/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.

  circ : [o * o] --> o 
  cons : [o * o] --> o 
  id : [] --> o 
  lift : [] --> o 
  msubst : [o * o] --> o 
  subst : [o * o] --> o 

  circ(cons(X, Y), Z) => cons(msubst(X, Z), circ(Y, Z)) 
  circ(cons(lift, X), cons(Y, Z)) => cons(Y, circ(X, Z)) 
  circ(cons(lift, X), cons(lift, Y)) => cons(lift, circ(X, Y)) 
  circ(circ(X, Y), Z) => circ(X, circ(Y, Z)) 
  circ(X, id) => X 
  circ(id, X) => X 
  circ(cons(lift, X), circ(cons(lift, Y), Z)) => circ(cons(lift, circ(X, Y)), Z) 
  subst(X, id) => X 
  msubst(X, id) => X 
  msubst(msubst(X, Y), Z) => msubst(X, circ(Y, 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] circ#(cons(X, Y), Z) =#> msubst#(X, Z)   
    1] circ#(cons(X, Y), Z) =#> circ#(Y, Z)   
    2] circ#(cons(lift, X), cons(Y, Z)) =#> circ#(X, Z)   
    3] circ#(cons(lift, X), cons(lift, Y)) =#> circ#(X, Y)   
    4] circ#(circ(X, Y), Z) =#> circ#(X, circ(Y, Z))   
    5] circ#(circ(X, Y), Z) =#> circ#(Y, Z)   
    6] circ#(cons(lift, X), circ(cons(lift, Y), Z)) =#> circ#(cons(lift, circ(X, Y)), Z)   
    7] circ#(cons(lift, X), circ(cons(lift, Y), Z)) =#> circ#(X, Y)   
    8] msubst#(msubst(X, Y), Z) =#> msubst#(X, circ(Y, Z))   
    9] msubst#(msubst(X, Y), Z) =#> circ#(Y, Z)   

  Rules R_0:

    circ(cons(X, Y), Z) => cons(msubst(X, Z), circ(Y, Z)) 
    circ(cons(lift, X), cons(Y, Z)) => cons(Y, circ(X, Z)) 
    circ(cons(lift, X), cons(lift, Y)) => cons(lift, circ(X, Y)) 
    circ(circ(X, Y), Z) => circ(X, circ(Y, Z)) 
    circ(X, id) => X 
    circ(id, X) => X 
    circ(cons(lift, X), circ(cons(lift, Y), Z)) => circ(cons(lift, circ(X, Y)), Z) 
    subst(X, id) => X 
    msubst(X, id) => X 
    msubst(msubst(X, Y), Z) => msubst(X, circ(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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_0, R_0) are:

  circ(cons(X, Y), Z) => cons(msubst(X, Z), circ(Y, Z)) 
  circ(cons(lift, X), cons(Y, Z)) => cons(Y, circ(X, Z)) 
  circ(cons(lift, X), cons(lift, Y)) => cons(lift, circ(X, Y)) 
  circ(circ(X, Y), Z) => circ(X, circ(Y, Z)) 
  circ(X, id) => X 
  circ(id, X) => X 
  circ(cons(lift, X), circ(cons(lift, Y), Z)) => circ(cons(lift, circ(X, Y)), Z) 
  msubst(X, id) => X 
  msubst(msubst(X, Y), Z) => msubst(X, circ(Y, Z)) 

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

  circ#(cons(X, Y), Z) >? msubst#(X, Z) 
  circ#(cons(X, Y), Z) >? circ#(Y, Z) 
  circ#(cons(lift, X), cons(Y, Z)) >? circ#(X, Z) 
  circ#(cons(lift, X), cons(lift, Y)) >? circ#(X, Y) 
  circ#(circ(X, Y), Z) >? circ#(X, circ(Y, Z)) 
  circ#(circ(X, Y), Z) >? circ#(Y, Z) 
  circ#(cons(lift, X), circ(cons(lift, Y), Z)) >? circ#(cons(lift, circ(X, Y)), Z) 
  circ#(cons(lift, X), circ(cons(lift, Y), Z)) >? circ#(X, Y) 
  msubst#(msubst(X, Y), Z) >? msubst#(X, circ(Y, Z)) 
  msubst#(msubst(X, Y), Z) >? circ#(Y, Z) 
  circ(cons(X, Y), Z) >= cons(msubst(X, Z), circ(Y, Z)) 
  circ(cons(lift, X), cons(Y, Z)) >= cons(Y, circ(X, Z)) 
  circ(cons(lift, X), cons(lift, Y)) >= cons(lift, circ(X, Y)) 
  circ(circ(X, Y), Z) >= circ(X, circ(Y, Z)) 
  circ(X, id) >= X 
  circ(id, X) >= X 
  circ(cons(lift, X), circ(cons(lift, Y), Z)) >= circ(cons(lift, circ(X, Y)), Z) 
  msubst(X, id) >= X 
  msubst(msubst(X, Y), Z) >= msubst(X, circ(Y, Z)) 

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

The following interpretation satisfies the requirements:

  circ = \y0y1.y1 + 2y0 + 2y0y1 
  circ# = \y0y1.2y0 + 2y0y1 
  cons = \y0y1.2 + y1 + 2y0 
  id = 3 
  lift = 0 
  msubst = \y0y1.2y0 + 2y0y1 + 2y1 
  msubst# = \y0y1.1 + y0 + y0y1 

Using this interpretation, the requirements translate to:

