/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:

    0 : [] --> a 
    comp : [b -> b * b -> b] --> b -> b 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    twice : [b -> b] --> b -> b 

  Rules:

    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    times(0, x) => 0 
    times(s(x), y) => plus(times(x, y), y) 
    comp(f, g) x => f (g x) 
    twice(f) => comp(f, f) 

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]):

  plus(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  times(0, X) >? 0 
  times(s(X), Y) >? plus(times(X, Y), Y) 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

about to try horpo

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

Argument functions:

  [[0]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, comp, plus, s, times, twice}, and the following precedence: times > plus > s > twice > comp > @_{o -> o}

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

  plus(_|_, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  times(_|_, X) >= _|_ 
  times(s(X), Y) > plus(times(X, Y), Y) 
  @_{o -> o}(comp(F, G), X) > @_{o -> o}(F, @_{o -> o}(G, X)) 
  twice(F) >= comp(F, F) 

With these choices, we have:

  1] plus(_|_, X) >= X  because [2], by (Star) 
  2] plus*(_|_, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] plus(s(X), Y) >= s(plus(X, Y))  because [5], by (Star) 
  5] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [6], by (Copy) 
  6] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [7] and [10], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] Y >= Y  by (Meta) 

  11] times(_|_, X) >= _|_  by (Bot) 

  12] times(s(X), Y) > plus(times(X, Y), Y)  because [13], by definition 
  13] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [14] and [19], by (Copy) 
  14] times*(s(X), Y) >= times(X, Y)  because times in Mul, [15] and [18], by (Stat) 
  15] s(X) > X  because [16], by definition 
  16] s*(X) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 
  18] Y >= Y  by (Meta) 
  19] times*(s(X), Y) >= Y  because [18], by (Select) 

  20] @_{o -> o}(comp(F, G), X) > @_{o -> o}(F, @_{o -> o}(G, X))  because [21], by definition 
  21] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [22], by (Select) 
  22] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [23] 
  23] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [24] and [26], by (Copy) 
  24] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [25], by (Select) 
  25] F >= F  by (Meta) 
  26] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [27] and [29], by (Copy) 
  27] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [28], by (Select) 
  28] G >= G  by (Meta) 
  29] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [30], by (Select) 
  30] @_{o -> o}*(comp(F, G), X) >= X  because [31], by (Select) 
  31] X >= X  by (Meta) 

  32] twice(F) >= comp(F, F)  because [33], by (Star) 
  33] twice*(F) >= comp(F, F)  because twice > comp, [34] and [34], by (Copy) 
  34] twice*(F) >= F  because [35], by (Select) 
  35] F >= F  by (Meta) 

We can thus remove the following rules:

  times(s(X), Y) => plus(times(X, Y), Y) 
  comp(F, G) X => F (G 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]):

  plus(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  times(0, X) >? 0 
  twice(F) >? comp(F, F) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  comp = \G0G1y2.G0(0) + G1(y2) 
  plus = \y0y1.3 + y1 + 3y0 
  s = \y0.3 + y0 
  times = \y0y1.3 + y1 + 3y0 
  twice = \G0y1.3 + 3y1 + G0(0) + 2G0(y1) 

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
  [[times(0, _x0)]] = 3 + x0 > 0 = [[0]] 
  [[twice(_F0)]] = \y0.3 + 3y0 + F0(0) + 2F0(y0) > \y0.F0(0) + F0(y0) = [[comp(_F0, _F0)]] 

We can thus remove the following rules:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  times(0, X) => 0 
  twice(F) => comp(F, F) 

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.