We consider the system Applicative_05__TreeMap.

  Alphabet:

    cons : [c * b] --> b 
    map : [c -> c * b] --> b 
    nil : [] --> b 
    node : [a * b] --> c 
    treemap : [a -> a] --> c -> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    treemap(f) node(x, y) => node(f x, map(treemap(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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  treemap(F) node(X, Y) >? node(F X, map(treemap(F), Y)) 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, cons, map, node, treemap}, and the following precedence: @_{o -> o} = map > node > treemap > 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}(treemap(F), node(X, Y)) > node(@_{o -> o}(F, X), map(treemap(F), Y)) 

With these choices, we have:

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

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

  13] @_{o -> o}(treemap(F), node(X, Y)) > node(@_{o -> o}(F, X), map(treemap(F), Y))  because [14], by definition 
  14] @_{o -> o}*(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y))  because @_{o -> o} > node, [15] and [22], by (Copy) 
  15] @_{o -> o}*(treemap(F), node(X, Y)) >= @_{o -> o}(F, X)  because @_{o -> o} in Mul, [16] and [19], by (Stat) 
  16] treemap(F) >= F  because [17], by (Star) 
  17] treemap*(F) >= F  because [18], by (Select) 
  18] F >= F  by (Meta) 
  19] node(X, Y) > X  because [20], by definition 
  20] node*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] @_{o -> o}*(treemap(F), node(X, Y)) >= map(treemap(F), Y)  because @_{o -> o} = map, @_{o -> o} in Mul, [23] and [25], by (Stat) 
  23] treemap(F) >= treemap(F)  because treemap in Mul and [24], by (Fun) 
  24] F >= F  by (Meta) 
  25] node(X, Y) > Y  because [26], by definition 
  26] node*(X, Y) >= Y  because [27], by (Select) 
  27] Y >= Y  by (Meta) 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  treemap(F) node(X, Y) => node(F X, map(treemap(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]):

  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.