/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.

  0 : [] --> o 
  active : [o] --> o 
  add : [o * o] --> o 
  and : [o * o] --> o 
  cons : [o * o] --> o 
  false : [] --> o 
  first : [o * o] --> o 
  from : [o] --> o 
  if : [o * o * o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  s : [o] --> o 
  top : [o] --> o 
  true : [] --> o 

  active(and(true, X)) => mark(X) 
  active(and(false, X)) => mark(false) 
  active(if(true, X, Y)) => mark(X) 
  active(if(false, X, Y)) => mark(Y) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(first(0, X)) => mark(nil) 
  active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(and(X, Y)) => and(active(X), Y) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(add(X, Y)) => add(active(X), Y) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  and(mark(X), Y) => mark(and(X, Y)) 
  if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
  add(mark(X), Y) => mark(add(X, Y)) 
  first(mark(X), Y) => mark(first(X, Y)) 
  first(X, mark(Y)) => mark(first(X, Y)) 
  proper(and(X, Y)) => and(proper(X), proper(Y)) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(from(X)) => from(proper(X)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  from(ok(X)) => ok(from(X)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(X)) 

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#(add(s(X), Y)) =#> s#(add(X, Y))   
    1] active#(add(s(X), Y)) =#> add#(X, Y)   
    2] active#(first(s(X), cons(Y, Z))) =#> cons#(Y, first(X, Z))   
    3] active#(first(s(X), cons(Y, Z))) =#> first#(X, Z)   
    4] active#(from(X)) =#> cons#(X, from(s(X)))   
    5] active#(from(X)) =#> from#(s(X))   
    6] active#(from(X)) =#> s#(X)   
    7] active#(and(X, Y)) =#> and#(active(X), Y)   
    8] active#(and(X, Y)) =#> active#(X)   
    9] active#(if(X, Y, Z)) =#> if#(active(X), Y, Z)   
    10] active#(if(X, Y, Z)) =#> active#(X)   
    11] active#(add(X, Y)) =#> add#(active(X), Y)   
    12] active#(add(X, Y)) =#> active#(X)   
    13] active#(first(X, Y)) =#> first#(active(X), Y)   
    14] active#(first(X, Y)) =#> active#(X)   
    15] active#(first(X, Y)) =#> first#(X, active(Y))   
    16] active#(first(X, Y)) =#> active#(Y)   
    17] and#(mark(X), Y) =#> and#(X, Y)   
    18] if#(mark(X), Y, Z) =#> if#(X, Y, Z)   
    19] add#(mark(X), Y) =#> add#(X, Y)   
    20] first#(mark(X), Y) =#> first#(X, Y)   
    21] first#(X, mark(Y)) =#> first#(X, Y)   
    22] proper#(and(X, Y)) =#> and#(proper(X), proper(Y))   
    23] proper#(and(X, Y)) =#> proper#(X)   
    24] proper#(and(X, Y)) =#> proper#(Y)   
    25] proper#(if(X, Y, Z)) =#> if#(proper(X), proper(Y), proper(Z))   
    26] proper#(if(X, Y, Z)) =#> proper#(X)   
    27] proper#(if(X, Y, Z)) =#> proper#(Y)   
    28] proper#(if(X, Y, Z)) =#> proper#(Z)   
    29] proper#(add(X, Y)) =#> add#(proper(X), proper(Y))   
    30] proper#(add(X, Y)) =#> proper#(X)   
    31] proper#(add(X, Y)) =#> proper#(Y)   
    32] proper#(s(X)) =#> s#(proper(X))   
    33] proper#(s(X)) =#> proper#(X)   
    34] proper#(first(X, Y)) =#> first#(proper(X), proper(Y))   
    35] proper#(first(X, Y)) =#> proper#(X)   
    36] proper#(first(X, Y)) =#> proper#(Y)   
    37] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y))   
    38] proper#(cons(X, Y)) =#> proper#(X)   
    39] proper#(cons(X, Y)) =#> proper#(Y)   
    40] proper#(from(X)) =#> from#(proper(X))   
    41] proper#(from(X)) =#> proper#(X)   
    42] and#(ok(X), ok(Y)) =#> and#(X, Y)   
    43] if#(ok(X), ok(Y), ok(Z)) =#> if#(X, Y, Z)   
    44] add#(ok(X), ok(Y)) =#> add#(X, Y)   
    45] s#(ok(X)) =#> s#(X)   
    46] first#(ok(X), ok(Y)) =#> first#(X, Y)   
    47] cons#(ok(X), ok(Y)) =#> cons#(X, Y)   
    48] from#(ok(X)) =#> from#(X)   
    49] top#(mark(X)) =#> top#(proper(X))   
    50] top#(mark(X)) =#> proper#(X)   
    51] top#(ok(X)) =#> top#(active(X))   
    52] top#(ok(X)) =#> active#(X)   

