/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.

  0 : [] --> o 
  add : [o * o] --> o 
  app : [o * o] --> o 
  concat : [o * o] --> o 
  cons : [o * o] --> o 
  false : [] --> o 
  leaf : [] --> o 
  less!6220leaves : [o * o] --> o 
  minus : [o * o] --> o 
  nil : [] --> o 
  quot : [o * o] --> o 
  reverse : [o] --> o 
  s : [o] --> o 
  shuffle : [o] --> o 
  true : [] --> o 

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 
  shuffle(nil) => nil 
  shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) 
  concat(leaf, X) => X 
  concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) 
  less!6220leaves(X, leaf) => false 
  less!6220leaves(leaf, cons(X, Y)) => true 
  less!6220leaves(cons(X, Y), cons(Z, U)) => less!6220leaves(concat(X, Y), concat(Z, U)) 

As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95].  Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows:

  0 : [] --> ze 
  add : [lb * ze] --> ze 
  app : [ze * ze] --> ze 
  concat : [ze * ze] --> ze 
  cons : [ze * ze] --> ze 
  false : [] --> af 
  leaf : [] --> ze 
  less!6220leaves : [ze * ze] --> af 
  minus : [ze * ze] --> ze 
  nil : [] --> ze 
  quot : [ze * ze] --> ze 
  reverse : [ze] --> ze 
  s : [ze] --> ze 
  shuffle : [ze] --> ze 
  true : [] --> af 

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] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y))   
    2] quot#(s(X), s(Y)) =#> minus#(X, Y)   
    3] app#(add(X, Y), Z) =#> app#(Y, Z)   
    4] reverse#(add(X, Y)) =#> app#(reverse(Y), add(X, nil))   
    5] reverse#(add(X, Y)) =#> reverse#(Y)   
    6] shuffle#(add(X, Y)) =#> shuffle#(reverse(Y))   
    7] shuffle#(add(X, Y)) =#> reverse#(Y)   
    8] concat#(cons(X, Y), Z) =#> concat#(Y, Z)   
    9] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> less!6220leaves#(concat(X, Y), concat(Z, U))   
    10] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> concat#(X, Y)   
    11] less!6220leaves#(cons(X, Y), cons(Z, U)) =#> concat#(Z, U)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
    app(nil, X) => X 
    app(add(X, Y), Z) => add(X, app(Y, Z)) 
    reverse(nil) => nil 
    reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 
    shuffle(nil) => nil 
    shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) 
    concat(leaf, X) => X 
    concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) 
    less!6220leaves(X, leaf) => false 
    less!6220leaves(leaf, cons(X, Y)) => true 
    less!6220leaves(cons(X, Y), cons(Z, U)) => less!6220leaves(concat(X, Y), concat(Z, U)) 

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 : 1, 2 
    * 2 : 0 
    * 3 : 3 
    * 4 : 3 
    * 5 : 4, 5 
    * 6 : 6, 7 
    * 7 : 4, 5 
    * 8 : 8 
    * 9 : 9, 10, 11 
    * 10 : 8 
    * 11 : 8 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

    app#(add(X, Y), Z) =#> app#(Y, Z)   

  P_4:

    reverse#(add(X, Y)) =#> reverse#(Y)   

  P_5:

    shuffle#(add(X, Y)) =#> shuffle#(reverse(Y))   

  P_6:

    concat#(cons(X, Y), Z) =#> concat#(Y, Z)   

  P_7:

    less!6220leaves#(cons(X, Y), cons(Z, U)) =#> less!6220leaves#(concat(X, Y), concat(Z, U))   

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), (P_6, R_0, m, f) and (P_7, 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), (P_6, R_0, minimal, formative) and (P_7, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_7, R_0, minimal, formative).

The formative rules of (P_7, R_0) are R_1 ::=

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 
  shuffle(nil) => nil 
  shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) 
  concat(leaf, X) => X 
  concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_7, R_0, minimal, formative) by (P_7, R_1, minimal, formative).

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), (P_6, R_0, minimal, formative) and (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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_7, R_1) are:

  concat(leaf, X) => X 
  concat(cons(X, Y), Z) => cons(X, concat(Y, Z)) 

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

  less!6220leaves#(cons(X, Y), cons(Z, U)) >? less!6220leaves#(concat(X, Y), concat(Z, U)) 
  concat(leaf, X) >= X 
  concat(cons(X, Y), Z) >= cons(X, concat(Y, Z)) 

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

The following interpretation satisfies the requirements:

  concat = \y0y1.y0 + y1 
  cons = \y0y1.3 + y0 + y1 
  leaf = 0 
  less!6220leaves# = \y0y1.3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[less!6220leaves#(cons(_x0, _x1), cons(_x2, _x3))]] = 18 + 3x0 + 3x1 + 3x2 + 3x3 > 3x0 + 3x1 + 3x2 + 3x3 = [[less!6220leaves#(concat(_x0, _x1), concat(_x2, _x3))]] 
  [[concat(leaf, _x0)]] = x0 >= x0 = [[_x0]] 
  [[concat(cons(_x0, _x1), _x2)]] = 3 + x0 + x1 + x2 >= 3 + x0 + x1 + x2 = [[cons(_x0, concat(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_7, R_1) by ({}, R_1).  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), (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(concat#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(concat#(cons(X, Y), Z)) = cons(X, Y) |> Y = nu(concat#(Y, Z)) 

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).

The formative rules of (P_5, R_0) are R_2 ::=

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 
  shuffle(nil) => nil 
  shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) 
  concat(leaf, X) => X 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_5, R_2, minimal, formative).

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_2, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_5, R_2) are:

  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 

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

  shuffle#(add(X, Y)) >? shuffle#(reverse(Y)) 
  app(nil, X) >= X 
  app(add(X, Y), Z) >= add(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(add(X, Y)) >= app(reverse(Y), add(X, nil)) 

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

The following interpretation satisfies the requirements:

  add = \y0y1.2 + y1 
  app = \y0y1.y0 + y1 
  nil = 0 
  reverse = \y0.y0 
  shuffle# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[shuffle#(add(_x0, _x1))]] = 4 + 2x1 > 2x1 = [[shuffle#(reverse(_x1))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(add(_x0, _x1), _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[add(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(add(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[app(reverse(_x1), add(_x0, nil))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_5, R_2) by ({}, R_2).  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(reverse#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(reverse#(add(X, Y))) = add(X, Y) |> Y = nu(reverse#(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(app#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(app#(add(X, Y), Z)) = add(X, Y) |> Y = nu(app#(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).

The formative rules of (P_2, R_0) are exactly R_2:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(add(X, Y)) => app(reverse(Y), add(X, nil)) 
  shuffle(nil) => nil 
  shuffle(add(X, Y)) => add(X, shuffle(reverse(Y))) 
  concat(leaf, X) => X 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_2, minimal, formative).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_2, R_2) are:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 

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

  quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  minus = \y0y1.y0 
  quot# = \y0y1.3y0 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[quot#(s(_x0), s(_x1))]] = 9 + 9x0 > 3x0 = [[quot#(minus(_x0, _x1), s(_x1))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 3 + 3x0 >= x0 = [[minus(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_2) by ({}, R_2).  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 +++

[FuhGieParSchSwi11]  C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski.  Proving Termination by Dependency Pairs and Inductive Theorem Proving.  In volume 47(2) of Journal of Automated Reasoning.  133--160, 2011.
[Gra95]  B. Gramlich.  Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems.  In volume 24(1-2) of Fundamentae Informaticae.  3--23, 1995.
[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.