We consider the system reverse.

  Alphabet:

    app : [list * list] --> list 
    cons : [nat * list] --> list 
    foldl : [list -> nat -> list * list * list] --> list 
    iconsc : [] --> list -> nat -> list 
    nil : [] --> list 
    reverse : [list] --> list 
    reverse1 : [list] --> list 

  Rules:

    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    foldl(f, x, nil) => x 
    foldl(f, x, cons(y, z)) => foldl(f, f x y, z) 
    iconsc => /\x./\y.cons(y, x) 
    reverse(x) => foldl(iconsc, nil, x) 
    reverse1(x) => foldl(/\y./\z.app(cons(z, nil), y), nil, x) 

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

We observe that the rules contain a first-order subset:

  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 

Moreover, the system is orthogonal.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = 2 + 2*x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2
 ||    POL(nil) = 1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
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:

  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  foldl(F, X, nil) => X 
  foldl(F, X, cons(Y, Z)) => foldl(F, F X Y, Z) 
  iconsc(X, Y) => (/\x./\y.cons(y, x)) X Y 
  reverse(X) => foldl(/\x./\y.iconsc(x, y), nil, X) 
  reverse1(X) => foldl(/\x./\y.app(cons(y, nil), x), nil, X) 

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

  Dependency Pairs P_0:

    0] foldl#(F, X, cons(Y, Z)) =#> foldl#(F, F X Y, Z)   
    1] reverse#(X) =#> foldl#(/\x./\y.iconsc(x, y), nil, X)   
    2] reverse#(X) =#> iconsc#(Y, Z)   
    3] reverse1#(X) =#> foldl#(/\x./\y.app(cons(y, nil), x), nil, X)   
    4] reverse1#(X) =#> app#(cons(Y, nil), Z)   

  Rules R_0:

    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    foldl(F, X, nil) => X 
    foldl(F, X, cons(Y, Z)) => foldl(F, F X Y, Z) 
    iconsc(X, Y) => (/\x./\y.cons(y, x)) X Y 
    reverse(X) => foldl(/\x./\y.iconsc(x, y), nil, X) 
    reverse1(X) => foldl(/\x./\y.app(cons(y, nil), x), nil, 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 : 0 
    * 2 :  
    * 3 : 0 
    * 4 :  

This graph has the following strongly connected components:

  P_1:

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

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, 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(foldl#) = 3 

Thus, we can orient the dependency pairs as follows:

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