  Rules R_0:

    active(and(true, X)) => mark(X) 
    active(and(false, X)) => mark(false) 
    active(if(true, X, Y)) => mark(X) 
    active(if(false, X, Y)) => mark(Y) 
    active(add(0, X)) => mark(X) 
    active(add(s(X), Y)) => mark(s(add(X, Y))) 
    active(first(0, X)) => mark(nil) 
    active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
    active(from(X)) => mark(cons(X, from(s(X)))) 
    active(and(X, Y)) => and(active(X), Y) 
    active(if(X, Y, Z)) => if(active(X), Y, Z) 
    active(add(X, Y)) => add(active(X), Y) 
    active(first(X, Y)) => first(active(X), Y) 
    active(first(X, Y)) => first(X, active(Y)) 
    and(mark(X), Y) => mark(and(X, Y)) 
    if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
    add(mark(X), Y) => mark(add(X, Y)) 
    first(mark(X), Y) => mark(first(X, Y)) 
    first(X, mark(Y)) => mark(first(X, Y)) 
    proper(and(X, Y)) => and(proper(X), proper(Y)) 
    proper(true) => ok(true) 
    proper(false) => ok(false) 
    proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
    proper(add(X, Y)) => add(proper(X), proper(Y)) 
    proper(0) => ok(0) 
    proper(s(X)) => s(proper(X)) 
    proper(first(X, Y)) => first(proper(X), proper(Y)) 
    proper(nil) => ok(nil) 
    proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
    proper(from(X)) => from(proper(X)) 
    and(ok(X), ok(Y)) => ok(and(X, Y)) 
    if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
    add(ok(X), ok(Y)) => ok(add(X, Y)) 
    s(ok(X)) => ok(s(X)) 
    first(ok(X), ok(Y)) => ok(first(X, Y)) 
    cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
    from(ok(X)) => ok(from(X)) 
    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 : 45 
    * 1 : 19, 44 
    * 2 : 47 
    * 3 : 20, 21, 46 
    * 4 : 47 
    * 5 : 48 
    * 6 : 45 
    * 7 : 17, 42 
    * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 9 : 18, 43 
    * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 11 : 19, 44 
    * 12 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 13 : 20, 21, 46 
    * 14 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 15 : 20, 21, 46 
    * 16 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 17 : 17, 42 
    * 18 : 18, 43 
    * 19 : 19, 44 
    * 20 : 20, 21, 46 
    * 21 : 20, 21, 46 
    * 22 : 17, 42 
    * 23 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 24 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 25 : 18, 43 
    * 26 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 27 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 28 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 29 : 19, 44 
    * 30 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 31 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 32 : 45 
    * 33 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 34 : 20, 21, 46 
    * 35 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 36 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 37 : 47 
    * 38 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 39 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 40 : 48 
    * 41 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 42 : 17, 42 
    * 43 : 18, 43 
    * 44 : 19, 44 
    * 45 : 45 
    * 46 : 20, 21, 46 
    * 47 : 47 
    * 48 : 48 
    * 49 : 49, 50, 51, 52 
    * 50 : 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41 
    * 51 : 49, 50, 51, 52 
    * 52 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 

