/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.

  and : [o * o] --> o 
  apply : [o * o] --> o 
  cons : [o * o] --> o 
  eq : [o * o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  lambda : [o * o] --> o 
  nil : [] --> o 
  ren : [o * o * o] --> o 
  true : [] --> o 
  var : [o] --> o 

  and(true, X) => X 
  and(false, X) => false 
  eq(nil, nil) => true 
  eq(cons(X, Y), nil) => false 
  eq(nil, cons(X, Y)) => false 
  eq(cons(X, Y), cons(Z, U)) => and(eq(X, Z), eq(Y, U)) 
  eq(var(X), var(Y)) => eq(X, Y) 
  eq(var(X), apply(Y, Z)) => false 
  eq(var(X), lambda(Y, Z)) => false 
  eq(apply(X, Y), var(Z)) => false 
  eq(apply(X, Y), apply(Z, U)) => and(eq(X, Z), eq(Y, U)) 
  eq(apply(X, Y), lambda(Z, X)) => false 
  eq(lambda(X, Y), var(Z)) => false 
  eq(lambda(X, Y), apply(Y, Z)) => false 
  eq(lambda(X, Y), lambda(Z, U)) => and(eq(X, Z), eq(Y, U)) 
  if(true, var(X), var(Y)) => var(X) 
  if(false, var(X), var(Y)) => var(Y) 
  ren(var(X), var(Y), var(Z)) => if(eq(X, Z), var(Y), var(Z)) 
  ren(X, Y, apply(Z, U)) => apply(ren(X, Y, Z), ren(X, Y, U)) 
  ren(X, Y, lambda(Z, U)) => lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) 

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, all):

  Dependency Pairs P_0:

    0] eq#(cons(X, Y), cons(Z, U)) =#> and#(eq(X, Z), eq(Y, U))   
    1] eq#(cons(X, Y), cons(Z, U)) =#> eq#(X, Z)   
    2] eq#(cons(X, Y), cons(Z, U)) =#> eq#(Y, U)   
    3] eq#(var(X), var(Y)) =#> eq#(X, Y)   
    4] eq#(apply(X, Y), apply(Z, U)) =#> and#(eq(X, Z), eq(Y, U))   
    5] eq#(apply(X, Y), apply(Z, U)) =#> eq#(X, Z)   
    6] eq#(apply(X, Y), apply(Z, U)) =#> eq#(Y, U)   
    7] eq#(lambda(X, Y), lambda(Z, U)) =#> and#(eq(X, Z), eq(Y, U))   
    8] eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(X, Z)   
    9] eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(Y, U)   
    10] ren#(var(X), var(Y), var(Z)) =#> if#(eq(X, Z), var(Y), var(Z))   
    11] ren#(var(X), var(Y), var(Z)) =#> eq#(X, Z)   
    12] ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, Z)   
    13] ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, U)   
    14] ren#(X, Y, lambda(Z, U)) =#> ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))   
    15] ren#(X, Y, lambda(Z, U)) =#> ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)   

  Rules R_0:

    and(true, X) => X 
    and(false, X) => false 
    eq(nil, nil) => true 
    eq(cons(X, Y), nil) => false 
    eq(nil, cons(X, Y)) => false 
    eq(cons(X, Y), cons(Z, U)) => and(eq(X, Z), eq(Y, U)) 
    eq(var(X), var(Y)) => eq(X, Y) 
    eq(var(X), apply(Y, Z)) => false 
    eq(var(X), lambda(Y, Z)) => false 
    eq(apply(X, Y), var(Z)) => false 
    eq(apply(X, Y), apply(Z, U)) => and(eq(X, Z), eq(Y, U)) 
    eq(apply(X, Y), lambda(Z, X)) => false 
    eq(lambda(X, Y), var(Z)) => false 
    eq(lambda(X, Y), apply(Y, Z)) => false 
    eq(lambda(X, Y), lambda(Z, U)) => and(eq(X, Z), eq(Y, U)) 
    if(true, var(X), var(Y)) => var(X) 
    if(false, var(X), var(Y)) => var(Y) 
    ren(var(X), var(Y), var(Z)) => if(eq(X, Z), var(Y), var(Z)) 
    ren(X, Y, apply(Z, U)) => apply(ren(X, Y, Z), ren(X, Y, U)) 
    ren(X, Y, lambda(Z, U)) => lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) 

Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, all).

