/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.

  !plus : [o * o] --> o 
  !times : [o * o] --> o 
  0 : [] --> o 
  I : [o] --> o 
  O : [o] --> o 

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
  !times(0, X) => 0 
  !times(X, 0) => 0 
  !times(O(X), Y) => O(!times(X, Y)) 
  !times(I(X), Y) => !plus(O(!times(X, Y)), 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] !plus#(O(X), O(Y)) =#> O#(!plus(X, Y))   
    1] !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    2] !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    3] !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    4] !plus#(I(X), I(Y)) =#> O#(!plus(!plus(X, Y), I(0)))   
    5] !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    6] !plus#(I(X), I(Y)) =#> !plus#(X, Y)   
    7] !times#(O(X), Y) =#> O#(!times(X, Y))   
    8] !times#(O(X), Y) =#> !times#(X, Y)   
    9] !times#(I(X), Y) =#> !plus#(O(!times(X, Y)), Y)   
    10] !times#(I(X), Y) =#> O#(!times(X, Y))   
    11] !times#(I(X), Y) =#> !times#(X, Y)   

  Rules R_0:

    O(0) => 0 
    !plus(0, X) => X 
    !plus(X, 0) => X 
    !plus(O(X), O(Y)) => O(!plus(X, Y)) 
    !plus(O(X), I(Y)) => I(!plus(X, Y)) 
    !plus(I(X), O(Y)) => I(!plus(X, Y)) 
    !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
    !times(0, X) => 0 
    !times(X, 0) => 0 
    !times(O(X), Y) => O(!times(X, Y)) 
    !times(I(X), Y) => !plus(O(!times(X, Y)), 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 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, 2, 3, 4, 5, 6 
    * 2 : 0, 1, 2, 3, 4, 5, 6 
    * 3 : 0, 1, 2, 3, 4, 5, 6 
    * 4 :  
    * 5 : 2, 4, 5, 6 
    * 6 : 0, 1, 2, 3, 4, 5, 6 
    * 7 :  
    * 8 : 7, 8, 9, 10, 11 
    * 9 : 0, 1, 2 
    * 10 :  
    * 11 : 7, 8, 9, 10, 11 

This graph has the following strongly connected components:

  P_1:

    !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    !plus#(I(X), I(Y)) =#> !plus#(X, Y)   

  P_2:

    !times#(O(X), Y) =#> !times#(X, Y)   
    !times#(I(X), Y) =#> !times#(X, Y)   

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 apply the subterm criterion with the following projection function:

  nu(!times#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(!times#(O(X), Y)) = O(X) |> X = nu(!times#(X, Y)) 
  nu(!times#(I(X), Y)) = I(X) |> X = nu(!times#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 

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

  !plus#(O(X), O(Y)) >? !plus#(X, Y) 
  !plus#(O(X), I(Y)) >? !plus#(X, Y) 
  !plus#(I(X), O(Y)) >? !plus#(X, Y) 
  !plus#(I(X), I(Y)) >? !plus#(!plus(X, Y), I(0)) 
  !plus#(I(X), I(Y)) >? !plus#(X, Y) 
  O(0) >= 0 
  !plus(0, X) >= X 
  !plus(X, 0) >= X 
  !plus(O(X), O(Y)) >= O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >= O(!plus(!plus(X, Y), I(0))) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  !plus# = \y0y1.2y0 + 2y1 
  0 = 0 
  I = \y0.2 + y0 
  O = \y0.y0 

Using this interpretation, the requirements translate to:

  [[!plus#(O(_x0), O(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(O(_x0), I(_x1))]] = 4 + 2x0 + 2x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), O(_x1))]] = 4 + 2x0 + 2x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), I(_x1))]] = 8 + 2x0 + 2x1 > 4 + 2x0 + 2x1 = [[!plus#(!plus(_x0, _x1), I(0))]] 
  [[!plus#(I(_x0), I(_x1))]] = 8 + 2x0 + 2x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[O(0)]] = 0 >= 0 = [[0]] 
  [[!plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(O(_x0), O(_x1))]] = x0 + x1 >= x0 + x1 = [[O(!plus(_x0, _x1))]] 
  [[!plus(O(_x0), I(_x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), O(_x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), I(_x1))]] = 4 + x0 + x1 >= 2 + x0 + x1 = [[O(!plus(!plus(_x0, _x1), I(0)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_3, R_0, minimal, formative), where P_3 consists of:

  !plus#(O(X), O(Y)) =#> !plus#(X, Y)   

Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(!plus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(!plus#(O(X), O(Y))) = O(X) |> X = nu(!plus#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, 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.