This graph has the following strongly connected components:

  P_1:

    active#(and(X, Y)) =#> active#(X)   
    active#(if(X, Y, Z)) =#> active#(X)   
    active#(add(X, Y)) =#> active#(X)   
    active#(first(X, Y)) =#> active#(X)   
    active#(first(X, Y)) =#> active#(Y)   

  P_2:

    and#(mark(X), Y) =#> and#(X, Y)   
    and#(ok(X), ok(Y)) =#> and#(X, Y)   

  P_3:

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

  P_4:

    add#(mark(X), Y) =#> add#(X, Y)   
    add#(ok(X), ok(Y)) =#> add#(X, Y)   

  P_5:

    first#(mark(X), Y) =#> first#(X, Y)   
    first#(X, mark(Y)) =#> first#(X, Y)   
    first#(ok(X), ok(Y)) =#> first#(X, Y)   

  P_6:

    proper#(and(X, Y)) =#> proper#(X)   
    proper#(and(X, Y)) =#> proper#(Y)   
    proper#(if(X, Y, Z)) =#> proper#(X)   
    proper#(if(X, Y, Z)) =#> proper#(Y)   
    proper#(if(X, Y, Z)) =#> proper#(Z)   
    proper#(add(X, Y)) =#> proper#(X)   
    proper#(add(X, Y)) =#> proper#(Y)   
    proper#(s(X)) =#> proper#(X)   
    proper#(first(X, Y)) =#> proper#(X)   
    proper#(first(X, Y)) =#> proper#(Y)   
    proper#(cons(X, Y)) =#> proper#(X)   
    proper#(cons(X, Y)) =#> proper#(Y)   
    proper#(from(X)) =#> proper#(X)   

  P_7:

    s#(ok(X)) =#> s#(X)   

  P_8:

    cons#(ok(X), ok(Y)) =#> cons#(X, Y)   

  P_9:

    from#(ok(X)) =#> from#(X)   

  P_10:

    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), (P_5, R_0, m, f), (P_6, R_0, m, f), (P_7, R_0, m, f), (P_8, R_0, m, f), (P_9, R_0, m, f) and (P_10, 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), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative) and (P_10, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_10, R_0, minimal, formative).

The formative rules of (P_10, R_0) are R_1 ::=

  active(and(true, X)) => mark(X) 
  active(and(false, X)) => mark(false) 
  active(if(true, X, Y)) => mark(X) 
  active(if(false, X, Y)) => mark(Y) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(first(0, X)) => mark(nil) 
  active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(and(X, Y)) => and(active(X), Y) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(add(X, Y)) => add(active(X), Y) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  and(mark(X), Y) => mark(and(X, Y)) 
  if(mark(X), Y, Z) => mark(if(X, Y, Z)) 
  add(mark(X), Y) => mark(add(X, Y)) 
  first(mark(X), Y) => mark(first(X, Y)) 
  first(X, mark(Y)) => mark(first(X, Y)) 
  proper(and(X, Y)) => and(proper(X), proper(Y)) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(from(X)) => from(proper(X)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  from(ok(X)) => ok(from(X)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_10, R_0, minimal, formative) by (P_10, 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), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative) and (P_10, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_10, 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(and(true, X)) >= mark(X) 
  active(and(false, X)) >= mark(false) 
  active(if(true, X, Y)) >= mark(X) 
  active(if(false, X, Y)) >= mark(Y) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(first(0, X)) >= mark(nil) 
  active(first(s(X), cons(Y, Z))) >= mark(cons(Y, first(X, Z))) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(and(X, Y)) >= and(active(X), Y) 
  active(if(X, Y, Z)) >= if(active(X), Y, Z) 
  active(add(X, Y)) >= add(active(X), Y) 
  active(first(X, Y)) >= first(active(X), Y) 
  active(first(X, Y)) >= first(X, active(Y)) 
  and(mark(X), Y) >= mark(and(X, Y)) 
  if(mark(X), Y, Z) >= mark(if(X, Y, Z)) 
  add(mark(X), Y) >= mark(add(X, Y)) 
  first(mark(X), Y) >= mark(first(X, Y)) 
  first(X, mark(Y)) >= mark(first(X, Y)) 
  proper(and(X, Y)) >= and(proper(X), proper(Y)) 
  proper(true) >= ok(true) 
  proper(false) >= ok(false) 
  proper(if(X, Y, Z)) >= if(proper(X), proper(Y), proper(Z)) 
  proper(add(X, Y)) >= add(proper(X), proper(Y)) 
  proper(0) >= ok(0) 
  proper(s(X)) >= s(proper(X)) 
  proper(first(X, Y)) >= first(proper(X), proper(Y)) 
  proper(nil) >= ok(nil) 
  proper(cons(X, Y)) >= cons(proper(X), proper(Y)) 
  proper(from(X)) >= from(proper(X)) 
  and(ok(X), ok(Y)) >= ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) >= ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) >= ok(add(X, Y)) 
  s(ok(X)) >= ok(s(X)) 
  first(ok(X), ok(Y)) >= ok(first(X, Y)) 
  cons(ok(X), ok(Y)) >= ok(cons(X, Y)) 
  from(ok(X)) >= ok(from(X)) 

