We consider the system AotoYamada_05__021.

  Alphabet:

    0 : [] --> a 
    cons : [a * b] --> b 
    double : [b] --> b 
    inc : [b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 
    plus : [a] --> a -> a 
    s : [a] --> a 
    times : [a] --> a -> a 

  Rules:

    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 
    inc(x) => map(plus(s(0)), x) 
    double(x) => map(times(s(s(0))), x) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(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:

  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) 

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:
 || 
 ||    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)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || times_2 > plus_2 > s_1
 || times_2 > 0 > s_1
 || 
 || 
 || Status:
 || plus_2: multiset status
 || 0: multiset status
 || s_1: multiset status
 || times_2: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    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)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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]).

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

  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) 
  inc(X) => map(/\x.plus(s(0), x), X) 
  double(X) => map(/\x.times(s(s(0)), x), X) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 

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

  Dependency Pairs P_0:

    0] inc#(X) =#> map#(/\x.plus(s(0), x), X)   
    1] inc#(X) =#> plus#(s(0), Y)   
    2] double#(X) =#> map#(/\x.times(s(s(0)), x), X)   
    3] double#(X) =#> times#(s(s(0)), Y)   
    4] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    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) 
    inc(X) => map(/\x.plus(s(0), x), X) 
    double(X) => map(/\x.times(s(s(0)), x), X) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(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 : 4 
    * 1 :  
    * 2 : 4 
    * 3 :  
    * 4 : 4 

This graph has the following strongly connected components:

  P_1:

    map#(F, cons(X, Y)) =#> map#(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(map#) = 2 

Thus, we can orient the dependency pairs as follows:

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