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


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> nat 
    build : [nat] --> list 
    collapse : [list] --> nat 
    cons : [nat -> nat * list] --> list 
    diff : [nat * nat] --> nat 
    gcd : [nat * nat] --> nat 
    min : [nat * nat] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 

  Rules:

    min(x, 0) => 0 
    min(0, x) => 0 
    min(s(x), s(y)) => s(min(x, y)) 
    diff(x, 0) => x 
    diff(0, x) => x 
    diff(s(x), s(y)) => diff(x, y) 
    gcd(s(x), 0) => s(x) 
    gcd(0, s(x)) => s(x) 
    gcd(s(x), s(y)) => gcd(diff(x, y), s(min(x, y))) 
    collapse(nil) => 0 
    collapse(cons(f, x)) => f collapse(x) 
    build(0) => nil 
    build(s(x)) => cons(/\y.gcd(y, x), build(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:

  min(X, 0) => 0 
  min(0, X) => 0 
  min(s(X), s(Y)) => s(min(X, Y)) 
  diff(X, 0) => X 
  diff(0, X) => X 
  diff(s(X), s(Y)) => diff(X, Y) 
  gcd(s(X), 0) => s(X) 
  gcd(0, s(X)) => s(X) 
  gcd(s(X), s(Y)) => gcd(diff(X, Y), s(min(X, Y))) 

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) DependencyPairsProof [EQUIVALENT]
 || (2) QDP
 || (3) DependencyGraphProof [EQUIVALENT]
 || (4) AND
 ||     (5) QDP
 ||         (6) UsableRulesProof [EQUIVALENT]
 ||         (7) QDP
 ||         (8) QDPSizeChangeProof [EQUIVALENT]
 ||         (9) YES
 ||     (10) QDP
 ||         (11) UsableRulesProof [EQUIVALENT]
 ||         (12) QDP
 ||         (13) QDPSizeChangeProof [EQUIVALENT]
 ||         (14) YES
 ||     (15) QDP
 ||         (16) QDPOrderProof [EQUIVALENT]
 ||         (17) QDP
 ||         (18) PisEmptyProof [EQUIVALENT]
 ||         (19) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(s(%X), s(%Y)) -> MIN(%X, %Y)
 ||    DIFF(s(%X), s(%Y)) -> DIFF(%X, %Y)
 ||    GCD(s(%X), s(%Y)) -> GCD(diff(%X, %Y), s(min(%X, %Y)))
 ||    GCD(s(%X), s(%Y)) -> DIFF(%X, %Y)
 ||    GCD(s(%X), s(%Y)) -> MIN(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (3) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
 || ----------------------------------------
 || 
 || (4)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (5)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    DIFF(s(%X), s(%Y)) -> DIFF(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (6) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    DIFF(s(%X), s(%Y)) -> DIFF(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *DIFF(s(%X), s(%Y)) -> DIFF(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (9)
 || YES
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(s(%X), s(%Y)) -> MIN(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (11) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (12)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(s(%X), s(%Y)) -> MIN(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (13) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *MIN(s(%X), s(%Y)) -> MIN(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (14)
 || YES
 || 
 || ----------------------------------------
 || 
 || (15)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    GCD(s(%X), s(%Y)) -> GCD(diff(%X, %Y), s(min(%X, %Y)))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (16) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    GCD(s(%X), s(%Y)) -> GCD(diff(%X, %Y), s(min(%X, %Y)))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( GCD_2(x_1, x_2) ) = max{0, 2x_1 + x_2 - 2}
 || POL( diff_2(x_1, x_2) ) = x_1 + x_2
 || POL( 0 ) = 0
 || POL( s_1(x_1) ) = 2x_1 + 2
 || POL( min_2(x_1, x_2) ) = x_1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 || 
 || 
 || ----------------------------------------
 || 
 || (17)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    min(%X, 0) -> 0
 ||    min(0, %X) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    diff(%X, 0) -> %X
 ||    diff(0, %X) -> %X
 ||    diff(s(%X), s(%Y)) -> diff(%X, %Y)
 ||    gcd(s(%X), 0) -> s(%X)
 ||    gcd(0, s(%X)) -> s(%X)
 ||    gcd(s(%X), s(%Y)) -> gcd(diff(%X, %Y), s(min(%X, %Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (18) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (19)
 || 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, formative):

  Dependency Pairs P_0:

    0] collapse#(cons(F, X)) =#> collapse#(X)   
    1] build#(s(X)) =#> gcd#(Y, X)   
    2] build#(s(X)) =#> build#(X)   

  Rules R_0:

    min(X, 0) => 0 
    min(0, X) => 0 
    min(s(X), s(Y)) => s(min(X, Y)) 
    diff(X, 0) => X 
    diff(0, X) => X 
    diff(s(X), s(Y)) => diff(X, Y) 
    gcd(s(X), 0) => s(X) 
    gcd(0, s(X)) => s(X) 
    gcd(s(X), s(Y)) => gcd(diff(X, Y), s(min(X, Y))) 
    collapse(nil) => 0 
    collapse(cons(F, X)) => F collapse(X) 
    build(0) => nil 
    build(s(X)) => cons(/\x.gcd(x, X), build(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 :  
    * 2 : 1, 2 

This graph has the following strongly connected components:

  P_1:

    collapse#(cons(F, X)) =#> collapse#(X)   

  P_2:

    build#(s(X)) =#> build#(X)   

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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We apply the subterm criterion with the following projection function:

  nu(build#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(build#(s(X))) = s(X) |> X = nu(build#(X)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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(collapse#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(collapse#(cons(F, X))) = cons(F, X) |> X = nu(collapse#(X)) 

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.