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

The following interpretation satisfies the requirements:

  0 = 1 
  active = \y0.y0 
  add = \y0y1.y1 + 2y0 
  and = \y0y1.1 + 2y0 + 2y1 
  cons = \y0y1.0 
  false = 0 
  first = \y0y1.y1 + 2y0 
  from = \y0.1 
  if = \y0y1y2.1 + y1 + 2y0 + 2y2 
  mark = \y0.1 + y0 
  nil = 0 
  ok = \y0.y0 
  proper = \y0.y0 
  s = \y0.1 
  top# = \y0.2y0 
  true = 0 

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(and(true, _x0))]] = 1 + 2x0 >= 1 + x0 = [[mark(_x0)]] 
  [[active(and(false, _x0))]] = 1 + 2x0 >= 1 = [[mark(false)]] 
  [[active(if(true, _x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x0 = [[mark(_x0)]] 
  [[active(if(false, _x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x1 = [[mark(_x1)]] 
  [[active(add(0, _x0))]] = 2 + x0 >= 1 + x0 = [[mark(_x0)]] 
  [[active(add(s(_x0), _x1))]] = 2 + x1 >= 2 = [[mark(s(add(_x0, _x1)))]] 
  [[active(first(0, _x0))]] = 2 + x0 >= 1 = [[mark(nil)]] 
  [[active(first(s(_x0), cons(_x1, _x2)))]] = 2 >= 1 = [[mark(cons(_x1, first(_x0, _x2)))]] 
  [[active(from(_x0))]] = 1 >= 1 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(and(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[and(active(_x0), _x1)]] 
  [[active(if(_x0, _x1, _x2))]] = 1 + x1 + 2x0 + 2x2 >= 1 + x1 + 2x0 + 2x2 = [[if(active(_x0), _x1, _x2)]] 
  [[active(add(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[add(active(_x0), _x1)]] 
  [[active(first(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[first(active(_x0), _x1)]] 
  [[active(first(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[first(_x0, active(_x1))]] 
  [[and(mark(_x0), _x1)]] = 3 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[mark(and(_x0, _x1))]] 
  [[if(mark(_x0), _x1, _x2)]] = 3 + x1 + 2x0 + 2x2 >= 2 + x1 + 2x0 + 2x2 = [[mark(if(_x0, _x1, _x2))]] 
  [[add(mark(_x0), _x1)]] = 2 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(add(_x0, _x1))]] 
  [[first(mark(_x0), _x1)]] = 2 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(first(_x0, _x1))]] 
  [[first(_x0, mark(_x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(first(_x0, _x1))]] 
  [[proper(and(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(true)]] = 0 >= 0 = [[ok(true)]] 
  [[proper(false)]] = 0 >= 0 = [[ok(false)]] 
  [[proper(if(_x0, _x1, _x2))]] = 1 + x1 + 2x0 + 2x2 >= 1 + x1 + 2x0 + 2x2 = [[if(proper(_x0), proper(_x1), proper(_x2))]] 
  [[proper(add(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[add(proper(_x0), proper(_x1))]] 
  [[proper(0)]] = 1 >= 1 = [[ok(0)]] 
  [[proper(s(_x0))]] = 1 >= 1 = [[s(proper(_x0))]] 
  [[proper(first(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[first(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] 
