We consider the system AotoYamada_05__013.

  Alphabet:

    append : [c * c] --> c 
    cons : [b * c] --> c 
    flatwith : [a -> b * b] --> c 
    flatwithsub : [a -> b * c] --> c 
    leaf : [a] --> b 
    nil : [] --> c 
    node : [c] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    flatwith(f, leaf(x)) => cons(f x, nil) 
    flatwith(f, node(x)) => flatwithsub(f, x) 
    flatwithsub(f, nil) => nil 
    flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(f, y)) 

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:

  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(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:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(append(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:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%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 [FuhKop19]).

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

  Dependency Pairs P_0:

    0] flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    1] flatwithsub#(F, cons(X, Y)) =#> append#(flatwith(F, X), flatwithsub(F, Y))   
    2] flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    3] flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    flatwith(F, leaf(X)) => cons(F X, nil) 
    flatwith(F, node(X)) => flatwithsub(F, X) 
    flatwithsub(F, nil) => nil 
    flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(F, Y)) 

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

This graph has the following strongly connected components:

  P_1:

    flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, 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).

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(flatwith#) = 2 
  nu(flatwithsub#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(flatwith#(F, node(X))) = node(X) |> X = nu(flatwithsub#(F, X)) 
  nu(flatwithsub#(F, cons(X, Y))) = cons(X, Y) |> X = nu(flatwith#(F, X)) 
  nu(flatwithsub#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(flatwithsub#(F, Y)) 

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.