We consider the system Applicative_05__TreeFlatten.

  Alphabet:

    append : [a * a] --> a 
    concat : [a] --> a 
    cons : [a * a] --> a 
    flatten : [] --> a -> a 
    map : [a -> a * a] --> a 
    nil : [] --> a 
    node : [a * a] --> a 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    flatten node(x, y) => cons(x, concat(map(flatten, y))) 
    concat(nil) => nil 
    concat(cons(x, y)) => append(x, concat(y)) 
    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 

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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 
  concat(cons(X, Y)) >? append(X, concat(Y)) 
  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 

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

Argument functions:

  [[nil]] = _|_ 

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

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

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

With these choices, we have:

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

  3] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [4], by (Star) 
  4] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [5] and [10], by (Copy) 
  5] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map = @_{o -> o}, map in Mul, [6] and [7], by (Stat) 
  6] F >= F  by (Meta) 
  7] cons(X, Y) > X  because [8], by definition 
  8] cons*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [6] and [11], 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] @_{o -> o}(flatten, node(X, Y)) >= cons(X, concat(map(flatten, Y)))  because [15], by (Star) 
  15] @_{o -> o}*(flatten, node(X, Y)) >= cons(X, concat(map(flatten, Y)))  because [16], by (Select) 
  16] flatten @_{o -> o}*(flatten, node(X, Y)) >= cons(X, concat(map(flatten, Y)))  because [17] 
  17] flatten*(@_{o -> o}*(flatten, node(X, Y))) >= cons(X, concat(map(flatten, Y)))  because flatten > cons, [18] and [23], by (Copy) 
  18] flatten*(@_{o -> o}*(flatten, node(X, Y))) >= X  because [19], by (Select) 
  19] @_{o -> o}*(flatten, node(X, Y)) >= X  because [20], by (Select) 
  20] node(X, Y) >= X  because [21], by (Star) 
  21] node*(X, Y) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 
  23] flatten*(@_{o -> o}*(flatten, node(X, Y))) >= concat(map(flatten, Y))  because [24], by (Select) 
  24] @_{o -> o}*(flatten, node(X, Y)) >= concat(map(flatten, Y))  because [25], by (Select) 
  25] node(X, Y) >= concat(map(flatten, Y))  because [26], by (Star) 
  26] node*(X, Y) >= concat(map(flatten, Y))  because node > concat and [27], by (Copy) 
  27] node*(X, Y) >= map(flatten, Y)  because node > map, [28] and [29], by (Copy) 
  28] node*(X, Y) >= flatten  because node > flatten, by (Copy) 
  29] node*(X, Y) >= Y  because [30], by (Select) 
  30] Y >= Y  by (Meta) 

  31] concat(_|_) >= _|_  by (Bot) 

  32] concat(cons(X, Y)) >= append(X, concat(Y))  because [33], by (Star) 
  33] concat*(cons(X, Y)) >= append(X, concat(Y))  because concat > append, [34] and [38], by (Copy) 
  34] concat*(cons(X, Y)) >= X  because [35], by (Select) 
  35] cons(X, Y) >= X  because [36], by (Star) 
  36] cons*(X, Y) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] concat*(cons(X, Y)) >= concat(Y)  because concat in Mul and [39], by (Stat) 
  39] cons(X, Y) > Y  because [40], by definition 
  40] cons*(X, Y) >= Y  because [41], by (Select) 
  41] Y >= Y  by (Meta) 

  42] append(_|_, X) >= X  because [43], by (Star) 
  43] append*(_|_, X) >= X  because [44], by (Select) 
  44] X >= X  by (Meta) 

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

We can thus remove the following rules:

  map(F, nil) => nil 
  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, cons(X, Y)) >? cons(F X, map(F, Y)) 
  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 
  concat(cons(X, Y)) >? append(X, concat(Y)) 
  append(nil, X) >? X 

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

Argument functions:

  [[concat(x_1)]] = x_1 
  [[map(x_1, x_2)]] = map(x_2, x_1) 

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

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

  map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 
  @_{o -> o}(flatten, node(X, Y)) >= cons(X, map(flatten, Y)) 
  nil >= nil 
  cons(X, Y) > append(X, Y) 
  append(nil, X) > X 

With these choices, we have:

  1] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [2], by definition 
  2] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [3] and [10], by (Copy) 
  3] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [4] and [6], by (Copy) 
  4] map*(F, cons(X, Y)) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] map*(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] map*(F, cons(X, Y)) >= map(F, Y)  because [11], [4] 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] map*(F, cons(X, Y)) >= Y  because [15], by (Select) 
  15] cons(X, Y) >= Y  because [12], by (Star) 

  16] @_{o -> o}(flatten, node(X, Y)) >= cons(X, map(flatten, Y))  because [17], by (Star) 
  17] @_{o -> o}*(flatten, node(X, Y)) >= cons(X, map(flatten, Y))  because [18], by (Select) 
  18] node(X, Y) >= cons(X, map(flatten, Y))  because [19], by (Star) 
  19] node*(X, Y) >= cons(X, map(flatten, Y))  because node > cons, [20] and [22], by (Copy) 
  20] node*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] node*(X, Y) >= map(flatten, Y)  because node > map, [23] and [24], by (Copy) 
  23] node*(X, Y) >= flatten  because node > flatten, by (Copy) 
  24] node*(X, Y) >= Y  because [25], by (Select) 
  25] Y >= Y  by (Meta) 

  26] nil >= nil  by (Fun) 

  27] cons(X, Y) > append(X, Y)  because [28], by definition 
  28] cons*(X, Y) >= append(X, Y)  because cons > append, [29] and [31], by (Copy) 
  29] cons*(X, Y) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 
  31] cons*(X, Y) >= Y  because [32], by (Select) 
  32] Y >= Y  by (Meta) 

  33] append(nil, X) > X  because [34], by definition 
  34] append*(nil, X) >= X  because [35], by (Select) 
  35] X >= X  by (Meta) 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  concat(cons(X, Y)) => append(X, concat(Y)) 
  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]):

  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

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

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

  @_{o -> o}(flatten, node(X, Y)) >= cons(X, concat(map(flatten, Y))) 
  concat(_|_) > _|_ 

With these choices, we have:

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

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

We can thus remove the following rules:

  concat(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]):

  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 

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

Argument functions:

  [[concat(x_1)]] = x_1 
  [[flatten]] = _|_ 

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

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

  @_{o -> o}(_|_, node(X, Y)) > cons(X, map(_|_, Y)) 

With these choices, we have:

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

We can thus remove the following rules:

  flatten node(X, Y) => cons(X, concat(map(flatten, 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.