  [[proper(cons(_x0, _x1))]] = 0 >= 0 = [[cons(proper(_x0), proper(_x1))]] 
  [[proper(from(_x0))]] = 1 >= 1 = [[from(proper(_x0))]] 
  [[and(ok(_x0), ok(_x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[ok(and(_x0, _x1))]] 
  [[if(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x1 + 2x0 + 2x2 >= 1 + x1 + 2x0 + 2x2 = [[ok(if(_x0, _x1, _x2))]] 
  [[add(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(add(_x0, _x1))]] 
  [[s(ok(_x0))]] = 1 >= 1 = [[ok(s(_x0))]] 
  [[first(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(first(_x0, _x1))]] 
  [[cons(ok(_x0), ok(_x1))]] = 0 >= 0 = [[ok(cons(_x0, _x1))]] 
  [[from(ok(_x0))]] = 1 >= 1 = [[ok(from(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_10, R_1, minimal, formative) by (P_11, R_1, minimal, formative), where P_11 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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative) and (P_11, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_11, R_1, minimal, formative).

The formative rules of (P_11, R_1) are R_2 ::=

  active(and(X, Y)) => and(active(X), Y) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(add(X, Y)) => add(active(X), Y) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  proper(and(X, Y)) => and(proper(X), proper(Y)) 
  proper(true) => ok(true) 
  proper(false) => ok(false) 
  proper(if(X, Y, Z)) => if(proper(X), proper(Y), proper(Z)) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(from(X)) => from(proper(X)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  from(ok(X)) => ok(from(X)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_11, R_1, minimal, formative) by (P_11, 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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative) and (P_11, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_11, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_11, R_2) are:

  active(and(X, Y)) => and(active(X), Y) 
  active(if(X, Y, Z)) => if(active(X), Y, Z) 
  active(add(X, Y)) => add(active(X), Y) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) => ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 

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

  top#(ok(X)) >? top#(active(X)) 
  active(and(X, Y)) >= and(active(X), Y) 
  active(if(X, Y, Z)) >= if(active(X), Y, Z) 
  active(add(X, Y)) >= add(active(X), Y) 
  active(first(X, Y)) >= first(active(X), Y) 
  active(first(X, Y)) >= first(X, active(Y)) 
  and(ok(X), ok(Y)) >= ok(and(X, Y)) 
  if(ok(X), ok(Y), ok(Z)) >= ok(if(X, Y, Z)) 
  add(ok(X), ok(Y)) >= ok(add(X, Y)) 
  first(ok(X), ok(Y)) >= ok(first(X, Y)) 

