/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 
  div : [o * o] --> o 
  quot : [o * o * o] --> o 
  s : [o] --> o 

  div(0, X) => 0 
  div(X, Y) => quot(X, Y, Y) 
  quot(0, s(X), Y) => 0 
  quot(s(X), s(Y), Z) => quot(X, Y, Z) 
  quot(X, 0, s(Y)) => s(div(X, s(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] div#(X, Y) =#> quot#(X, Y, Y)   
    1] quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z)   
    2] quot#(X, 0, s(Y)) =#> div#(X, s(Y))   

  Rules R_0:

    div(0, X) => 0 
    div(X, Y) => quot(X, Y, Y) 
    quot(0, s(X), Y) => 0 
    quot(s(X), s(Y), Z) => quot(X, Y, Z) 
    quot(X, 0, s(Y)) => s(div(X, s(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 apply the subterm criterion with the following projection function:

  nu(div#) = 1 
  nu(quot#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(div#(X, Y)) = X = X = nu(quot#(X, Y, Y)) 
  nu(quot#(s(X), s(Y), Z)) = s(X) |> X = nu(quot#(X, Y, Z)) 
  nu(quot#(X, 0, s(Y))) = X = X = nu(div#(X, s(Y))) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_0, R_0, minimal, f) by (P_1, R_0, minimal, f), where P_1 contains:

  div#(X, Y) =#> quot#(X, Y, Y)   
  quot#(X, 0, s(Y)) =#> div#(X, s(Y))   

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:

  div#(X, Y) >? quot#(X, Y, Y) 
  quot#(X, 0, s(Y)) >? div#(X, s(Y)) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  div# = \y0y1.1 + 2y1 
  quot# = \y0y1y2.2y1 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[div#(_x0, _x1)]] = 1 + 2x1 > 2x1 = [[quot#(_x0, _x1, _x1)]] 
  [[quot#(_x0, 0, s(_x1))]] = 6 > 1 = [[div#(_x0, s(_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.