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:

  [[node(x_1, x_2)]] = node(x_2, x_1) 

We choose Lex = {cons, node} and Mul = {@_{o -> o}, map, nil, treemap}, and the following precedence: nil > treemap > node > @_{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, nil) > nil 
  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, nil) > nil  because [2], by definition 
  2] map*(F, nil) >= nil  because [3], by (Select) 
  3] nil >= nil  by (Fun) 

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

  15] @_{o -> o}(treemap(F), node(X, Y)) > node(@_{o -> o}(F, X), map(treemap(F), Y))  because [16], by definition 
  16] @_{o -> o}*(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y))  because [17], by (Select) 
  17] treemap(F) @_{o -> o}*(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y))  because [18] 
  18] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= node(@_{o -> o}(F, X), map(treemap(F), Y))  because treemap > node, [19] and [27], by (Copy) 
  19] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= @_{o -> o}(F, X)  because treemap > @_{o -> o}, [20] and [22], by (Copy) 
  20] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= F  because [21], by (Select) 
  21] F >= F  by (Meta) 
  22] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= X  because [23], by (Select) 
  23] @_{o -> o}*(treemap(F), node(X, Y)) >= X  because [24], by (Select) 
  24] node(X, Y) >= X  because [25], by (Star) 
  25] node*(X, Y) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 
  27] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= map(treemap(F), Y)  because [28], by (Select) 
  28] @_{o -> o}*(treemap(F), node(X, Y)) >= map(treemap(F), Y)  because @_{o -> o} = map, @_{o -> o} in Mul, [29] and [31], by (Stat) 
  29] treemap(F) >= treemap(F)  because treemap in Mul and [30], by (Fun) 
  30] F >= F  by (Meta) 
  31] node(X, Y) > Y  because [32], by definition 
  32] node*(X, Y) >= Y  because [33], by (Select) 
  33] Y >= Y  by (Meta) 

We can thus remove the following rules:

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

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  cons = Lam[y0;y1].3 + y0 + y1 
  map = Lam[G0;y1].3*y1 + G0(y1) + 2*y1*G0(y1) 

Using this interpretation, the requirements translate to:

  [[map(_F0, cons(_x1, _x2))]] = 9 + 3*x1 + 3*x2 + 2*x1*F0(3 + x1 + x2) + 2*x2*F0(3 + x1 + x2) + 7*F0(3 + x1 + x2) > 3 + x1 + 3*x2 + F0(x1) + F0(x2) + 2*x2*F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, 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.