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, 7, 8, 9 
    * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 4 :  
    * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 7 :  
    * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 9 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 10 :  
    * 11 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 12 : 10, 11, 12, 13, 14, 15 
    * 13 : 10, 11, 12, 13, 14, 15 
    * 14 : 10, 11, 12, 13, 14, 15 
    * 15 : 10, 11, 12, 13, 14, 15 

This graph has the following strongly connected components:

  P_1:

    eq#(cons(X, Y), cons(Z, U)) =#> eq#(X, Z)   
    eq#(cons(X, Y), cons(Z, U)) =#> eq#(Y, U)   
    eq#(var(X), var(Y)) =#> eq#(X, Y)   
    eq#(apply(X, Y), apply(Z, U)) =#> eq#(X, Z)   
    eq#(apply(X, Y), apply(Z, U)) =#> eq#(Y, U)   
    eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(X, Z)   
    eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(Y, U)   

  P_2:

    ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, Z)   
    ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, U)   
    ren#(X, Y, lambda(Z, U)) =#> ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))   
    ren#(X, Y, lambda(Z, U)) =#> ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), 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) and (P_2, R_0, m, f).

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

We consider the dependency pair problem (P_2, R_0, minimal, all).

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:

  ren#(X, Y, apply(Z, U)) >? ren#(X, Y, Z) 
  ren#(X, Y, apply(Z, U)) >? ren#(X, Y, U) 
  ren#(X, Y, lambda(Z, U)) >? ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)) 
  ren#(X, Y, lambda(Z, U)) >? ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U) 
  and(true, X) >= X 
  and(false, X) >= false 
  eq(nil, nil) >= true 
  eq(cons(X, Y), nil) >= false 
  eq(nil, cons(X, Y)) >= false 
  eq(cons(X, Y), cons(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  eq(var(X), var(Y)) >= eq(X, Y) 
  eq(var(X), apply(Y, Z)) >= false 
  eq(var(X), lambda(Y, Z)) >= false 
  eq(apply(X, Y), var(Z)) >= false 
  eq(apply(X, Y), apply(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  eq(apply(X, Y), lambda(Z, X)) >= false 
  eq(lambda(X, Y), var(Z)) >= false 
  eq(lambda(X, Y), apply(Y, Z)) >= false 
  eq(lambda(X, Y), lambda(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  if(true, var(X), var(Y)) >= var(X) 
  if(false, var(X), var(Y)) >= var(Y) 
  ren(var(X), var(Y), var(Z)) >= if(eq(X, Z), var(Y), var(Z)) 
  ren(X, Y, apply(Z, U)) >= apply(ren(X, Y, Z), ren(X, Y, U)) 
  ren(X, Y, lambda(Z, U)) >= lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) 

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

We consider usable_rules with respect to the following argument filtering:

  if(x_1,x_2,x_3) = if(x_2x_3) 

This leaves the following ordering requirements:

  ren#(X, Y, apply(Z, U)) >= ren#(X, Y, Z) 
  ren#(X, Y, apply(Z, U)) > ren#(X, Y, U) 
  ren#(X, Y, lambda(Z, U)) >= ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)) 
  ren#(X, Y, lambda(Z, U)) >= ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U) 
  if(true, var(X), var(Y)) >= var(X) 
  if(false, var(X), var(Y)) >= var(Y) 
  ren(var(X), var(Y), var(Z)) >= if(eq(X, Z), var(Y), var(Z)) 
  ren(X, Y, apply(Z, U)) >= apply(ren(X, Y, Z), ren(X, Y, U)) 
  ren(X, Y, lambda(Z, U)) >= lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) 

