/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
We consider the system theBenchmark.

  Alphabet:

    !facplus : [b * b] --> b 
    !factimes : [a * b] --> b 
    cons : [d * e] --> e 
    false : [] --> c 
    filter : [d -> c * e] --> e 
    filter2 : [c * d -> c * d * e] --> e 
    map : [d -> d * e] --> e 
    nil : [] --> e 
    true : [] --> c 

  Rules:

    !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

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

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

  !factimes(X, !facplus(Y, Z)) >? !facplus(!factimes(X, Y), !factimes(X, Z)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >? filter(F, Y) 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) 
  [[nil]] = _|_ 

We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {!facplus, !factimes, cons, false, map, true}, and the following precedence: map > false > @_{o -> o} = filter = filter2 > !factimes > !facplus > cons > true

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  !factimes(X, !facplus(Y, Z)) > !facplus(!factimes(X, Y), !factimes(X, Z)) 
  map(F, _|_) >= _|_ 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, _|_) >= _|_ 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) > filter(F, Y) 

With these choices, we have:

  1] !factimes(X, !facplus(Y, Z)) > !facplus(!factimes(X, Y), !factimes(X, Z))  because [2], by definition 
  2] !factimes*(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z))  because !factimes > !facplus, [3] and [8], by (Copy) 
  3] !factimes*(X, !facplus(Y, Z)) >= !factimes(X, Y)  because !factimes in Mul, [4] and [5], by (Stat) 
  4] X >= X  by (Meta) 
  5] !facplus(Y, Z) > Y  because [6], by definition 
  6] !facplus*(Y, Z) >= Y  because [7], by (Select) 
  7] Y >= Y  by (Meta) 
  8] !factimes*(X, !facplus(Y, Z)) >= !factimes(X, Z)  because !factimes in Mul, [4] and [9], by (Stat) 
  9] !facplus(Y, Z) > Z  because [10], by definition 
  10] !facplus*(Y, Z) >= Z  because [11], by (Select) 
  11] Z >= Z  by (Meta) 

  12] map(F, _|_) >= _|_  by (Bot) 

  13] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [14], by (Star) 
  14] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [15] and [22], by (Copy) 
  15] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [16] and [18], by (Copy) 
  16] map*(F, cons(X, Y)) >= F  because [17], by (Select) 
  17] F >= F  by (Meta) 
  18] map*(F, cons(X, Y)) >= X  because [19], by (Select) 
  19] cons(X, Y) >= X  because [20], by (Star) 
  20] cons*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [23] and [24], by (Stat) 
  23] F >= F  by (Meta) 
  24] cons(X, Y) > Y  because [25], by definition 
  25] cons*(X, Y) >= Y  because [26], by (Select) 
  26] Y >= Y  by (Meta) 

  27] filter(F, _|_) >= _|_  by (Bot) 

  28] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [29], by (Star) 
  29] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [30], [31], [34], [38], [39] and [41], by (Stat) 
  30] F >= F  by (Meta) 
  31] cons(X, Y) > Y  because [32], by definition 
  32] cons*(X, Y) >= Y  because [33], by (Select) 
  33] Y >= Y  by (Meta) 
  34] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter = @_{o -> o}, [30], [35], [38] and [39], by (Stat) 
  35] cons(X, Y) > X  because [36], by definition 
  36] cons*(X, Y) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] filter*(F, cons(X, Y)) >= F  because [30], by (Select) 
  39] filter*(F, cons(X, Y)) >= X  because [40], by (Select) 
  40] cons(X, Y) >= X  because [36], by (Star) 
  41] filter*(F, cons(X, Y)) >= Y  because [42], by (Select) 
  42] cons(X, Y) >= Y  because [32], by (Star) 

  43] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [44], by (Star) 
  44] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [45] and [47], by (Copy) 
  45] filter2*(true, F, X, Y) >= X  because [46], by (Select) 
  46] X >= X  by (Meta) 
  47] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [48], [49], [50] and [51], by (Stat) 
  48] F >= F  by (Meta) 
  49] Y >= Y  by (Meta) 
  50] filter2*(true, F, X, Y) >= F  because [48], by (Select) 
  51] filter2*(true, F, X, Y) >= Y  because [49], by (Select) 

  52] filter2(false, F, X, Y) > filter(F, Y)  because [53], by definition 
  53] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [54], [55], [56] and [57], by (Stat) 
  54] F >= F  by (Meta) 
  55] Y >= Y  by (Meta) 
  56] filter2*(false, F, X, Y) >= F  because [54], by (Select) 
  57] filter2*(false, F, X, Y) >= Y  because [55], by (Select) 

We can thus remove the following rules:

  !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) 
  filter2(false, F, X, Y) => filter(F, Y) 

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

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

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + y0 + y1 
  filter = \G0y1.3y1 + G0(y1) + 3y1G0(y1) 
  filter2 = \y0G1y2y3.1 + y0 + 2y2 + 3y3 + G1(y3) + 2G1(0) + 3y3G1(y3) 
  map = \G0y1.3y1 + G0(0) + 2y1G0(y1) + 2G0(y1) 
  nil = 2 
  true = 3 

Using this interpretation, the requirements translate to:

  [[map(_F0, nil)]] = 6 + F0(0) + 6F0(2) > 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + F0(0) + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 8F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 6 + 7F0(2) > 2 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 10F0(3 + x1 + x2) > 1 + 3x1 + 3x2 + F0(x1) + F0(x2) + 2F0(0) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x1 + 3x2 + F0(x2) + 2F0(0) + 3x2F0(x2) > 3 + x1 + 3x2 + F0(x2) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 

We can thus remove the following rules:

  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  filter(F, nil) => nil 
  filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.