/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.

  X : [] --> o 
  active : [o] --> o 
  c : [] --> o 
  check : [o] --> o 
  f : [o] --> o 
  found : [o] --> o 
  mark : [o] --> o 
  match : [o * o] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  start : [o] --> o 
  top : [o] --> o 

  active(f(Y)) => mark(Y) 
  top(active(c)) => top(mark(c)) 
  top(mark(Y)) => top(check(Y)) 
  check(f(Y)) => f(check(Y)) 
  check(Y) => start(match(f(X), Y)) 
  match(f(Y), f(Z)) => f(match(Y, Z)) 
  match(X, Y) => proper(Y) 
  proper(c) => ok(c) 
  proper(f(Y)) => f(proper(Y)) 
  f(ok(Y)) => ok(f(Y)) 
  start(ok(Y)) => found(Y) 
  f(found(Y)) => found(f(Y)) 
  top(found(Y)) => top(active(Y)) 
  active(f(Y)) => f(active(Y)) 
  f(mark(Y)) => mark(f(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] top#(active(c)) =#> top#(mark(c))   
    1] top#(mark(Y)) =#> top#(check(Y))   
    2] top#(mark(Y)) =#> check#(Y)   
    3] check#(f(Y)) =#> f#(check(Y))   
    4] check#(f(Y)) =#> check#(Y)   
    5] check#(Y) =#> start#(match(f(X), Y))   
    6] check#(Y) =#> match#(f(X), Y)   
    7] check#(Y) =#> f#(X)   
    8] match#(f(Y), f(Z)) =#> f#(match(Y, Z))   
    9] match#(f(Y), f(Z)) =#> match#(Y, Z)   
    10] match#(X, Y) =#> proper#(Y)   
    11] proper#(f(Y)) =#> f#(proper(Y))   
    12] proper#(f(Y)) =#> proper#(Y)   
    13] f#(ok(Y)) =#> f#(Y)   
    14] f#(found(Y)) =#> f#(Y)   
    15] top#(found(Y)) =#> top#(active(Y))   
    16] top#(found(Y)) =#> active#(Y)   
    17] active#(f(Y)) =#> f#(active(Y))   
    18] active#(f(Y)) =#> active#(Y)   
    19] f#(mark(Y)) =#> f#(Y)   

  Rules R_0:

    active(f(Y)) => mark(Y) 
    top(active(c)) => top(mark(c)) 
    top(mark(Y)) => top(check(Y)) 
    check(f(Y)) => f(check(Y)) 
    check(Y) => start(match(f(X), Y)) 
    match(f(Y), f(Z)) => f(match(Y, Z)) 
    match(X, Y) => proper(Y) 
    proper(c) => ok(c) 
    proper(f(Y)) => f(proper(Y)) 
    f(ok(Y)) => ok(f(Y)) 
    start(ok(Y)) => found(Y) 
    f(found(Y)) => found(f(Y)) 
    top(found(Y)) => top(active(Y)) 
    active(f(Y)) => f(active(Y)) 
    f(mark(Y)) => mark(f(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, 2 
    * 1 : 1, 2, 15, 16 
    * 2 : 3, 4, 5, 6, 7 
    * 3 : 13, 14, 19 
    * 4 : 3, 4, 5, 6, 7 
    * 5 :  
    * 6 : 8, 9 
    * 7 :  
    * 8 : 13, 14, 19 
    * 9 : 8, 9, 10 
    * 10 : 11, 12 
    * 11 : 13, 14, 19 
    * 12 : 11, 12 
    * 13 : 13, 14, 19 
    * 14 : 13, 14, 19 
    * 15 : 0, 1, 2, 15, 16 
    * 16 : 17, 18 
    * 17 : 13, 14, 19 
    * 18 : 17, 18 
    * 19 : 13, 14, 19 

This graph has the following strongly connected components:

  P_1:

    top#(active(c)) =#> top#(mark(c))   
    top#(mark(Y)) =#> top#(check(Y))   
    top#(found(Y)) =#> top#(active(Y))   

  P_2:

    check#(f(Y)) =#> check#(Y)   

  P_3:

    match#(f(Y), f(Z)) =#> match#(Y, Z)   

  P_4:

    proper#(f(Y)) =#> proper#(Y)   