The following interpretation satisfies the requirements:

  and = \y0y1.0 
  apply = \y0y1.1 + y0 + y1 
  cons = \y0y1.0 
  eq = \y0y1.0 
  false = 0 
  if = \y0y1y2.0 
  lambda = \y0y1.2y1 
  nil = 0 
  ren = \y0y1y2.y2 
  ren# = \y0y1y2.2y1 + 2y2 
  true = 0 
  var = \y0.0 

Using this interpretation, the requirements translate to:

  [[ren#(_x0, _x1, apply(_x2, _x3))]] = 2 + 2x1 + 2x2 + 2x3 > 2x1 + 2x2 = [[ren#(_x0, _x1, _x2)]] 
  [[ren#(_x0, _x1, apply(_x2, _x3))]] = 2 + 2x1 + 2x2 + 2x3 > 2x1 + 2x3 = [[ren#(_x0, _x1, _x3)]] 
  [[ren#(_x0, _x1, lambda(_x2, _x3))]] = 2x1 + 4x3 >= 2x1 + 2x3 = [[ren#(_x0, _x1, ren(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3))]] 
  [[ren#(_x0, _x1, lambda(_x2, _x3))]] = 2x1 + 4x3 >= 2x3 = [[ren#(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3)]] 
  [[if(true, var(_x0), var(_x1))]] = 0 >= 0 = [[var(_x0)]] 
  [[if(false, var(_x0), var(_x1))]] = 0 >= 0 = [[var(_x1)]] 
  [[ren(var(_x0), var(_x1), var(_x2))]] = 0 >= 0 = [[if(eq(_x0, _x2), var(_x1), var(_x2))]] 
  [[ren(_x0, _x1, apply(_x2, _x3))]] = 1 + x2 + x3 >= 1 + x2 + x3 = [[apply(ren(_x0, _x1, _x2), ren(_x0, _x1, _x3))]] 
  [[ren(_x0, _x1, lambda(_x2, _x3))]] = 2x3 >= 2x3 = [[lambda(var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), ren(_x0, _x1, ren(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, minimal, all) by (P_3, R_0, minimal, all), where P_3 consists of:

  ren#(X, Y, lambda(Z, U)) =#> ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))   
  ren#(X, Y, lambda(Z, U)) =#> ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, all) and (P_3, R_0, minimal, all) is finite.

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

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:

  ren#(X, Y, lambda(Z, U)) >? ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)) 
  ren#(X, Y, lambda(Z, U)) >? ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U) 
  and(true, X) >= X 
  and(false, X) >= false 
  eq(nil, nil) >= true 
  eq(cons(X, Y), nil) >= false 
  eq(nil, cons(X, Y)) >= false 
  eq(cons(X, Y), cons(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  eq(var(X), var(Y)) >= eq(X, Y) 
  eq(var(X), apply(Y, Z)) >= false 
  eq(var(X), lambda(Y, Z)) >= false 
  eq(apply(X, Y), var(Z)) >= false 
  eq(apply(X, Y), apply(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  eq(apply(X, Y), lambda(Z, X)) >= false 
  eq(lambda(X, Y), var(Z)) >= false 
  eq(lambda(X, Y), apply(Y, Z)) >= false 
  eq(lambda(X, Y), lambda(Z, U)) >= and(eq(X, Z), eq(Y, U)) 
  if(true, var(X), var(Y)) >= var(X) 
  if(false, var(X), var(Y)) >= var(Y) 
  ren(var(X), var(Y), var(Z)) >= if(eq(X, Z), var(Y), var(Z)) 
  ren(X, Y, apply(Z, U)) >= apply(ren(X, Y, Z), ren(X, Y, U)) 
  ren(X, Y, lambda(Z, U)) >= lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.y1 
  apply = \y0y1.0 
  cons = \y0y1.0 
  eq = \y0y1.0 
  false = 0 
  if = \y0y1y2.0 
  lambda = \y0y1.2 + 2y1 
  nil = 0 
  ren = \y0y1y2.y2 
  ren# = \y0y1y2.y2 + 2y1 
  true = 0 
  var = \y0.0 

