We consider the system fuhkop11frocos.

  Alphabet:

    append : [list * list] --> list 
    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    mirror : [list] --> list 
    nil : [] --> list 
    reverse : [list] --> list 
    shuffle : [list] --> list 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    reverse(nil) => nil 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    mirror(nil) => nil 
    mirror(cons(x, y)) => append(cons(x, mirror(y)), cons(x, nil)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(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]):

  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  reverse(nil) >? nil 
  shuffle(nil) >? nil 
  shuffle(cons(X, Y)) >? cons(X, shuffle(reverse(Y))) 
  mirror(nil) >? nil 
  mirror(cons(X, Y)) >? append(cons(X, mirror(Y)), cons(X, nil)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 

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

Argument functions:

  [[nil]] = _|_ 
  [[reverse(x_1)]] = x_1 

We choose Lex = {shuffle} and Mul = {@_{o -> o}, append, cons, map, mirror}, and the following precedence: mirror > append > shuffle > map > cons > @_{o -> o}

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

  append(_|_, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  _|_ >= _|_ 
  shuffle(_|_) >= _|_ 
  shuffle(cons(X, Y)) > cons(X, shuffle(Y)) 
  mirror(_|_) >= _|_ 
  mirror(cons(X, Y)) > append(cons(X, mirror(Y)), cons(X, _|_)) 
  map(F, _|_) > _|_ 
  map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 

With these choices, we have:

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

  4] append(cons(X, Y), Z) >= cons(X, append(Y, Z))  because [5], by (Star) 
  5] append*(cons(X, Y), Z) >= cons(X, append(Y, Z))  because append > cons, [6] and [10], by (Copy) 
  6] append*(cons(X, Y), Z) >= 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] append*(cons(X, Y), Z) >= append(Y, Z)  because append in Mul, [11] and [14], by (Stat) 
  11] cons(X, Y) > Y  because [12], by definition 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 
  14] Z >= Z  by (Meta) 

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

  16] shuffle(_|_) >= _|_  by (Bot) 

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

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

  30] mirror(cons(X, Y)) > append(cons(X, mirror(Y)), cons(X, _|_))  because [31], by definition 
  31] mirror*(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, _|_))  because mirror > append, [32] and [41], by (Copy) 
  32] mirror*(cons(X, Y)) >= cons(X, mirror(Y))  because mirror > cons, [33] and [37], by (Copy) 
  33] mirror*(cons(X, Y)) >= X  because [34], by (Select) 
  34] cons(X, Y) >= X  because [35], by (Star) 
  35] cons*(X, Y) >= X  because [36], by (Select) 
  36] X >= X  by (Meta) 
  37] mirror*(cons(X, Y)) >= mirror(Y)  because mirror in Mul and [38], by (Stat) 
  38] cons(X, Y) > Y  because [39], by definition 
  39] cons*(X, Y) >= Y  because [40], by (Select) 
  40] Y >= Y  by (Meta) 
  41] mirror*(cons(X, Y)) >= cons(X, _|_)  because [42], by (Select) 
  42] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [43] and [44], by (Fun) 
  43] X >= X  by (Meta) 
  44] Y >= _|_  by (Bot) 

  45] map(F, _|_) > _|_  because [46], by definition 
  46] map*(F, _|_) >= _|_  by (Bot) 

  47] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [48], by definition 
  48] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [49] and [56], by (Copy) 
  49] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [50] and [52], by (Copy) 
  50] map*(F, cons(X, Y)) >= F  because [51], by (Select) 
  51] F >= F  by (Meta) 
  52] map*(F, cons(X, Y)) >= X  because [53], by (Select) 
  53] cons(X, Y) >= X  because [54], by (Star) 
  54] cons*(X, Y) >= X  because [55], by (Select) 
  55] X >= X  by (Meta) 
  56] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [57] and [58], by (Stat) 
  57] F >= F  by (Meta) 
  58] cons(X, Y) > Y  because [59], by definition 
  59] cons*(X, Y) >= Y  because [60], by (Select) 
  60] Y >= Y  by (Meta) 

We can thus remove the following rules:

  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
  mirror(cons(X, Y)) => append(cons(X, mirror(Y)), cons(X, nil)) 
  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]):