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

The following interpretation satisfies the requirements:

  active = \y0.2 + 2y0 
  add = \y0y1.3 + 2y0 + 2y1 
  and = \y0y1.3 + 3y1 
  first = \y0y1.1 + y0 + y1 
  if = \y0y1y2.3 + y1 + 2y0 + 2y2 
  ok = \y0.3 + 3y0 
  top# = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[top#(ok(_x0))]] = 9 + 9x0 > 6 + 6x0 = [[top#(active(_x0))]] 
  [[active(and(_x0, _x1))]] = 8 + 6x1 >= 3 + 3x1 = [[and(active(_x0), _x1)]] 
  [[active(if(_x0, _x1, _x2))]] = 8 + 2x1 + 4x0 + 4x2 >= 7 + x1 + 2x2 + 4x0 = [[if(active(_x0), _x1, _x2)]] 
  [[active(add(_x0, _x1))]] = 8 + 4x0 + 4x1 >= 7 + 2x1 + 4x0 = [[add(active(_x0), _x1)]] 
  [[active(first(_x0, _x1))]] = 4 + 2x0 + 2x1 >= 3 + x1 + 2x0 = [[first(active(_x0), _x1)]] 
  [[active(first(_x0, _x1))]] = 4 + 2x0 + 2x1 >= 3 + x0 + 2x1 = [[first(_x0, active(_x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 12 + 9x1 >= 12 + 9x1 = [[ok(and(_x0, _x1))]] 
  [[if(ok(_x0), ok(_x1), ok(_x2))]] = 18 + 3x1 + 6x0 + 6x2 >= 12 + 3x1 + 6x0 + 6x2 = [[ok(if(_x0, _x1, _x2))]] 
  [[add(ok(_x0), ok(_x1))]] = 15 + 6x0 + 6x1 >= 12 + 6x0 + 6x1 = [[ok(add(_x0, _x1))]] 
  [[first(ok(_x0), ok(_x1))]] = 7 + 3x0 + 3x1 >= 6 + 3x0 + 3x1 = [[ok(first(_x0, _x1))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_11, 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), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_9, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_9, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(from#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(from#(ok(X))) = ok(X) |> X = nu(from#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_9, 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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative) and (P_8, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_8, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(cons#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(cons#(ok(X), ok(Y))) = ok(X) |> X = nu(cons#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_8, 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), (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).

We apply the subterm criterion with the following projection function:

  nu(s#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(s#(ok(X))) = ok(X) |> X = nu(s#(X)) 

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), (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(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(and(X, Y))) = and(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(and(X, Y))) = and(X, Y) |> Y = nu(proper#(Y)) 
  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)) 
  nu(proper#(add(X, Y))) = add(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(add(X, Y))) = add(X, Y) |> Y = nu(proper#(Y)) 
  nu(proper#(s(X))) = s(X) |> X = nu(proper#(X)) 
  nu(proper#(first(X, Y))) = first(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(first(X, Y))) = first(X, Y) |> Y = nu(proper#(Y)) 
  nu(proper#(cons(X, Y))) = cons(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(cons(X, Y))) = cons(X, Y) |> Y = nu(proper#(Y)) 
  nu(proper#(from(X))) = from(X) |> X = nu(proper#(X)) 

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(first#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by (P_12, R_0, minimal, f), where P_12 contains:

  first#(X, mark(Y)) =#> first#(X, Y)   

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_12, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_12, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(first#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(first#(X, mark(Y))) = mark(Y) |> Y = nu(first#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_12, 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(add#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(add#(mark(X), Y)) = mark(X) |> X = nu(add#(X, Y)) 
  nu(add#(ok(X), ok(Y))) = ok(X) |> X = nu(add#(X, 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(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#(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 ({}, 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(and#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(and#(mark(X), Y)) = mark(X) |> X = nu(and#(X, Y)) 
  nu(and#(ok(X), ok(Y))) = ok(X) |> X = nu(and#(X, 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).

We apply the subterm criterion with the following projection function:

  nu(active#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(active#(and(X, Y))) = and(X, Y) |> X = nu(active#(X)) 
  nu(active#(if(X, Y, Z))) = if(X, Y, Z) |> X = nu(active#(X)) 
  nu(active#(add(X, Y))) = add(X, Y) |> X = nu(active#(X)) 
  nu(active#(first(X, Y))) = first(X, Y) |> X = nu(active#(X)) 
  nu(active#(first(X, Y))) = first(X, Y) |> 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.