We consider the system Applicative_05__Ex6Folding.

  Alphabet:

    0 : [] --> c 
    1 : [] --> c 
    add : [] --> c -> a -> c 
    cons : [a * b] --> b 
    fold : [c -> a -> c * b * c] --> c 
    mul : [] --> c -> a -> c 
    nil : [] --> b 
    prod : [b] --> c 
    sum : [b] --> c 

  Rules:

    fold(f, nil, x) => x 
    fold(f, cons(x, y), z) => fold(f, y, f z x) 
    sum(x) => fold(add, x, 0) 
    fold(mul, x, 1) => prod(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 [FuhKop19]).

In order to do so, we start by eta-expanding the system, which gives:

  fold(F, nil, X) => X 
  fold(F, cons(X, Y), Z) => fold(F, Y, F Z X) 
  sum(X) => fold(/\x./\y.add(x, y), X, 0) 
  fold(/\x./\y.mul(x, y), X, 1) => prod(X) 

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

  Dependency Pairs P_0:

    0] fold#(F, cons(X, Y), Z) =#> fold#(F, Y, F Z X)   
    1] sum#(X) =#> fold#(/\x./\y.add(x, y), X, 0)   

  Rules R_0:

    fold(F, nil, X) => X 
    fold(F, cons(X, Y), Z) => fold(F, Y, F Z X) 
    sum(X) => fold(/\x./\y.add(x, y), X, 0) 
    fold(/\x./\y.mul(x, y), X, 1) => prod(X) 

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

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

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 

This graph has the following strongly connected components:

  P_1:

    fold#(F, cons(X, Y), Z) =#> fold#(F, Y, F Z X)   

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).

Thus, the original system is terminating if (P_1, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(fold#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(fold#(F, cons(X, Y), Z)) = cons(X, Y) |> Y = nu(fold#(F, Y, F Z X)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, 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 +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[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.