We consider the system 05height.

  Alphabet:

    cons : [t * f] --> f 
    heightf : [f] --> N 
    heightt : [t] --> N 
    leaf : [] --> t 
    max : [N * N] --> N 
    nil : [] --> f 
    node : [f] --> t 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    heightf(nil) => z 
    heightf(cons(x, y)) => max(heightt(x), heightf(y)) 
    heightt(leaf) => z 
    heightt(node(x)) => s(heightf(x)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] heightf#(cons(X, Y)) =#> heightt#(X)   
    1] heightf#(cons(X, Y)) =#> heightf#(Y)   
    2] heightt#(node(X)) =#> heightf#(X)   

  Rules R_0:

    heightf(nil) => z 
    heightf(cons(X, Y)) => max(heightt(X), heightf(Y)) 
    heightt(leaf) => z 
    heightt(node(X)) => s(heightf(X)) 

Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  heightf#(cons(X, Y)) >? heightt#(X) 
  heightf#(cons(X, Y)) >? heightf#(Y) 
  heightt#(node(X)) >? heightf#(X) 
  heightf(nil) >= z 
  heightf(cons(X, Y)) >= max(heightt(X), heightf(Y)) 
  heightt(leaf) >= z 
  heightt(node(X)) >= s(heightf(X)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y0 + 2y1 
  heightf = \y0.0 
  heightf# = \y0.2y0 
  heightt = \y0.0 
  heightt# = \y0.2y0 
  leaf = 3 
  max = \y0y1.0 
  nil = 3 
  node = \y0.3 + 2y0 
  s = \y0.0 
  z = 0 

Using this interpretation, the requirements translate to:

  [[heightf#(cons(_x0, _x1))]] = 6 + 4x0 + 4x1 > 2x0 = [[heightt#(_x0)]] 
  [[heightf#(cons(_x0, _x1))]] = 6 + 4x0 + 4x1 > 2x1 = [[heightf#(_x1)]] 
  [[heightt#(node(_x0))]] = 6 + 4x0 > 2x0 = [[heightf#(_x0)]] 
  [[heightf(nil)]] = 0 >= 0 = [[z]] 
  [[heightf(cons(_x0, _x1))]] = 0 >= 0 = [[max(heightt(_x0), heightf(_x1))]] 
  [[heightt(leaf)]] = 0 >= 0 = [[z]] 
  [[heightt(node(_x0))]] = 0 >= 0 = [[s(heightf(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.