  [[circ#(cons(_x0, _x1), _x2)]] = 4 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 4x2 > 1 + x0 + x0x2 = [[msubst#(_x0, _x2)]] 
  [[circ#(cons(_x0, _x1), _x2)]] = 4 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 4x2 > 2x1 + 2x1x2 = [[circ#(_x1, _x2)]] 
  [[circ#(cons(lift, _x0), cons(_x1, _x2))]] = 12 + 2x0x2 + 4x0x1 + 4x2 + 6x0 + 8x1 > 2x0 + 2x0x2 = [[circ#(_x0, _x2)]] 
  [[circ#(cons(lift, _x0), cons(lift, _x1))]] = 12 + 2x0x1 + 4x1 + 6x0 > 2x0 + 2x0x1 = [[circ#(_x0, _x1)]] 
  [[circ#(circ(_x0, _x1), _x2)]] = 2x1 + 2x1x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 >= 2x0 + 2x0x2 + 4x0x1 + 4x0x1x2 = [[circ#(_x0, circ(_x1, _x2))]] 
  [[circ#(circ(_x0, _x1), _x2)]] = 2x1 + 2x1x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 >= 2x1 + 2x1x2 = [[circ#(_x1, _x2)]] 
  [[circ#(cons(lift, _x0), circ(cons(lift, _x1), _x2))]] = 20 + 4x0x1 + 4x0x1x2 + 8x1 + 8x1x2 + 10x0 + 10x0x2 + 20x2 > 4 + 2x1 + 2x1x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 + 4x2 = [[circ#(cons(lift, circ(_x0, _x1)), _x2)]] 
  [[circ#(cons(lift, _x0), circ(cons(lift, _x1), _x2))]] = 20 + 4x0x1 + 4x0x1x2 + 8x1 + 8x1x2 + 10x0 + 10x0x2 + 20x2 > 2x0 + 2x0x1 = [[circ#(_x0, _x1)]] 
  [[msubst#(msubst(_x0, _x1), _x2)]] = 1 + 2x0 + 2x0x1 + 2x0x1x2 + 2x0x2 + 2x1 + 2x1x2 >= 1 + x0 + 2x0x1 + 2x0x1x2 + x0x2 = [[msubst#(_x0, circ(_x1, _x2))]] 
  [[msubst#(msubst(_x0, _x1), _x2)]] = 1 + 2x0 + 2x0x1 + 2x0x1x2 + 2x0x2 + 2x1 + 2x1x2 > 2x1 + 2x1x2 = [[circ#(_x1, _x2)]] 
  [[circ(cons(_x0, _x1), _x2)]] = 4 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 5x2 >= 2 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 5x2 = [[cons(msubst(_x0, _x2), circ(_x1, _x2))]] 
  [[circ(cons(lift, _x0), cons(_x1, _x2))]] = 14 + 2x0x2 + 4x0x1 + 5x2 + 6x0 + 10x1 >= 2 + x2 + 2x0 + 2x0x2 + 2x1 = [[cons(_x1, circ(_x0, _x2))]] 
  [[circ(cons(lift, _x0), cons(lift, _x1))]] = 14 + 2x0x1 + 5x1 + 6x0 >= 2 + x1 + 2x0 + 2x0x1 = [[cons(lift, circ(_x0, _x1))]] 
  [[circ(circ(_x0, _x1), _x2)]] = x2 + 2x1 + 2x1x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 >= x2 + 2x0 + 2x0x2 + 2x1 + 2x1x2 + 4x0x1 + 4x0x1x2 = [[circ(_x0, circ(_x1, _x2))]] 
  [[circ(_x0, id)]] = 3 + 8x0 >= x0 = [[_x0]] 
  [[circ(id, _x0)]] = 6 + 7x0 >= x0 = [[_x0]] 
  [[circ(cons(lift, _x0), circ(cons(lift, _x1), _x2))]] = 24 + 4x0x1 + 4x0x1x2 + 10x0 + 10x0x2 + 10x1 + 10x1x2 + 25x2 >= 4 + 2x1 + 2x1x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 + 5x2 = [[circ(cons(lift, circ(_x0, _x1)), _x2)]] 
  [[msubst(_x0, id)]] = 6 + 8x0 >= x0 = [[_x0]] 
  [[msubst(msubst(_x0, _x1), _x2)]] = 2x2 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 + 4x1 + 4x1x2 >= 2x0 + 2x0x2 + 2x2 + 4x0x1 + 4x0x1x2 + 4x1 + 4x1x2 = [[msubst(_x0, circ(_x1, _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:

  circ#(circ(X, Y), Z) =#> circ#(X, circ(Y, Z))   
  circ#(circ(X, Y), Z) =#> circ#(Y, Z)   
  msubst#(msubst(X, Y), Z) =#> msubst#(X, circ(Y, 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 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 : 0, 1 
    * 2 : 2 

This graph has the following strongly connected components:

  P_2:

    circ#(circ(X, Y), Z) =#> circ#(X, circ(Y, Z))   
    circ#(circ(X, Y), Z) =#> circ#(Y, Z)   

  P_3:

    msubst#(msubst(X, Y), Z) =#> msubst#(X, circ(Y, Z))   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_0, m, f) by (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (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 apply the subterm criterion with the following projection function:

  nu(msubst#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(msubst#(msubst(X, Y), Z)) = msubst(X, Y) |> X = nu(msubst#(X, circ(Y, Z))) 

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.

Thus, the original system is terminating if (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(circ#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(circ#(circ(X, Y), Z)) = circ(X, Y) |> X = nu(circ#(X, circ(Y, Z))) 
  nu(circ#(circ(X, Y), Z)) = circ(X, Y) |> Y = nu(circ#(Y, Z)) 

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.

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.