/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.

  active : [o] --> o 
  c : [] --> o 
  f : [o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  mark : [o] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  top : [o] --> o 
  true : [] --> o 

  active(f(X)) => mark(if(X, c, f(true))) 
  active(if(true, X, Y)) => mark(X) 
  active(if(false, X, Y)) => mark(Y) 
  active(f(X)) => f(active(X)) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(if(X, Y, Z)) => if(X, active(Y), Z) 
  f(mark(X)) => mark(f(X)) 
  if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
  if(X, mark(Y), Z) => mark(if(X, Y, Z)) 
  proper(f(X)) => f(proper(X)) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(c) => ok(c) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  f(ok(X)) => ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(X)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  active(f(X)) >? mark(if(X, c, f(true))) 
  active(if(true, X, Y)) >? mark(X) 
  active(if(false, X, Y)) >? mark(Y) 
  active(f(X)) >? f(active(X)) 
  active(if(X, Y, Z)) >? if(active(X), Y, Z) 
  active(if(X, Y, Z)) >? if(X, active(Y), Z) 
  f(mark(X)) >? mark(f(X)) 
  if(mark(X), Y, Z) >? mark(if(X, Y, Z)) 
  if(X, mark(Y), Z) >? mark(if(X, Y, Z)) 
  proper(f(X)) >? f(proper(X)) 
  proper(if(X, Y, Z)) >? if(proper(X), proper(Y), proper(Z)) 
  proper(c) >? ok(c) 
  proper(true) >? ok(true) 
  proper(false) >? ok(false) 
  f(ok(X)) >? ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) >? ok(if(X, Y, Z)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  c = 0 
  f = \y0.2y0 
  false = 1 
  if = \y0y1y2.y2 + 2y0 + 2y1 
  mark = \y0.y0 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[active(f(_x0))]] = 2x0 >= 2x0 = [[mark(if(_x0, c, f(true)))]] 
  [[active(if(true, _x0, _x1))]] = x1 + 2x0 >= x0 = [[mark(_x0)]] 
  [[active(if(false, _x0, _x1))]] = 2 + x1 + 2x0 > x1 = [[mark(_x1)]] 
  [[active(f(_x0))]] = 2x0 >= 2x0 = [[f(active(_x0))]] 
  [[active(if(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[if(active(_x0), _x1, _x2)]] 
  [[active(if(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[if(_x0, active(_x1), _x2)]] 
  [[f(mark(_x0))]] = 2x0 >= 2x0 = [[mark(f(_x0))]] 
  [[if(mark(_x0), _x1, _x2)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[mark(if(_x0, _x1, _x2))]] 
  [[if(_x0, mark(_x1), _x2)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[mark(if(_x0, _x1, _x2))]] 
  [[proper(f(_x0))]] = 2x0 >= 2x0 = [[f(proper(_x0))]] 
  [[proper(if(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[if(proper(_x0), proper(_x1), proper(_x2))]] 
  [[proper(c)]] = 0 >= 0 = [[ok(c)]] 
  [[proper(true)]] = 0 >= 0 = [[ok(true)]] 
  [[proper(false)]] = 1 >= 1 = [[ok(false)]] 
  [[f(ok(_x0))]] = 2x0 >= 2x0 = [[ok(f(_x0))]] 
  [[if(ok(_x0), ok(_x1), ok(_x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[ok(if(_x0, _x1, _x2))]] 
  [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(if(false, X, Y)) => mark(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] active#(f(X)) =#> if#(X, c, f(true))   
    1] active#(f(X)) =#> f#(true)   
    2] active#(f(X)) =#> f#(active(X))   
    3] active#(f(X)) =#> active#(X)   
    4] active#(if(X, Y, Z)) =#> if#(active(X), Y, Z)   
    5] active#(if(X, Y, Z)) =#> active#(X)   
    6] active#(if(X, Y, Z)) =#> if#(X, active(Y), Z)   
    7] active#(if(X, Y, Z)) =#> active#(Y)   
    8] f#(mark(X)) =#> f#(X)   
    9] if#(mark(X), Y, Z) =#> if#(X, Y, Z)   
    10] if#(X, mark(Y), Z) =#> if#(X, Y, Z)   
    11] proper#(f(X)) =#> f#(proper(X))   
    12] proper#(f(X)) =#> proper#(X)   
    13] proper#(if(X, Y, Z)) =#> if#(proper(X), proper(Y), proper(Z))   
    14] proper#(if(X, Y, Z)) =#> proper#(X)   
    15] proper#(if(X, Y, Z)) =#> proper#(Y)   
    16] proper#(if(X, Y, Z)) =#> proper#(Z)   
    17] f#(ok(X)) =#> f#(X)   
    18] if#(ok(X), ok(Y), ok(Z)) =#> if#(X, Y, Z)   
    19] top#(mark(X)) =#> top#(proper(X))   
    20] top#(mark(X)) =#> proper#(X)   
    21] top#(ok(X)) =#> top#(active(X))   
    22] top#(ok(X)) =#> active#(X)   

