We consider the system Applicative_05__Ex5Folding.

  Alphabet:

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

  Rules:

    fold(f, x) nil => x 
    fold(f, x) cons(y, z) => f y (fold(f, x) z) 
    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    times(0, x) => 0 
    times(s(x), y) => plus(times(x, y), y) 
    sum => fold(add, 0) 
    prod => fold(mul, s(0)) 

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]).

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

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

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

  Dependency Pairs P_0:

    0] fold#(F, X, cons(Y, Z)) =#> fold#(F, X, Z)   
    1] plus#(s(X), Y) =#> plus#(X, Y)   
    2] times#(s(X), Y) =#> plus#(times(X, Y), Y)   
    3] times#(s(X), Y) =#> times#(X, Y)   
    4] sum#(X) =#> fold#(/\x./\y.add(x, y), 0, X)   
    5] prod#(X) =#> fold#(/\x./\y.mul(x, y), s(0), X)   

  Rules R_0:

    fold(F, X, nil) => X 
    fold(F, X, cons(Y, Z)) => F Y fold(F, X, Z) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(times(X, Y), Y) 
    sum(X) => fold(/\x./\y.add(x, y), 0, X) 
    prod(X) => fold(/\x./\y.mul(x, y), s(0), 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 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 : 1 
    * 2 : 1 
    * 3 : 2, 3 
    * 4 : 0 
    * 5 : 0 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

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

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), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_3, R_0, static, formative) is finite.

We consider the dependency pair problem (P_3, R_0, static, formative).

We apply the subterm criterion with the following projection function:

  nu(times#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(times#(s(X), Y)) = s(X) |> X = nu(times#(X, Y)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_3, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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

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

We apply the subterm criterion with the following projection function:

  nu(plus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(plus#(s(X), Y)) = s(X) |> X = nu(plus#(X, Y)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by ({}, R_0, static, f).  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, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(fold#) = 3 

Thus, we can orient the dependency pairs as follows:

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

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.