/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 
  1 : [] --> o 
  plus : [o * o] --> o 
  times : [o * o] --> o 

  times(X, plus(Y, 1)) => plus(times(X, plus(Y, times(1, 0))), X) 
  times(X, 1) => X 
  plus(X, 0) => X 
  times(X, 0) => 0 

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] times#(X, plus(Y, 1)) =#> plus#(times(X, plus(Y, times(1, 0))), X)   
    1] times#(X, plus(Y, 1)) =#> times#(X, plus(Y, times(1, 0)))   
    2] times#(X, plus(Y, 1)) =#> plus#(Y, times(1, 0))   
    3] times#(X, plus(Y, 1)) =#> times#(1, 0)   

  Rules R_0:

    times(X, plus(Y, 1)) => plus(times(X, plus(Y, times(1, 0))), X) 
    times(X, 1) => X 
    plus(X, 0) => X 
    times(X, 0) => 0 

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

This graph has the following strongly connected components:

  P_1:

    times#(X, plus(Y, 1)) =#> times#(X, plus(Y, times(1, 0)))   

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 will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  times#(X, plus(Y, 1)) >? times#(X, plus(Y, times(1, 0))) 
  times(X, plus(Y, 1)) >= plus(times(X, plus(Y, times(1, 0))), X) 
  times(X, 1) >= X 
  plus(X, 0) >= X 
  times(X, 0) >= 0 

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

We consider usable_rules with respect to the following argument filtering:


This leaves the following ordering requirements:

  times#(X, plus(Y, 1)) > times#(X, plus(Y, times(1, 0))) 
  plus(X, 0) >= X 
  times(X, 0) >= 0 

The following interpretation satisfies the requirements:

  0 = 0 
  1 = 2 
  plus = \y0y1.y0 + 2y1 
  times = \y0y1.0 
  times# = \y0y1.y1 

Using this interpretation, the requirements translate to:

  [[times#(_x0, plus(_x1, 1))]] = 4 + x1 > x1 = [[times#(_x0, plus(_x1, times(1, 0)))]] 
  [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[times(_x0, 0)]] = 0 >= 0 = [[0]] 

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.