/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> o 
    a : [] --> o 
    f : [o -> o] --> o 
    g : [o] --> o 
    h : [o * o] --> o 
    s : [o] --> o 

  Rules:

    a => f(/\x.g(x)) 
    f(/\x.y) => a 
    g(x) => h(x, x) 
    h(0, x) => x 
    h(s(x), 0) => g(x) 

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

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a >? f(/\x.g(x)) 
  f(/\x.X) >? a 
  g(X) >? h(X, X) 
  h(0, X) >? X 
  h(s(X), 0) >? g(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 3 
  a = 0 
  f = \G0.G0(0) 
  g = \y0.3y0 
  h = \y0y1.y0 + 2y1 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[a]] = 0 >= 0 = [[f(/\x.g(x))]] 
  [[f(/\x._x0)]] = x0 >= 0 = [[a]] 
  [[g(_x0)]] = 3x0 >= 3x0 = [[h(_x0, _x0)]] 
  [[h(0, _x0)]] = 3 + 2x0 > x0 = [[_x0]] 
  [[h(s(_x0), 0)]] = 9 + 3x0 > 3x0 = [[g(_x0)]] 

We can thus remove the following rules:

  h(0, X) => X 
  h(s(X), 0) => g(X) 

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

  g(X) => h(X, X) 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-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:
 || 
 ||    g(%X) -> h(%X, %X)
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(g(x_1)) = 1 + 2*x_1
 ||    POL(h(x_1, x_2)) = x_1 + x_2
 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    g(%X) -> h(%X, %X)
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%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 in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, all):

  Dependency Pairs P_0:

    0] a# =#> f#(/\x.g(x))   
    1] a# =#> g#(X)   
    2] f#(/\x.X) =#> a#   

  Rules R_0:

    a => f(/\x.g(x)) 
    f(/\x.X) => a 
    g(X) => h(X, X) 

Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, all).

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 : 0, 1 

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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.
[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.