/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:

    a : [] --> a 
    b : [] --> a 
    cons : [e * f] --> f 
    f : [a] --> b 
    false : [] --> d 
    filter : [e -> d * f] --> f 
    filter2 : [d * e -> d * e * f] --> f 
    g : [a] --> c 
    map : [e -> e * f] --> f 
    nil : [] --> f 
    true : [] --> d 

  Rules:

    f(a) => f(b) 
    g(b) => g(a) 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, y) 

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]):

  f(a) >? f(b) 
  g(b) >? g(a) 
  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) 

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

The following interpretation satisfies the requirements:

  a = 0 
  b = 0 
  cons = \y0y1.2 + y1 + 2y0 
  f = \y0.y0 
  false = 3 
  filter = \G0y1.2 + 2y1 + G0(0) + y1G0(y1) 
  filter2 = \y0G1y2y3.1 + y0 + 2y3 + 3y2 + G1(y2) + y3G1(y3) 
  g = \y0.2y0 
  map = \G0y1.1 + 3y1 + G0(0) + 2y1G0(y1) 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f(a)]] = 0 >= 0 = [[f(b)]] 
  [[g(b)]] = 0 >= 0 = [[g(a)]] 
  [[map(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 7 + 3x2 + 6x1 + F0(0) + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) > 3 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 1 + 2x2 + 4x1 + 2F0(x1) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) >= 4 + 2x1 + 2x2 + F0(0) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) > 2 + 2x2 + F0(0) + x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  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(false, F, X, Y) => filter(F, Y) 

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

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

  f(a) >? f(b) 
  g(b) >? g(a) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 

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

The following interpretation satisfies the requirements:

  a = 0 
  b = 0 
  cons = \y0y1.y0 + y1 
  f = \y0.2y0 
  filter = \G0y1.y1 + G0(0) 
  filter2 = \y0G1y2y3.3 + 3y0 + 3y2 + 3y3 + G1(y0) + 2G1(0) + 2G1(y3) 
  g = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f(a)]] = 0 >= 0 = [[f(b)]] 
  [[g(b)]] = 0 >= 0 = [[g(a)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 12 + 3x1 + 3x2 + F0(3) + 2F0(0) + 2F0(x2) > x1 + x2 + F0(0) = [[cons(_x1, filter(_F0, _x2))]] 

We can thus remove the following rules:

  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 

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

  f(a) => f(b) 
  g(b) => g(a) 

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) RFCMatchBoundsTRSProof [EQUIVALENT]
 || (2) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    f(a) -> f(b)
 ||    g(b) -> g(a)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) RFCMatchBoundsTRSProof (EQUIVALENT)
 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R.
 || The following rules were used to construct the certificate:
 || 
 ||    f(a) -> f(b)
 ||    g(b) -> g(a)
 || 
 || The certificate found is represented by the following graph.
 || The certificate consists of the following enumerated nodes:
 || 1, 2, 5, 6
 || 
 || Node 1 is start node and node 2 is final node.
 || 
 || Those nodes are connected through the following edges:
 || 
 || * 1 to 5 labelled f_1(0)* 1 to 6 labelled g_1(0)* 2 to 2 labelled #_1(0)* 5 to 2 labelled b(0)* 6 to 2 labelled a(0)
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || 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, formative):

  Dependency Pairs P_0:


  Rules R_0:

    f(a) => f(b) 
    g(b) => g(a) 

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:


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.