  P_5:

    f#(ok(Y)) =#> f#(Y)   
    f#(found(Y)) =#> f#(Y)   
    f#(mark(Y)) =#> f#(Y)   

  P_6:

    active#(f(Y)) =#> active#(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), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f), (P_5, R_0, m, f) and (P_6, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(active#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(active#(f(Y))) = f(Y) |> Y = nu(active#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, 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 each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(ok(Y))) = ok(Y) |> Y = nu(f#(Y)) 
  nu(f#(found(Y))) = found(Y) |> Y = nu(f#(Y)) 
  nu(f#(mark(Y))) = mark(Y) |> Y = nu(f#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, 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 each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(f(Y))) = f(Y) |> Y = nu(proper#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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 each of (P_1, R_0, minimal, formative), (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(match#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(match#(f(Y), f(Z))) = f(Y) |> Y = nu(match#(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 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(check#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(check#(f(Y))) = f(Y) |> Y = nu(check#(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).

The formative rules of (P_1, R_0) are R_1 ::=

  active(f(Y)) => mark(Y) 
  check(f(Y)) => f(check(Y)) 
  check(Y) => start(match(f(X), Y)) 
  match(f(Y), f(Z)) => f(match(Y, Z)) 
  match(X, Y) => proper(Y) 
  proper(c) => ok(c) 
  proper(f(Y)) => f(proper(Y)) 
  f(ok(Y)) => ok(f(Y)) 
  start(ok(Y)) => found(Y) 
  f(found(Y)) => found(f(Y)) 
  active(f(Y)) => f(active(Y)) 
  f(mark(Y)) => mark(f(Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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:

  top#(active(c)) >? top#(mark(c)) 
  top#(mark(Y)) >? top#(check(Y)) 
  top#(found(Y)) >? top#(active(Y)) 
  active(f(Y)) >= mark(Y) 
  check(f(Y)) >= f(check(Y)) 
  check(Y) >= start(match(f(X), Y)) 
  match(f(Y), f(Z)) >= f(match(Y, Z)) 
  match(X, Y) >= proper(Y) 
  proper(c) >= ok(c) 
  proper(f(Y)) >= f(proper(Y)) 
  f(ok(Y)) >= ok(f(Y)) 
  start(ok(Y)) >= found(Y) 
  f(found(Y)) >= found(f(Y)) 
  active(f(Y)) >= f(active(Y)) 
  f(mark(Y)) >= mark(f(Y)) 

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

The following interpretation satisfies the requirements:

  X = 2 
  active = \y0.2y0 
  c = 2 
  check = \y0.0 
  f = \y0.0 
  found = \y0.2y0 
  mark = \y0.0 
  match = \y0y1.y0 
  ok = \y0.y0 
  proper = \y0.2 
  start = \y0.2y0 
  top# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[top#(active(c))]] = 4 > 0 = [[top#(mark(c))]] 
  [[top#(mark(_x0))]] = 0 >= 0 = [[top#(check(_x0))]] 
  [[top#(found(_x0))]] = 2x0 >= 2x0 = [[top#(active(_x0))]] 
  [[active(f(_x0))]] = 0 >= 0 = [[mark(_x0)]] 
  [[check(f(_x0))]] = 0 >= 0 = [[f(check(_x0))]] 
  [[check(_x0)]] = 0 >= 0 = [[start(match(f(X), _x0))]] 
  [[match(f(_x0), f(_x1))]] = 0 >= 0 = [[f(match(_x0, _x1))]] 
  [[match(X, _x0)]] = 2 >= 2 = [[proper(_x0)]] 
  [[proper(c)]] = 2 >= 2 = [[ok(c)]] 
  [[proper(f(_x0))]] = 2 >= 0 = [[f(proper(_x0))]] 
  [[f(ok(_x0))]] = 0 >= 0 = [[ok(f(_x0))]] 
  [[start(ok(_x0))]] = 2x0 >= 2x0 = [[found(_x0)]] 
  [[f(found(_x0))]] = 0 >= 0 = [[found(f(_x0))]] 
  [[active(f(_x0))]] = 0 >= 0 = [[f(active(_x0))]] 
  [[f(mark(_x0))]] = 0 >= 0 = [[mark(f(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_7, R_1, minimal, formative), where P_7 consists of:

  top#(mark(Y)) =#> top#(check(Y))   
  top#(found(Y)) =#> top#(active(Y))   