Using this interpretation, the requirements translate to:

  [[ren#(_x0, _x1, lambda(_x2, _x3))]] = 2 + 2x1 + 2x3 > x3 + 2x1 = [[ren#(_x0, _x1, ren(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3))]] 
  [[ren#(_x0, _x1, lambda(_x2, _x3))]] = 2 + 2x1 + 2x3 > x3 = [[ren#(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3)]] 
  [[and(true, _x0)]] = x0 >= x0 = [[_x0]] 
  [[and(false, _x0)]] = x0 >= 0 = [[false]] 
  [[eq(nil, nil)]] = 0 >= 0 = [[true]] 
  [[eq(cons(_x0, _x1), nil)]] = 0 >= 0 = [[false]] 
  [[eq(nil, cons(_x0, _x1))]] = 0 >= 0 = [[false]] 
  [[eq(cons(_x0, _x1), cons(_x2, _x3))]] = 0 >= 0 = [[and(eq(_x0, _x2), eq(_x1, _x3))]] 
  [[eq(var(_x0), var(_x1))]] = 0 >= 0 = [[eq(_x0, _x1)]] 
  [[eq(var(_x0), apply(_x1, _x2))]] = 0 >= 0 = [[false]] 
  [[eq(var(_x0), lambda(_x1, _x2))]] = 0 >= 0 = [[false]] 
  [[eq(apply(_x0, _x1), var(_x2))]] = 0 >= 0 = [[false]] 
  [[eq(apply(_x0, _x1), apply(_x2, _x3))]] = 0 >= 0 = [[and(eq(_x0, _x2), eq(_x1, _x3))]] 
  [[eq(apply(_x0, _x1), lambda(_x2, _x0))]] = 0 >= 0 = [[false]] 
  [[eq(lambda(_x0, _x1), var(_x2))]] = 0 >= 0 = [[false]] 
  [[eq(lambda(_x0, _x1), apply(_x1, _x2))]] = 0 >= 0 = [[false]] 
  [[eq(lambda(_x0, _x1), lambda(_x2, _x3))]] = 0 >= 0 = [[and(eq(_x0, _x2), eq(_x1, _x3))]] 
  [[if(true, var(_x0), var(_x1))]] = 0 >= 0 = [[var(_x0)]] 
  [[if(false, var(_x0), var(_x1))]] = 0 >= 0 = [[var(_x1)]] 
  [[ren(var(_x0), var(_x1), var(_x2))]] = 0 >= 0 = [[if(eq(_x0, _x2), var(_x1), var(_x2))]] 
  [[ren(_x0, _x1, apply(_x2, _x3))]] = 0 >= 0 = [[apply(ren(_x0, _x1, _x2), ren(_x0, _x1, _x3))]] 
  [[ren(_x0, _x1, lambda(_x2, _x3))]] = 2 + 2x3 >= 2 + 2x3 = [[lambda(var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), ren(_x0, _x1, ren(_x2, var(cons(_x0, cons(_x1, cons(lambda(_x2, _x3), nil)))), _x3)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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, all) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, all).

We apply the subterm criterion with the following projection function:

  nu(eq#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(eq#(cons(X, Y), cons(Z, U))) = cons(X, Y) |> X = nu(eq#(X, Z)) 
  nu(eq#(cons(X, Y), cons(Z, U))) = cons(X, Y) |> Y = nu(eq#(Y, U)) 
  nu(eq#(var(X), var(Y))) = var(X) |> X = nu(eq#(X, Y)) 
  nu(eq#(apply(X, Y), apply(Z, U))) = apply(X, Y) |> X = nu(eq#(X, Z)) 
  nu(eq#(apply(X, Y), apply(Z, U))) = apply(X, Y) |> Y = nu(eq#(Y, U)) 
  nu(eq#(lambda(X, Y), lambda(Z, U))) = lambda(X, Y) |> X = nu(eq#(X, Z)) 
  nu(eq#(lambda(X, Y), lambda(Z, U))) = lambda(X, Y) |> Y = nu(eq#(Y, U)) 

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.