  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  reverse(nil) >? nil 
  shuffle(nil) >? nil 
  mirror(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 
  [[reverse(x_1)]] = x_1 

We choose Lex = {} and Mul = {append, cons, mirror, shuffle}, and the following precedence: append > cons > shuffle > mirror

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

  append(_|_, X) > X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  _|_ >= _|_ 
  shuffle(_|_) >= _|_ 
  mirror(_|_) >= _|_ 

With these choices, we have:

  1] append(_|_, X) > X  because [2], by definition 
  2] append*(_|_, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] append(cons(X, Y), Z) >= cons(X, append(Y, Z))  because [5], by (Star) 
  5] append*(cons(X, Y), Z) >= cons(X, append(Y, Z))  because append > cons, [6] and [10], by (Copy) 
  6] append*(cons(X, Y), Z) >= 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] append*(cons(X, Y), Z) >= append(Y, Z)  because append in Mul, [11] and [14], by (Stat) 
  11] cons(X, Y) > Y  because [12], by definition 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 
  14] Z >= Z  by (Meta) 

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

  16] shuffle(_|_) >= _|_  by (Bot) 

  17] mirror(_|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  append(nil, X) => X 

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

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

  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  reverse(nil) >? nil 
  shuffle(nil) >? nil 
  mirror(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {append, cons, mirror, reverse, shuffle}, and the following precedence: append > shuffle > cons > reverse > mirror

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

  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(_|_) >= _|_ 
  shuffle(_|_) >= _|_ 
  mirror(_|_) > _|_ 

With these choices, we have:

  1] append(cons(X, Y), Z) >= cons(X, append(Y, Z))  because [2], by (Star) 
  2] append*(cons(X, Y), Z) >= cons(X, append(Y, Z))  because append > cons, [3] and [7], by (Copy) 
  3] append*(cons(X, Y), Z) >= X  because [4], by (Select) 
  4] cons(X, Y) >= X  because [5], by (Star) 
  5] cons*(X, Y) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 
  7] append*(cons(X, Y), Z) >= append(Y, Z)  because append in Mul, [8] and [11], by (Stat) 
  8] cons(X, Y) > Y  because [9], by definition 
  9] cons*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 
  11] Z >= Z  by (Meta) 

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

  13] shuffle(_|_) >= _|_  by (Bot) 

  14] mirror(_|_) > _|_  because [15], by definition 
  15] mirror*(_|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  mirror(nil) => nil 

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

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

  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  reverse(nil) >? nil 
  shuffle(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {append, cons, reverse, shuffle}, and the following precedence: append > reverse > shuffle > cons

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

  append(cons(X, Y), Z) > cons(X, append(Y, Z)) 
  reverse(_|_) >= _|_ 
  shuffle(_|_) >= _|_ 

With these choices, we have:

  1] append(cons(X, Y), Z) > cons(X, append(Y, Z))  because [2], by definition 
  2] append*(cons(X, Y), Z) >= cons(X, append(Y, Z))  because append > cons, [3] and [7], by (Copy) 
  3] append*(cons(X, Y), Z) >= X  because [4], by (Select) 
  4] cons(X, Y) >= X  because [5], by (Star) 
  5] cons*(X, Y) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 
  7] append*(cons(X, Y), Z) >= append(Y, Z)  because append in Mul, [8] and [11], by (Stat) 
  8] cons(X, Y) > Y  because [9], by definition 
  9] cons*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 
  11] Z >= Z  by (Meta) 

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

  13] shuffle(_|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 

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

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

  reverse(nil) >? nil 
  shuffle(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {reverse, shuffle}, and the following precedence: reverse > shuffle

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

  reverse(_|_) > _|_ 
  shuffle(_|_) >= _|_ 

With these choices, we have:

  1] reverse(_|_) > _|_  because [2], by definition 
  2] reverse*(_|_) >= _|_  by (Bot) 

  3] shuffle(_|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  reverse(nil) => nil 

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

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

  shuffle(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {shuffle}, and the following precedence: shuffle

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

  shuffle(_|_) > _|_ 

With these choices, we have:

  1] shuffle(_|_) > _|_  because [2], by definition 
  2] shuffle*(_|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  shuffle(nil) => nil 

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.