/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
We consider the system theBenchmark.

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

  !minus : [o * o] --> o 
  0 : [] --> o 
  f : [o] --> o 
  g : [o] --> o 
  s : [o] --> o 

  !minus(X, 0) => X 
  !minus(0, s(X)) => 0 
  !minus(s(X), s(Y)) => !minus(X, Y) 
  f(0) => 0 
  f(s(X)) => !minus(s(X), g(f(X))) 
  g(0) => s(0) 
  g(s(X)) => !minus(s(X), f(g(X))) 

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] !minus#(s(X), s(Y)) =#> !minus#(X, Y)   
    1] f#(s(X)) =#> !minus#(s(X), g(f(X)))   
    2] f#(s(X)) =#> g#(f(X))   
    3] f#(s(X)) =#> f#(X)   
    4] g#(s(X)) =#> !minus#(s(X), f(g(X)))   
    5] g#(s(X)) =#> f#(g(X))   
    6] g#(s(X)) =#> g#(X)   

  Rules R_0:

    !minus(X, 0) => X 
    !minus(0, s(X)) => 0 
    !minus(s(X), s(Y)) => !minus(X, Y) 
    f(0) => 0 
    f(s(X)) => !minus(s(X), g(f(X))) 
    g(0) => s(0) 
    g(s(X)) => !minus(s(X), f(g(X))) 

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 : 0 
    * 2 : 4, 5, 6 
    * 3 : 1, 2, 3 
    * 4 : 0 
    * 5 : 1, 2, 3 
    * 6 : 4, 5, 6 

This graph has the following strongly connected components:

  P_1:

    !minus#(s(X), s(Y)) =#> !minus#(X, Y)   

  P_2:

    f#(s(X)) =#> g#(f(X))   
    f#(s(X)) =#> f#(X)   
    g#(s(X)) =#> f#(g(X))   
    g#(s(X)) =#> g#(X)   

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) and (P_2, R_0, m, f).

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

  f#(s(X)) >? g#(f(X)) 
  f#(s(X)) >? f#(X) 
  g#(s(X)) >? f#(g(X)) 
  g#(s(X)) >? g#(X) 
  !minus(X, 0) >= X 
  !minus(0, s(X)) >= 0 
  !minus(s(X), s(Y)) >= !minus(X, Y) 
  f(0) >= 0 
  f(s(X)) >= !minus(s(X), g(f(X))) 
  g(0) >= s(0) 
  g(s(X)) >= !minus(s(X), f(g(X))) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 
  0 = 0 
  f = \y0.y0 
  f# = \y0.2y0 
  g = \y0.1 + y0 
  g# = \y0.1 + 2y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[g#(f(_x0))]] 
  [[f#(s(_x0))]] = 2 + 2x0 > 2x0 = [[f#(_x0)]] 
  [[g#(s(_x0))]] = 3 + 2x0 > 2 + 2x0 = [[f#(g(_x0))]] 
  [[g#(s(_x0))]] = 3 + 2x0 > 1 + 2x0 = [[g#(_x0)]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[!minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[!minus(_x0, _x1)]] 
  [[f(0)]] = 0 >= 0 = [[0]] 
  [[f(s(_x0))]] = 1 + x0 >= 1 + x0 = [[!minus(s(_x0), g(f(_x0)))]] 
  [[g(0)]] = 1 >= 1 = [[s(0)]] 
  [[g(s(_x0))]] = 2 + x0 >= 1 + x0 = [[!minus(s(_x0), f(g(_x0)))]] 

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(!minus#) = 1 

Thus, we can orient the dependency pairs as follows:

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