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


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  choose : [o * o * o * o] --> o 
  cons : [o * o] --> o 
  insert : [o * o] --> o 
  nil : [] --> o 
  s : [o] --> o 
  sort : [o] --> o 

  sort(nil) => nil 
  sort(cons(X, Y)) => insert(X, sort(Y)) 
  insert(X, nil) => cons(X, nil) 
  insert(X, cons(Y, Z)) => choose(X, cons(Y, Z), X, Y) 
  choose(X, cons(Y, Z), U, 0) => cons(X, cons(Y, Z)) 
  choose(X, cons(Y, Z), 0, s(U)) => cons(Y, insert(X, Z)) 
  choose(X, cons(Y, Z), s(U), s(V)) => choose(X, cons(Y, Z), U, V) 

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] sort#(cons(X, Y)) =#> insert#(X, sort(Y))   
    1] sort#(cons(X, Y)) =#> sort#(Y)   
    2] insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y)   
    3] choose#(X, cons(Y, Z), 0, s(U)) =#> insert#(X, Z)   
    4] choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V)   

  Rules R_0:

    sort(nil) => nil 
    sort(cons(X, Y)) => insert(X, sort(Y)) 
    insert(X, nil) => cons(X, nil) 
    insert(X, cons(Y, Z)) => choose(X, cons(Y, Z), X, Y) 
    choose(X, cons(Y, Z), U, 0) => cons(X, cons(Y, Z)) 
    choose(X, cons(Y, Z), 0, s(U)) => cons(Y, insert(X, Z)) 
    choose(X, cons(Y, Z), s(U), s(V)) => choose(X, cons(Y, Z), U, V) 

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 : 2 
    * 1 : 0, 1 
    * 2 : 3, 4 
    * 3 : 2 
    * 4 : 3, 4 

This graph has the following strongly connected components:

  P_1:

    sort#(cons(X, Y)) =#> sort#(Y)   

  P_2:

    insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y)   
    choose#(X, cons(Y, Z), 0, s(U)) =#> insert#(X, Z)   
    choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V)   

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(choose#) = 2 
  nu(insert#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(insert#(X, cons(Y, Z))) = cons(Y, Z) = cons(Y, Z) = nu(choose#(X, cons(Y, Z), X, Y)) 
  nu(choose#(X, cons(Y, Z), 0, s(U))) = cons(Y, Z) |> Z = nu(insert#(X, Z)) 
  nu(choose#(X, cons(Y, Z), s(U), s(V))) = cons(Y, Z) = cons(Y, Z) = nu(choose#(X, cons(Y, Z), U, V)) 

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:

  insert#(X, cons(Y, Z)) =#> choose#(X, cons(Y, Z), X, Y)   
  choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V)   

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 
    * 1 : 1 

This graph has the following strongly connected components:

  P_4:

    choose#(X, cons(Y, Z), s(U), s(V)) =#> choose#(X, cons(Y, Z), U, V)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_3, R_0, m, f) by (P_4, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(choose#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(choose#(X, cons(Y, Z), s(U), s(V))) = s(U) |> U = nu(choose#(X, cons(Y, Z), U, V)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, 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, 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(sort#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sort#(cons(X, Y))) = cons(X, Y) |> Y = nu(sort#(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.