  Rules R_0:

    active(f(X)) => mark(if(X, c, f(true))) 
    active(if(true, X, Y)) => mark(X) 
    active(f(X)) => f(active(X)) 
    active(if(X, Y, Z)) => if(active(X), Y, Z) 
    active(if(X, Y, Z)) => if(X, active(Y), Z) 
    f(mark(X)) => mark(f(X)) 
    if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
    if(X, mark(Y), Z) => mark(if(X, Y, Z)) 
    proper(f(X)) => f(proper(X)) 
    proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
    proper(c) => ok(c) 
    proper(true) => ok(true) 
    proper(false) => ok(false) 
    f(ok(X)) => ok(f(X)) 
    if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
    top(mark(X)) => top(proper(X)) 
    top(ok(X)) => top(active(X)) 

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 : 9 
    * 1 :  
    * 2 : 8, 17 
    * 3 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 4 : 9, 10, 18 
    * 5 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 6 : 9, 10, 18 
    * 7 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 8 : 8, 17 
    * 9 : 9, 10, 18 
    * 10 : 9, 10, 18 
    * 11 : 8, 17 
    * 12 : 11, 12, 13, 14, 15, 16 
    * 13 : 9, 10, 18 
    * 14 : 11, 12, 13, 14, 15, 16 
    * 15 : 11, 12, 13, 14, 15, 16 
    * 16 : 11, 12, 13, 14, 15, 16 
    * 17 : 8, 17 
    * 18 : 9, 10, 18 
    * 19 : 19, 20, 21, 22 
    * 20 : 11, 12, 13, 14, 15, 16 
    * 21 : 19, 20, 21, 22 
    * 22 : 0, 1, 2, 3, 4, 5, 6, 7 

This graph has the following strongly connected components:

  P_1:

    active#(f(X)) =#> active#(X)   
    active#(if(X, Y, Z)) =#> active#(X)   
    active#(if(X, Y, Z)) =#> active#(Y)   

  P_2:

    f#(mark(X)) =#> f#(X)   
    f#(ok(X)) =#> f#(X)   

  P_3:

    if#(mark(X), Y, Z) =#> if#(X, Y, Z)   
    if#(X, mark(Y), Z) =#> if#(X, Y, Z)   
    if#(ok(X), ok(Y), ok(Z)) =#> if#(X, Y, Z)   

  P_4:

    proper#(f(X)) =#> proper#(X)   
    proper#(if(X, Y, Z)) =#> proper#(X)   
    proper#(if(X, Y, Z)) =#> proper#(Y)   
    proper#(if(X, Y, Z)) =#> proper#(Z)   

  P_5:

    top#(mark(X)) =#> top#(proper(X))   
    top#(ok(X)) =#> top#(active(X))   

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) and (P_5, 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) 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_1 ::=

  active(f(X)) => mark(if(X, c, f(true))) 
  active(if(true, X, Y)) => mark(X) 
  active(f(X)) => f(active(X)) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(if(X, Y, Z)) => if(X, active(Y), Z) 
  f(mark(X)) => mark(f(X)) 
  if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
  if(X, mark(Y), Z) => mark(if(X, Y, Z)) 
  proper(f(X)) => f(proper(X)) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(c) => ok(c) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  f(ok(X)) => ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_5, 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) and (P_5, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_5, 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(X)) >? top#(proper(X)) 
  top#(ok(X)) >? top#(active(X)) 
  active(f(X)) >= mark(if(X, c, f(true))) 
  active(if(true, X, Y)) >= mark(X) 
  active(f(X)) >= f(active(X)) 
  active(if(X, Y, Z)) >= if(active(X), Y, Z) 
  active(if(X, Y, Z)) >= if(X, active(Y), Z) 
  f(mark(X)) >= mark(f(X)) 
  if(mark(X), Y, Z) >= mark(if(X, Y, Z)) 
  if(X, mark(Y), Z) >= mark(if(X, Y, Z)) 
  proper(f(X)) >= f(proper(X)) 
  proper(if(X, Y, Z)) >= if(proper(X), proper(Y), proper(Z)) 
  proper(c) >= ok(c) 
  proper(true) >= ok(true) 
  proper(false) >= ok(false) 
  f(ok(X)) >= ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) >= ok(if(X, Y, Z)) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  c = 0 
  f = \y0.1 + y0 
  false = 0 
  if = \y0y1y2.y0 + y1 
  mark = \y0.1 + y0 
  ok = \y0.y0 
  proper = \y0.y0 
  top# = \y0.2y0 
  true = 1 

Using this interpretation, the requirements translate to:

