We consider the system Applicative_first_order_05__30.

  Alphabet:

    !faccolon : [a * a] --> a 
    C : [] --> 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 
    true : [] --> b 

  Rules:

    !faccolon(!faccolon(!faccolon(!faccolon(C, x), y), z), u) => !faccolon(!faccolon(x, z), !faccolon(!faccolon(!faccolon(x, y), z), u)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

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

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  !faccolon(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >? !faccolon(!faccolon(X, Z), !faccolon(!faccolon(!faccolon(X, Y), Z), U)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >? filter(F, Y) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[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) 

We choose Lex = {!faccolon, filter, filter2} and Mul = {@_{o -> o}, C, cons, false, map, nil, true}, and the following precedence: true > map > false > C > filter = filter2 > @_{o -> o} > cons > nil > !faccolon

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

  !faccolon(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) > !faccolon(!faccolon(X, Z), !faccolon(!faccolon(!faccolon(X, Y), Z), U)) 
  map(F, nil) > nil 
  map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, nil) >= nil 
  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] !faccolon(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) > !faccolon(!faccolon(X, Z), !faccolon(!faccolon(!faccolon(X, Y), Z), U))  because [2], by definition 
  2] !faccolon*(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >= !faccolon(!faccolon(X, Z), !faccolon(!faccolon(!faccolon(X, Y), Z), U))  because [3], [14] and [18], by (Stat) 
  3] !faccolon(!faccolon(!faccolon(C, X), Y), Z) > !faccolon(X, Z)  because [4], by definition 
  4] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= !faccolon(X, Z)  because [5], [10] and [12], by (Stat) 
  5] !faccolon(!faccolon(C, X), Y) > X  because [6], by definition 
  6] !faccolon*(!faccolon(C, X), Y) >= X  because [7], by (Select) 
  7] !faccolon(C, X) >= X  because [8], by (Star) 
  8] !faccolon*(C, X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= X  because [11], by (Select) 
  11] !faccolon(!faccolon(C, X), Y) >= X  because [6], by (Star) 
  12] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= Z  because [13], by (Select) 
  13] Z >= Z  by (Meta) 
  14] !faccolon*(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >= !faccolon(X, Z)  because [15], by (Select) 
  15] !faccolon(!faccolon(!faccolon(C, X), Y), Z) >= !faccolon(X, Z)  because [16] and [17], by (Fun) 
  16] !faccolon(!faccolon(C, X), Y) >= X  because [6], by (Star) 
  17] Z >= Z  by (Meta) 
  18] !faccolon*(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >= !faccolon(!faccolon(!faccolon(X, Y), Z), U)  because [19], [30] and [35], by (Stat) 
  19] !faccolon(!faccolon(!faccolon(C, X), Y), Z) > !faccolon(!faccolon(X, Y), Z)  because [20], by definition 
  20] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= !faccolon(!faccolon(X, Y), Z)  because [21], [27] and [12], by (Stat) 
  21] !faccolon(!faccolon(C, X), Y) > !faccolon(X, Y)  because [22], by definition 
  22] !faccolon*(!faccolon(C, X), Y) >= !faccolon(X, Y)  because [23], [6] and [25], by (Stat) 
  23] !faccolon(C, X) > X  because [24], by definition 
  24] !faccolon*(C, X) >= X  because [9], by (Select) 
  25] !faccolon*(!faccolon(C, X), Y) >= Y  because [26], by (Select) 
  26] Y >= Y  by (Meta) 
  27] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= !faccolon(X, Y)  because [5], [10] and [28], by (Stat) 
  28] !faccolon*(!faccolon(!faccolon(C, X), Y), Z) >= Y  because [29], by (Select) 
  29] !faccolon(!faccolon(C, X), Y) >= Y  because [25], by (Star) 
  30] !faccolon*(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >= !faccolon(!faccolon(X, Y), Z)  because [31], by (Select) 
  31] !faccolon(!faccolon(!faccolon(C, X), Y), Z) >= !faccolon(!faccolon(X, Y), Z)  because [32] and [17], by (Fun) 
  32] !faccolon(!faccolon(C, X), Y) >= !faccolon(X, Y)  because [33] and [34], by (Fun) 
  33] !faccolon(C, X) >= X  because [24], by (Star) 
  34] Y >= Y  by (Meta) 
  35] !faccolon*(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) >= U  because [36], by (Select) 
  36] U >= U  by (Meta) 

  37] map(F, nil) > nil  because [38], by definition 
  38] map*(F, nil) >= nil  because map > nil, by (Copy) 

  39] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [40], by definition 
  40] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [41] and [48], by (Copy) 
  41] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [42] and [44], by (Copy) 
  42] map*(F, cons(X, Y)) >= F  because [43], by (Select) 
  43] F >= F  by (Meta) 
  44] map*(F, cons(X, Y)) >= X  because [45], by (Select) 
  45] cons(X, Y) >= X  because [46], by (Star) 
  46] cons*(X, Y) >= X  because [47], by (Select) 
  47] X >= X  by (Meta) 
  48] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [49] and [50], by (Stat) 
  49] F >= F  by (Meta) 
  50] cons(X, Y) > Y  because [51], by definition 
  51] cons*(X, Y) >= Y  because [52], by (Select) 
  52] Y >= Y  by (Meta) 

  53] filter(F, nil) >= nil  because [54], by (Star) 
  54] filter*(F, nil) >= nil  because [55], by (Select) 
  55] nil >= nil  by (Fun) 

