We consider the system Applicative_05__Ex3Lists.

  Alphabet:

    append : [b * b] --> b 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    append(append(x, y), z) => append(x, append(y, z)) 
    map(f, append(x, y)) => append(map(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)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  append(append(X, Y), Z) >? append(X, append(Y, Z)) 
  map(F, append(X, Y)) >? append(map(F, X), map(F, Y)) 

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

Argument functions:

  [[nil]] = _|_ 

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

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

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 [11], [14] and [16], 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] append*(cons(X, Y), Z) >= Y  because [15], by (Select) 
  15] cons(X, Y) >= Y  because [12], by (Star) 
  16] append*(cons(X, Y), Z) >= Z  because [17], by (Select) 
  17] Z >= Z  by (Meta) 

  18] map(F, _|_) >= _|_  by (Bot) 

  19] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [20], by definition 
  20] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [21] and [26], by (Copy) 
  21] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map = @_{o -> o}, map in Mul, [22] and [23], by (Stat) 
  22] F >= F  by (Meta) 
  23] cons(X, Y) > X  because [24], by definition 
  24] cons*(X, Y) >= X  because [25], by (Select) 
  25] X >= X  by (Meta) 
  26] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [22] and [27], by (Stat) 
  27] cons(X, Y) > Y  because [28], by definition 
  28] cons*(X, Y) >= Y  because [29], by (Select) 
  29] Y >= Y  by (Meta) 

  30] append(append(X, Y), Z) > append(X, append(Y, Z))  because [31], by definition 
  31] append*(append(X, Y), Z) >= append(X, append(Y, Z))  because [32], [35] and [37], by (Stat) 
  32] append(X, Y) > X  because [33], by definition 
  33] append*(X, Y) >= X  because [34], by (Select) 
  34] X >= X  by (Meta) 
  35] append*(append(X, Y), Z) >= X  because [36], by (Select) 
  36] append(X, Y) >= X  because [33], by (Star) 
  37] append*(append(X, Y), Z) >= append(Y, Z)  because [38], [41] and [43], by (Stat) 
  38] append(X, Y) > Y  because [39], by definition 
  39] append*(X, Y) >= Y  because [40], by (Select) 
  40] Y >= Y  by (Meta) 
  41] append*(append(X, Y), Z) >= Y  because [42], by (Select) 
  42] append(X, Y) >= Y  because [39], by (Star) 
  43] append*(append(X, Y), Z) >= Z  because [44], by (Select) 
  44] Z >= Z  by (Meta) 

  45] map(F, append(X, Y)) > append(map(F, X), map(F, Y))  because [46], by definition 
  46] map*(F, append(X, Y)) >= append(map(F, X), map(F, Y))  because map > append, [47] and [52], by (Copy) 
  47] map*(F, append(X, Y)) >= map(F, X)  because map in Mul, [48] and [49], by (Stat) 
  48] F >= F  by (Meta) 
  49] append(X, Y) > X  because [50], by definition 
  50] append*(X, Y) >= X  because [51], by (Select) 
  51] X >= X  by (Meta) 
  52] map*(F, append(X, Y)) >= map(F, Y)  because map in Mul, [48] and [53], by (Stat) 
  53] append(X, Y) > Y  because [54], by definition 
  54] append*(X, Y) >= Y  because [55], by (Select) 
  55] Y >= Y  by (Meta) 

We can thus remove the following rules:

  append(nil, X) => X 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  append(append(X, Y), Z) => append(X, append(Y, Z)) 
  map(F, append(X, Y)) => append(map(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(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  map(F, nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

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

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)) 
  map(F, _|_) >= _|_ 

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] map(F, _|_) >= _|_  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]):

  map(F, nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

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

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

  map(F, _|_) > _|_ 

With these choices, we have:

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

We can thus remove the following rules:

  map(F, 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.