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


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

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

  0 : [] --> o 
  L : [o] --> o 
  N : [o * o] --> o 
  max : [o] --> o 
  s : [o] --> o 

  max(L(X)) => X 
  max(N(L(0), L(X))) => X 
  max(N(L(s(X)), L(s(Y)))) => s(max(N(L(X), L(Y)))) 
  max(N(L(X), N(Y, Z))) => max(N(L(X), L(max(N(Y, Z))))) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  max(L(X)) >? X 
  max(N(L(0), L(X))) >? X 
  max(N(L(s(X)), L(s(Y)))) >? s(max(N(L(X), L(Y)))) 
  max(N(L(X), N(Y, Z))) >? max(N(L(X), L(max(N(Y, Z))))) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  L = \y0.y0 
  N = \y0y1.y0 + y1 
  max = \y0.y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[max(L(_x0))]] = x0 >= x0 = [[_x0]] 
  [[max(N(L(0), L(_x0)))]] = 3 + x0 > x0 = [[_x0]] 
  [[max(N(L(s(_x0)), L(s(_x1))))]] = 2 + x0 + x1 > 1 + x0 + x1 = [[s(max(N(L(_x0), L(_x1))))]] 
  [[max(N(L(_x0), N(_x1, _x2)))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[max(N(L(_x0), L(max(N(_x1, _x2)))))]] 

We can thus remove the following rules:

  max(N(L(0), L(X))) => X 
  max(N(L(s(X)), L(s(Y)))) => s(max(N(L(X), L(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] max#(N(L(X), N(Y, Z))) =#> max#(N(L(X), L(max(N(Y, Z)))))   
    1] max#(N(L(X), N(Y, Z))) =#> max#(N(Y, Z))   

  Rules R_0:

    max(L(X)) => X 
    max(N(L(X), N(Y, Z))) => max(N(L(X), L(max(N(Y, Z))))) 

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 

This graph has the following strongly connected components:

  P_1:

    max#(N(L(X), N(Y, Z))) =#> max#(N(Y, Z))   

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 apply the subterm criterion with the following projection function:

  nu(max#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(max#(N(L(X), N(Y, Z)))) = N(L(X), N(Y, Z)) |> N(Y, Z) = nu(max#(N(Y, Z))) 

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.