/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:

    abs : [] --> a -> b -> Arrab 
    app : [] --> Arrab -> a -> b 
    box : [] --> a -> Boxa 
    unbox : [] --> Boxa -> a 

  Rules:

    app (abs (/\x.f x)) y => f y 
    abs (/\x.app y x) => y 
    unbox (box x) => x 
    box (unbox x) => x 

Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.

We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:

  Alphabet:

    abs : [a -> b] --> Arrab 
    app : [Arrab * a] --> b 
    box : [a] --> Boxa 
    unbox : [Boxa] --> a 
    ~AP1 : [a -> b * a] --> b 

  Rules:

    app(abs(/\x.~AP1(F, x)), X) => ~AP1(F, X) 
    abs(/\x.app(X, x)) => X 
    unbox(box(X)) => X 
    box(unbox(X)) => X 
    app(abs(/\x.app(X, x)), Y) => app(X, Y) 
    ~AP1(F, X) => F X 

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

  unbox(box(X)) => X 
  box(unbox(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:
 || 
 ||    unbox(box(%X)) -> %X
 ||    box(unbox(%X)) -> %X
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(box(x_1)) = 2 + x_1
 ||    POL(unbox(x_1)) = 1 + 2*x_1
 ||    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:
 || 
 ||    unbox(box(%X)) -> %X
 ||    box(unbox(%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 and accessible arguments in [Kop13]).

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

  Dependency Pairs P_0:

    0] app#(abs(/\x.~AP1(F, x)), X) =#> ~AP1#(F, X)   
    1] app#(abs(/\x.app(X, x)), Y) =#> app#(X, Y)   

  Rules R_0:

    app(abs(/\x.~AP1(F, x)), X) => ~AP1(F, X) 
    abs(/\x.app(X, x)) => X 
    unbox(box(X)) => X 
    box(unbox(X)) => X 
    app(abs(/\x.app(X, x)), Y) => app(X, Y) 
    ~AP1(F, X) => F X 

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

We consider the dependency pair problem (P_0, R_0, static, 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 : 0, 1 

This graph has the following strongly connected components:

  P_1:

    app#(abs(/\x.app(X, x)), Y) =#> app#(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).

Thus, the original system is terminating if (P_1, R_0, static, all) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(app#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(app#(abs(/\x.app(X, x)), Y)) = abs(/\y.app(X, y)) |> X = nu(app#(X, Y)) 

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 +++

[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
[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.