/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 : [] --> c 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    rev : [d] --> d 
    rev1 : [c * d] --> c 
    rev2 : [c * d] --> d 
    s : [a] --> c 
    true : [] --> b 

  Rules:

    rev(nil) => nil 
    rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) 
    rev1(0, nil) => 0 
    rev1(s(x), nil) => s(x) 
    rev1(x, cons(y, z)) => rev1(y, z) 
    rev2(x, nil) => nil 
    rev2(x, cons(y, z)) => rev(cons(x, rev2(y, z))) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(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:

  rev(nil) => nil 
  rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
  rev1(0, nil) => 0 
  rev1(s(X), nil) => s(X) 
  rev1(X, cons(Y, Z)) => rev1(Y, Z) 
  rev2(X, nil) => nil 
  rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 

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) Overlay + Local Confluence [EQUIVALENT]
 || (2) QTRS
 || (3) DependencyPairsProof [EQUIVALENT]
 || (4) QDP
 || (5) DependencyGraphProof [EQUIVALENT]
 || (6) AND
 ||     (7) QDP
 ||         (8) UsableRulesProof [EQUIVALENT]
 ||         (9) QDP
 ||         (10) QReductionProof [EQUIVALENT]
 ||         (11) QDP
 ||         (12) QDPSizeChangeProof [EQUIVALENT]
 ||         (13) YES
 ||     (14) QDP
 ||         (15) UsableRulesProof [EQUIVALENT]
 ||         (16) QDP
 ||         (17) QDPOrderProof [EQUIVALENT]
 ||         (18) QDP
 ||         (19) DependencyGraphProof [EQUIVALENT]
 ||         (20) TRUE
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    rev(nil) -> nil
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) Overlay + Local Confluence (EQUIVALENT)
 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    rev(nil) -> nil
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || 
 || ----------------------------------------
 || 
 || (3) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV(cons(%X, %Y)) -> REV1(%X, %Y)
 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    rev(nil) -> nil
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    rev(nil) -> nil
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (9)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (10) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (12) 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:
 || *REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
 || The graph contains the following edges 2 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (13)
 || YES
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    rev(nil) -> nil
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (15) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (16)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (17) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(REV(x_1)) = 1 + x_1
 ||    POL(REV2(x_1, x_2)) = 1 + x_2
 ||    POL(cons(x_1, x_2)) = 1 + x_2
 ||    POL(nil) = 0
 ||    POL(rev(x_1)) = x_1
 ||    POL(rev1(x_1, x_2)) = 0
 ||    POL(rev2(x_1, x_2)) = x_2
 ||    POL(s(x_1)) = 0
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    rev2(%X, nil) -> nil
 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
 ||    rev1(0, nil) -> 0
 ||    rev1(s(%X), nil) -> s(%X)
 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
 || 
 || The set Q consists of the following terms:
 || 
 ||    rev(nil)
 ||    rev(cons(x0, x1))
 ||    rev1(0, nil)
 ||    rev1(s(x0), nil)
 ||    rev1(x0, cons(x1, x2))
 ||    rev2(x0, nil)
 ||    rev2(x0, cons(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (19) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (20)
 || 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] map#(F, cons(X, Y)) =#> map#(F, Y)   
    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    rev(nil) => nil 
    rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
    rev1(0, nil) => 0 
    rev1(s(X), nil) => s(X) 
    rev1(X, cons(Y, Z)) => rev1(Y, Z) 
    rev2(X, nil) => nil 
    rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, 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 : 2, 3 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(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) and (P_2, R_0, m, f).

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

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

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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 :  
    * 1 :  

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

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(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 [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.