We consider the system Applicative_05__BTreeMember.

  Alphabet:

    0 : [] --> a 
    eq : [a * a] --> c 
    false : [] --> c 
    fork : [b * a * b] --> b 
    if : [c * c * c] --> c 
    lt : [a * a] --> c 
    member : [a * b] --> c 
    null : [] --> b 
    s : [a] --> a 
    true : [] --> c 

  Rules:

    lt(s(x), s(y)) => lt(x, y) 
    lt(0, s(x)) => true 
    lt(x, 0) => false 
    eq(x, x) => true 
    eq(s(x), 0) => false 
    eq(0, s(x)) => false 
    member(x, null) => false 
    member(x, fork(y, z, u)) => if(lt(x, z), member(x, y), if(eq(x, z), true, member(x, u))) 

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, all):

  Dependency Pairs P_0:

    0] lt#(s(X), s(Y)) =#> lt#(X, Y)   
    1] member#(X, fork(Y, Z, U)) =#> lt#(X, Z)   
    2] member#(X, fork(Y, Z, U)) =#> member#(X, Y)   
    3] member#(X, fork(Y, Z, U)) =#> eq#(X, Z)   
    4] member#(X, fork(Y, Z, U)) =#> member#(X, U)   

  Rules R_0:

    lt(s(X), s(Y)) => lt(X, Y) 
    lt(0, s(X)) => true 
    lt(X, 0) => false 
    eq(X, X) => true 
    eq(s(X), 0) => false 
    eq(0, s(X)) => false 
    member(X, null) => false 
    member(X, fork(Y, Z, U)) => if(lt(X, Z), member(X, Y), if(eq(X, Z), true, member(X, U))) 

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

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

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 : 0 
    * 2 : 1, 2, 3, 4 
    * 3 :  
    * 4 : 1, 2, 3, 4 

This graph has the following strongly connected components:

  P_1:

    lt#(s(X), s(Y)) =#> lt#(X, Y)   

  P_2:

    member#(X, fork(Y, Z, U)) =#> member#(X, Y)   
    member#(X, fork(Y, Z, U)) =#> member#(X, U)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, all) and (P_2, R_0, static, all) is finite.

We consider the dependency pair problem (P_2, R_0, static, all).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  member#(X, fork(Y, Z, U)) >? member#(X, Y) 
  member#(X, fork(Y, Z, U)) >? member#(X, U) 

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

The following interpretation satisfies the requirements:

  fork = \y0y1y2.3 + y0 + y2 
  member# = \y0y1.y1 

Using this interpretation, the requirements translate to:

  [[member#(_x0, fork(_x1, _x2, _x3))]] = 3 + x1 + x3 > x1 = [[member#(_x0, _x1)]] 
  [[member#(_x0, fork(_x1, _x2, _x3))]] = 3 + x1 + x3 > x3 = [[member#(_x0, _x3)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, static, all) is finite.

We consider the dependency pair problem (P_1, R_0, static, all).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  lt#(s(X), s(Y)) >? lt#(X, Y) 

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

The following interpretation satisfies the requirements:

  lt# = \y0y1.y0 + y1 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[lt#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[lt#(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.