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


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


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> nat 
    cons : [nat * list] --> list 
    false : [] --> bool 
    gcd : [nat * nat] --> nat 
    gcdlists : [list * list] --> list 
    if : [bool * nat * nat] --> nat 
    le : [nat * nat] --> bool 
    minus : [nat * nat] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 
    true : [] --> bool 
    zipWith : [nat -> nat -> nat * list * list] --> list 

  Rules:

    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    gcd(0, x) => 0 
    gcd(s(x), 0) => 0 
    gcd(s(x), s(y)) => if(le(y, x), s(x), s(y)) 
    if(true, s(x), s(y)) => gcd(minus(x, y), s(y)) 
    if(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 
    zipWith(f, x, nil) => nil 
    zipWith(f, nil, x) => nil 
    zipWith(f, cons(x, y), cons(z, u)) => cons(f x z, zipWith(f, y, u)) 
    gcdlists(x, y) => zipWith(/\z./\u.gcd(z, u), x, 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:

  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  gcd(0, X) => 0 
  gcd(s(X), 0) => 0 
  gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
  if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
  if(false, s(X), s(Y)) => gcd(minus(Y, X), s(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) 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) DependencyGraphProof [EQUIVALENT]
 ||         (19) TRUE
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
 ||    GCD(s(%X), s(%Y)) -> LE(%Y, %X)
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(true, s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 ||    IF(false, s(%X), s(%Y)) -> MINUS(%Y, %X)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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 3 less nodes.
 || ----------------------------------------
 || 
 || (4)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (5)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%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:
 || *MINUS(s(%X), s(%Y)) -> MINUS(%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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%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:
 || *LE(s(%X), s(%Y)) -> LE(%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)) -> IF(le(%Y, %X), s(%X), s(%Y))
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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.
 || 
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(GCD(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(IF(x_1, x_2, x_3)) = 1 + x_2 + x_3
 ||    POL(false) = 0
 ||    POL(le(x_1, x_2)) = 0
 ||    POL(minus(x_1, x_2)) = x_1
 ||    POL(s(x_1)) = 1 + x_1
 ||    POL(true) = 0
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (17)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (18) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (19)
 || TRUE
 || 
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, formative):

  Dependency Pairs P_0:

    0] zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   
    1] gcdlists#(X, Y) =#> zipWith#(/\x./\y.gcd(x, y), X, Y)   
    2] gcdlists#(X, Y) =#> gcd#(Z, U)   

  Rules R_0:

    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    gcd(0, X) => 0 
    gcd(s(X), 0) => 0 
    gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
    if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
    if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 
    zipWith(F, X, nil) => nil 
    zipWith(F, nil, X) => nil 
    zipWith(F, cons(X, Y), cons(Z, U)) => cons(F X Z, zipWith(F, Y, U)) 
    gcdlists(X, Y) => zipWith(/\x./\y.gcd(x, y), X, Y) 

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

We consider the dependency pair problem (P_0, R_0, minimal, 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 : 0 
    * 2 :  

This graph has the following strongly connected components:

  P_1:

    zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   

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, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(zipWith#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(zipWith#(F, cons(X, Y), cons(Z, U))) = cons(X, Y) |> Y = nu(zipWith#(F, Y, U)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, 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.
[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.