Fun) 

  41] filter2(F, X) >= filter(F, X)  because filter2 = filter, filter2 in Mul, [11] and [12], by (Fun) 

  42] filter2(F, X) >= filter(F, X)  because filter2 = filter, filter2 in Mul, [43] and [44], by (Fun) 
  43] F >= F  by (Meta) 
  44] 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_5, R_1, static, formative) by (P_6, R_1, static, formative), where P_6 consists of:

  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if (P_6, R_1, static, formative) is finite.

We consider the dependency pair problem (P_6, R_1, static, formative).

The formative rules of (P_6, R_1) are R_2 ::=

  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_6, R_1, static, formative) by (P_6, R_2, static, formative).

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:

  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, 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:

  cons = \y0y1.1 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  filter2# = \y0G1y2y3.2 + 2y3 + 2y3G1(y3) + 2G1(y3) 
  filter# = \G0y1.1 + 2y1 + 2y1G0(y1) 
  map = \G0y1.y1 + 2y1G0(y1) + 2G0(0) 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 3 + 2x2 + 2x2F0(1 + x2) + 2F0(1 + x2) > 2 + 2x2 + 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2 + 2x2 + 2x2F0(x2) + 2F0(x2) > 1 + 2x2 + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 + 2x2F0(1 + x2) + 2F0(0) + 2F0(1 + x2) >= 1 + x2 + 2x2F0(x2) + 2F0(0) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[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 a dependency pair problem (P_6, 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.