  [[top#(mark(_x0))]] = 2 + 2x0 > 2x0 = [[top#(proper(_x0))]] 
  [[top#(ok(_x0))]] = 2x0 >= 2x0 = [[top#(active(_x0))]] 
  [[active(f(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(if(_x0, c, f(true)))]] 
  [[active(if(true, _x0, _x1))]] = 1 + x0 >= 1 + x0 = [[mark(_x0)]] 
  [[active(f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(active(_x0))]] 
  [[active(if(_x0, _x1, _x2))]] = x0 + x1 >= x0 + x1 = [[if(active(_x0), _x1, _x2)]] 
  [[active(if(_x0, _x1, _x2))]] = x0 + x1 >= x0 + x1 = [[if(_x0, active(_x1), _x2)]] 
  [[f(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(f(_x0))]] 
  [[if(mark(_x0), _x1, _x2)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(if(_x0, _x1, _x2))]] 
  [[if(_x0, mark(_x1), _x2)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(if(_x0, _x1, _x2))]] 
  [[proper(f(_x0))]] = 1 + x0 >= 1 + x0 = [[f(proper(_x0))]] 
  [[proper(if(_x0, _x1, _x2))]] = x0 + x1 >= x0 + x1 = [[if(proper(_x0), proper(_x1), proper(_x2))]] 
  [[proper(c)]] = 0 >= 0 = [[ok(c)]] 
  [[proper(true)]] = 1 >= 1 = [[ok(true)]] 
  [[proper(false)]] = 0 >= 0 = [[ok(false)]] 
  [[f(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(f(_x0))]] 
  [[if(ok(_x0), ok(_x1), ok(_x2))]] = x0 + x1 >= x0 + x1 = [[ok(if(_x0, _x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_1, minimal, formative) by (P_6, R_1, minimal, formative), where P_6 consists of:

  top#(ok(X)) =#> top#(active(X))   

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_6, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_1, minimal, formative).

The formative rules of (P_6, R_1) are R_2 ::=

  active(f(X)) => f(active(X)) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(if(X, Y, Z)) => if(X, active(Y), Z) 
  proper(f(X)) => f(proper(X)) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(c) => ok(c) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  f(ok(X)) => ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_6, R_1, minimal, formative) by (P_6, 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_6, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_6, R_2) are:

  active(f(X)) => f(active(X)) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(if(X, Y, Z)) => if(X, active(Y), Z) 
  f(ok(X)) => ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 

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

  top#(ok(X)) >? top#(active(X)) 
  active(f(X)) >= f(active(X)) 
  active(if(X, Y, Z)) >= if(active(X), Y, Z) 
  active(if(X, Y, Z)) >= if(X, active(Y), Z) 
  f(ok(X)) >= ok(f(X)) 
  if(ok(X), ok(Y), ok(Z)) >= ok(if(X, Y, Z)) 

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

The following interpretation satisfies the requirements:

  active = \y0.2y0 
  f = \y0.y0 
  if = \y0y1y2.2 + 2y1 
  ok = \y0.3 + 2y0 
  top# = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[top#(ok(_x0))]] = 9 + 6x0 > 6x0 = [[top#(active(_x0))]] 
  [[active(f(_x0))]] = 2x0 >= 2x0 = [[f(active(_x0))]] 
  [[active(if(_x0, _x1, _x2))]] = 4 + 4x1 >= 2 + 2x1 = [[if(active(_x0), _x1, _x2)]] 
  [[active(if(_x0, _x1, _x2))]] = 4 + 4x1 >= 2 + 4x1 = [[if(_x0, active(_x1), _x2)]] 
  [[f(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(f(_x0))]] 
  [[if(ok(_x0), ok(_x1), ok(_x2))]] = 8 + 4x1 >= 7 + 4x1 = [[ok(if(_x0, _x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(f(X))) = f(X) |> X = nu(proper#(X)) 
  nu(proper#(if(X, Y, Z))) = if(X, Y, Z) |> X = nu(proper#(X)) 
  nu(proper#(if(X, Y, Z))) = if(X, Y, Z) |> Y = nu(proper#(Y)) 
  nu(proper#(if(X, Y, Z))) = if(X, Y, Z) |> Z = nu(proper#(Z)) 

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(if#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(if#(mark(X), Y, Z)) = mark(X) |> X = nu(if#(X, Y, Z)) 
  nu(if#(X, mark(Y), Z)) = X = X = nu(if#(X, Y, Z)) 
  nu(if#(ok(X), ok(Y), ok(Z))) = ok(X) |> X = nu(if#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_7, R_0, minimal, f), where P_7 contains:

  if#(X, mark(Y), Z) =#> if#(X, Y, Z)   

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

We apply the subterm criterion with the following projection function:

  nu(if#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(if#(X, mark(Y), Z)) = mark(Y) |> Y = nu(if#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, 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(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(mark(X))) = mark(X) |> X = nu(f#(X)) 
  nu(f#(ok(X))) = ok(X) |> X = nu(f#(X)) 

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 apply the subterm criterion with the following projection function:

  nu(active#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(active#(f(X))) = f(X) |> X = nu(active#(X)) 
  nu(active#(if(X, Y, Z))) = if(X, Y, Z) |> X = nu(active#(X)) 
  nu(active#(if(X, Y, Z))) = if(X, Y, Z) |> Y = nu(active#(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 +++

[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.