/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.

  a : [o] --> o 
  b : [o] --> o 
  p : [o * o] --> o 

  p(a(X), p(a(a(a(Y))), Z)) => p(a(Z), p(a(a(b(X))), Z)) 

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] p#(a(X), p(a(a(a(Y))), Z)) =#> p#(a(Z), p(a(a(b(X))), Z))   
    1] p#(a(X), p(a(a(a(Y))), Z)) =#> p#(a(a(b(X))), Z)   

  Rules R_0:

    p(a(X), p(a(a(a(Y))), Z)) => p(a(Z), p(a(a(b(X))), 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 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:

  p#(a(X), p(a(a(a(Y))), Z)) >? p#(a(Z), p(a(a(b(X))), Z)) 
  p#(a(X), p(a(a(a(Y))), Z)) >? p#(a(a(b(X))), Z) 
  p(a(X), p(a(a(a(Y))), Z)) >= p(a(Z), p(a(a(b(X))), Z)) 

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

The following interpretation satisfies the requirements:

  a = \y0.3 
  b = \y0.0 
  p = \y0y1.3 + 2y1 
  p# = \y0y1.y1 

Using this interpretation, the requirements translate to:

  [[p#(a(_x0), p(a(a(a(_x1))), _x2))]] = 3 + 2x2 >= 3 + 2x2 = [[p#(a(_x2), p(a(a(b(_x0))), _x2))]] 
  [[p#(a(_x0), p(a(a(a(_x1))), _x2))]] = 3 + 2x2 > x2 = [[p#(a(a(b(_x0))), _x2)]] 
  [[p(a(_x0), p(a(a(a(_x1))), _x2))]] = 9 + 4x2 >= 9 + 4x2 = [[p(a(_x2), p(a(a(b(_x0))), _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of:

  p#(a(X), p(a(a(a(Y))), Z)) =#> p#(a(Z), p(a(a(b(X))), Z))   

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:

  p#(a(X), p(a(a(a(Y))), Z)) >? p#(a(Z), p(a(a(b(X))), Z)) 
  p(a(X), p(a(a(a(Y))), Z)) >= p(a(Z), p(a(a(b(X))), Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[b(x_1)]] = b 

We choose Lex = {} and Mul = {a, b, p, p#}, and the following precedence: a = p > p# > b

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  p#(a(X), p(a(a(a(Y))), Z)) > p#(a(Z), p(a(a(b)), Z)) 
  p(a(X), p(a(a(a(Y))), Z)) >= p(a(Z), p(a(a(b)), Z)) 

With these choices, we have:

  1] p#(a(X), p(a(a(a(Y))), Z)) > p#(a(Z), p(a(a(b)), Z))  because [2], by definition 
  2] p#*(a(X), p(a(a(a(Y))), Z)) >= p#(a(Z), p(a(a(b)), Z))  because [3], by (Select) 
  3] p(a(a(a(Y))), Z) >= p#(a(Z), p(a(a(b)), Z))  because [4], by (Star) 
  4] p*(a(a(a(Y))), Z) >= p#(a(Z), p(a(a(b)), Z))  because p > p#, [5] and [7], by (Copy) 
  5] p*(a(a(a(Y))), Z) >= a(Z)  because p = a, p in Mul and [6], by (Stat) 
  6] Z >= Z  by (Meta) 
  7] p*(a(a(a(Y))), Z) >= p(a(a(b)), Z)  because p in Mul, [8] and [6], by (Stat) 
  8] a(a(a(Y))) > a(a(b))  because [9], by definition 
  9] a*(a(a(Y))) >= a(a(b))  because a in Mul and [10], by (Stat) 
  10] a(a(Y)) > a(b)  because [11], by definition 
  11] a*(a(Y)) >= a(b)  because a in Mul and [12], by (Stat) 
  12] a(Y) > b  because [13], by definition 
  13] a*(Y) >= b  because a > b, by (Copy) 

  14] p(a(X), p(a(a(a(Y))), Z)) >= p(a(Z), p(a(a(b)), Z))  because [15], by (Star) 
  15] p*(a(X), p(a(a(a(Y))), Z)) >= p(a(Z), p(a(a(b)), Z))  because p in Mul, [16] and [17], by (Stat) 
  16] p(a(a(a(Y))), Z) > a(Z)  because [5], by definition 
  17] p(a(a(a(Y))), Z) > p(a(a(b)), Z)  because [7], by definition 

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.