Thus, the original system is terminating if (P_7, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_7, R_1, 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:

  top#(mark(Y)) >? top#(check(Y)) 
  top#(found(Y)) >? top#(active(Y)) 
  active(f(Y)) >= mark(Y) 
  check(f(Y)) >= f(check(Y)) 
  check(Y) >= start(match(f(X), Y)) 
  match(f(Y), f(Z)) >= f(match(Y, Z)) 
  match(X, Y) >= proper(Y) 
  proper(c) >= ok(c) 
  proper(f(Y)) >= f(proper(Y)) 
  f(ok(Y)) >= ok(f(Y)) 
  start(ok(Y)) >= found(Y) 
  f(found(Y)) >= found(f(Y)) 
  active(f(Y)) >= f(active(Y)) 
  f(mark(Y)) >= mark(f(Y)) 

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

The following interpretation satisfies the requirements:

  X = 0 
  active = \y0.y0 
  c = 0 
  check = \y0.y0 
  f = \y0.1 + y0 
  found = \y0.y0 
  mark = \y0.1 + y0 
  match = \y0y1.y1 
  ok = \y0.y0 
  proper = \y0.y0 
  start = \y0.y0 
  top# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[top#(mark(_x0))]] = 2 + 2x0 > 2x0 = [[top#(check(_x0))]] 
  [[top#(found(_x0))]] = 2x0 >= 2x0 = [[top#(active(_x0))]] 
  [[active(f(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(_x0)]] 
  [[check(f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(check(_x0))]] 
  [[check(_x0)]] = x0 >= x0 = [[start(match(f(X), _x0))]] 
  [[match(f(_x0), f(_x1))]] = 1 + x1 >= 1 + x1 = [[f(match(_x0, _x1))]] 
  [[match(X, _x0)]] = x0 >= x0 = [[proper(_x0)]] 
  [[proper(c)]] = 0 >= 0 = [[ok(c)]] 
  [[proper(f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(proper(_x0))]] 
  [[f(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(f(_x0))]] 
  [[start(ok(_x0))]] = x0 >= x0 = [[found(_x0)]] 
  [[f(found(_x0))]] = 1 + x0 >= 1 + x0 = [[found(f(_x0))]] 
  [[active(f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(active(_x0))]] 
  [[f(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(f(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_1, minimal, formative) by (P_8, R_1, minimal, formative), where P_8 consists of:

  top#(found(Y)) =#> top#(active(Y))   

Thus, the original system is terminating if (P_8, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_8, R_1, minimal, formative).

The formative rules of (P_8, R_1) are R_2 ::=

  check(f(Y)) => f(check(Y)) 
  check(Y) => start(match(f(X), Y)) 
  match(f(Y), f(Z)) => f(match(Y, Z)) 
  match(X, Y) => proper(Y) 
  proper(c) => ok(c) 
  proper(f(Y)) => f(proper(Y)) 
  f(ok(Y)) => ok(f(Y)) 
  start(ok(Y)) => found(Y) 
  f(found(Y)) => found(f(Y)) 
  active(f(Y)) => f(active(Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_8, R_1, minimal, formative) by (P_8, R_2, minimal, formative).

Thus, the original system is terminating if (P_8, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_8, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_8, R_2) are:

  f(ok(Y)) => ok(f(Y)) 
  f(found(Y)) => found(f(Y)) 
  active(f(Y)) => f(active(Y)) 

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

  top#(found(Y)) >? top#(active(Y)) 
  f(ok(Y)) >= ok(f(Y)) 
  f(found(Y)) >= found(f(Y)) 
  active(f(Y)) >= f(active(Y)) 

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

The following interpretation satisfies the requirements:

  active = \y0.0 
  f = \y0.2y0 
  found = \y0.3 + 3y0 
  ok = \y0.2 
  top# = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[top#(found(_x0))]] = 9 + 9x0 > 0 = [[top#(active(_x0))]] 
  [[f(ok(_x0))]] = 4 >= 2 = [[ok(f(_x0))]] 
  [[f(found(_x0))]] = 6 + 6x0 >= 3 + 6x0 = [[found(f(_x0))]] 
  [[active(f(_x0))]] = 0 >= 0 = [[f(active(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_8, R_2) by ({}, R_2).  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.