  56] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [57], by (Star) 
  57] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [58], [61], [62], [64] and [68], by (Stat) 
  58] cons(X, Y) > Y  because [59], by definition 
  59] cons*(X, Y) >= Y  because [60], by (Select) 
  60] Y >= Y  by (Meta) 
  61] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [62] and [64], by (Copy) 
  62] filter*(F, cons(X, Y)) >= F  because [63], by (Select) 
  63] F >= F  by (Meta) 
  64] filter*(F, cons(X, Y)) >= X  because [65], by (Select) 
  65] cons(X, Y) >= X  because [66], by (Star) 
  66] cons*(X, Y) >= X  because [67], by (Select) 
  67] X >= X  by (Meta) 
  68] filter*(F, cons(X, Y)) >= Y  because [69], by (Select) 
  69] cons(X, Y) >= Y  because [59], by (Star) 

  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, [75], [76], [77] and [78], by (Stat) 
  75] F >= F  by (Meta) 
  76] Y >= Y  by (Meta) 
  77] filter2*(true, F, X, Y) >= F  because [75], by (Select) 
  78] filter2*(true, F, X, Y) >= Y  because [76], by (Select) 

  79] filter2(false, F, X, Y) >= filter(F, Y)  because [80], by (Star) 
  80] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [81], [82], [83] and [84], by (Stat) 
  81] F >= F  by (Meta) 
  82] Y >= Y  by (Meta) 
  83] filter2*(false, F, X, Y) >= F  because [81], by (Select) 
  84] filter2*(false, F, X, Y) >= Y  because [82], by (Select) 

We can thus remove the following rules:

  !faccolon(!faccolon(!faccolon(!faccolon(C, X), Y), Z), U) => !faccolon(!faccolon(X, Z), !faccolon(!faccolon(!faccolon(X, Y), Z), U)) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >? filter(F, Y) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

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

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

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

  filter(F, _|_) > _|_ 
  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] filter(F, _|_) > _|_  because [2], by definition 
  2] filter*(F, _|_) >= _|_  by (Bot) 

  3] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [4], by (Star) 
  4] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [5], [6], [9], [10], [11] and [15], by (Stat) 
  5] F >= F  by (Meta) 
  6] cons(X, Y) > Y  because [7], by definition 
  7] cons*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 
  9] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [10] and [11], by (Copy) 
  10] filter*(F, cons(X, Y)) >= F  because [5], by (Select) 
  11] filter*(F, cons(X, Y)) >= X  because [12], by (Select) 
  12] cons(X, Y) >= X  because [13], by (Star) 
  13] cons*(X, Y) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 
  15] filter*(F, cons(X, Y)) >= Y  because [16], by (Select) 
  16] cons(X, Y) >= Y  because [7], by (Star) 

  17] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [18], by (Star) 
  18] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [19] and [21], by (Copy) 
  19] filter2*(true, F, X, Y) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 
  21] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [22], [23], [24] and [25], by (Stat) 
  22] F >= F  by (Meta) 
  23] Y >= Y  by (Meta) 
  24] filter2*(true, F, X, Y) >= F  because [22], by (Select) 
  25] filter2*(true, F, X, Y) >= Y  because [23], by (Select) 

  26] filter2(false, F, X, Y) > filter(F, Y)  because [27], by definition 
  27] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [28], [29], [30] and [31], by (Stat) 
  28] F >= F  by (Meta) 
  29] Y >= Y  by (Meta) 
  30] filter2*(false, F, X, Y) >= F  because [28], by (Select) 
  31] filter2*(false, F, X, Y) >= Y  because [29], by (Select) 

We can thus remove the following rules:

  filter(F, nil) => nil 
  filter2(false, F, X, Y) => filter(F, Y) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

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

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) 

We choose Lex = {filter, filter2} and Mul = {@_{o -> o}, cons, true}, and the following precedence: filter = filter2 > cons > @_{o -> o} > true

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

  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) > cons(X, filter(F, Y)) 

With these choices, we have:

  1] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [2], by (Star) 
  2] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [3], [4], [7], [8], [9] and [13], by (Stat) 
  3] F >= F  by (Meta) 
  4] cons(X, Y) > Y  because [5], by definition 
  5] cons*(X, Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 
  7] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [8] and [9], by (Copy) 
  8] filter*(F, cons(X, Y)) >= F  because [3], by (Select) 
  9] filter*(F, cons(X, Y)) >= X  because [10], by (Select) 
  10] cons(X, Y) >= X  because [11], by (Star) 
  11] cons*(X, Y) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 
  13] filter*(F, cons(X, Y)) >= Y  because [14], by (Select) 
  14] cons(X, Y) >= Y  because [5], by (Star) 

  15] filter2(true, F, X, Y) > cons(X, filter(F, Y))  because [16], by definition 
  16] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [17] and [19], by (Copy) 
  17] filter2*(true, F, X, Y) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 
  19] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [20], [21], [22] and [23], by (Stat) 
  20] F >= F  by (Meta) 
  21] Y >= Y  by (Meta) 
  22] filter2*(true, F, X, Y) >= F  because [20], by (Select) 
  23] filter2*(true, F, X, Y) >= Y  because [21], by (Select) 

We can thus remove the following rules:

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

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {@_{o -> o}, cons, filter, filter2}, and the following precedence: filter > filter2 > cons > @_{o -> o}

With these choices, we have:

  1] filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y)  because [2], by definition 
  2] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter > filter2, [3], [4], [6] and [10], by (Copy) 
  3] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [4] and [6], by (Copy) 
  4] filter*(F, cons(X, Y)) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] filter*(F, cons(X, Y)) >= X  because [7], by (Select) 
  7] cons(X, Y) >= X  because [8], by (Star) 
  8] cons*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] filter*(F, cons(X, Y)) >= Y  because [11], by (Select) 
  11] cons(X, Y) >= Y  because [12], by (Star) 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 

We can thus remove the following rules:

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

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.