/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 
  ack : [o * o] --> o 
  plus : [o * o] --> o 
  s : [o] --> o 

  plus(s(s(X)), Y) => s(plus(X, s(Y))) 
  plus(X, s(s(Y))) => s(plus(s(X), Y)) 
  plus(s(0), X) => s(X) 
  plus(0, X) => X 
  ack(0, X) => s(X) 
  ack(s(X), 0) => ack(X, s(0)) 
  ack(s(X), s(Y)) => ack(X, plus(Y, ack(s(X), Y))) 

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] plus#(s(s(X)), Y) =#> plus#(X, s(Y))   
    1] plus#(X, s(s(Y))) =#> plus#(s(X), Y)   
    2] ack#(s(X), 0) =#> ack#(X, s(0))   
    3] ack#(s(X), s(Y)) =#> ack#(X, plus(Y, ack(s(X), Y)))   
    4] ack#(s(X), s(Y)) =#> plus#(Y, ack(s(X), Y))   
    5] ack#(s(X), s(Y)) =#> ack#(s(X), Y)   

  Rules R_0:

    plus(s(s(X)), Y) => s(plus(X, s(Y))) 
    plus(X, s(s(Y))) => s(plus(s(X), Y)) 
    plus(s(0), X) => s(X) 
    plus(0, X) => X 
    ack(0, X) => s(X) 
    ack(s(X), 0) => ack(X, s(0)) 
    ack(s(X), s(Y)) => ack(X, plus(Y, ack(s(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 
    * 1 : 0, 1 
    * 2 : 3, 4, 5 
    * 3 : 2, 3, 4, 5 
    * 4 : 0, 1 
    * 5 : 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_1:

    plus#(s(s(X)), Y) =#> plus#(X, s(Y))   
    plus#(X, s(s(Y))) =#> plus#(s(X), Y)   

  P_2:

    ack#(s(X), 0) =#> ack#(X, s(0))   
    ack#(s(X), s(Y)) =#> ack#(X, plus(Y, ack(s(X), Y)))   
    ack#(s(X), s(Y)) =#> ack#(s(X), 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(ack#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(ack#(s(X), 0)) = s(X) |> X = nu(ack#(X, s(0))) 
  nu(ack#(s(X), s(Y))) = s(X) |> X = nu(ack#(X, plus(Y, ack(s(X), Y)))) 
  nu(ack#(s(X), s(Y))) = s(X) = s(X) = nu(ack#(s(X), 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:

  ack#(s(X), s(Y)) =#> ack#(s(X), 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 apply the subterm criterion with the following projection function:

  nu(ack#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(ack#(s(X), s(Y))) = s(Y) |> Y = nu(ack#(s(X), Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(s(s(X)), Y) >? plus#(X, s(Y)) 
  plus#(X, s(s(Y))) >? plus#(s(X), Y) 

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

The following interpretation satisfies the requirements:

  plus# = \y0y1.2y0 + 2y1 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(s(_x0)), _x1)]] = 8 + 2x0 + 2x1 > 4 + 2x0 + 2x1 = [[plus#(_x0, s(_x1))]] 
  [[plus#(_x0, s(s(_x1)))]] = 8 + 2x0 + 2x1 > 4 + 2x0 + 2x1 = [[plus#(s(_x0), _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  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.