/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 
  log : [o] --> o 
  min : [o * o] --> o 
  quot : [o * o] --> o 
  s : [o] --> o 

  min(X, 0) => X 
  min(s(X), s(Y)) => min(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) 
  log(s(0)) => 0 
  log(s(s(X))) => s(log(s(quot(X, s(s(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] min#(s(X), s(Y)) =#> min#(X, Y)   
    1] quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y))   
    2] quot#(s(X), s(Y)) =#> min#(X, Y)   
    3] log#(s(s(X))) =#> log#(s(quot(X, s(s(0)))))   
    4] log#(s(s(X))) =#> quot#(X, s(s(0)))   

  Rules R_0:

    min(X, 0) => X 
    min(s(X), s(Y)) => min(X, Y) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) 
    log(s(0)) => 0 
    log(s(s(X))) => s(log(s(quot(X, s(s(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 : 0 
    * 1 : 1, 2 
    * 2 : 0 
    * 3 : 3, 4 
    * 4 : 1, 2 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y))   

  P_3:

    log#(s(s(X))) =#> log#(s(quot(X, s(s(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), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_0) are:

  min(X, 0) => X 
  min(s(X), s(Y)) => min(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) 

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

  log#(s(s(X))) >? log#(s(quot(X, s(s(0))))) 
  min(X, 0) >= X 
  min(s(X), s(Y)) >= min(X, Y) 
  quot(0, s(X)) >= 0 
  quot(s(X), s(Y)) >= s(quot(min(X, Y), s(Y))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  log# = \y0.2y0 
  min = \y0y1.y0 
  quot = \y0y1.y0 
  s = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[log#(s(s(_x0)))]] = 6 + 8x0 > 2 + 4x0 = [[log#(s(quot(_x0, s(s(0)))))]] 
  [[min(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[min(s(_x0), s(_x1))]] = 1 + 2x0 >= x0 = [[min(_x0, _x1)]] 
  [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[quot(s(_x0), s(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[s(quot(min(_x0, _x1), s(_x1)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_2, R_0) are:

  min(X, 0) => X 
  min(s(X), s(Y)) => min(X, Y) 

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

  quot#(s(X), s(Y)) >? quot#(min(X, Y), s(Y)) 
  min(X, 0) >= X 
  min(s(X), s(Y)) >= min(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  min = \y0y1.y0 
  quot# = \y0y1.3y0 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[quot#(s(_x0), s(_x1))]] = 9 + 9x0 > 3x0 = [[quot#(min(_x0, _x1), s(_x1))]] 
  [[min(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[min(s(_x0), s(_x1))]] = 3 + 3x0 >= x0 = [[min(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  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(min#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(min#(s(X), s(Y))) = s(X) |> X = nu(min#(X, 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.