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 apply the subterm criterion with the following projection function:

  nu(eq#) = 1 
  nu(lt#) = 1 
  nu(member#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(lt#(s(X), s(Y))) = s(X) |> X = nu(lt#(X, Y)) 
  nu(member#(X, fork(Y, Z, U))) = X = X = nu(lt#(X, Z)) 
  nu(member#(X, fork(Y, Z, U))) = X = X = nu(member#(X, Y)) 
  nu(member#(X, fork(Y, Z, U))) = X = X = nu(eq#(X, Z)) 
  nu(member#(X, fork(Y, Z, U))) = X = X = nu(member#(X, U)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_0, R_0, static, f) by (P_1, R_0, static, f), where P_1 contains:

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

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 apply the subterm criterion with the following projection function:

  nu(eq#) = 2 
  nu(lt#) = 2 
  nu(member#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(member#(X, fork(Y, Z, U))) = fork(Y, Z, U) |> Z = nu(lt#(X, Z)) 
  nu(member#(X, fork(Y, Z, U))) = fork(Y, Z, U) |> Y = nu(member#(X, Y)) 
  nu(member#(X, fork(Y, Z, U))) = fork(Y, Z, U) |> Z = nu(eq#(X, Z)) 
  nu(member#(X, fork(Y, Z, U))) = fork(Y, Z, U) |> U = nu(member#(X, U)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  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.