/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 : [] --> nat 
    false : [] --> bool 
    h : [nat -> bool * nat -> nat * nat] --> nat 
    if : [bool * nat * nat] --> nat 
    s : [nat] --> nat 
    true : [] --> bool 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    h(f, g, 0) => 0 
    h(f, g, s(x)) => g h(f, g, if(f x, x, 0)) 

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

We observe that the rules contain a first-order subset:

  if(true, X, Y) => X 
  if(false, X, Y) => Y 

Moreover, the system is orthogonal.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(false) = 2
 ||    POL(if(x_1, x_2, x_3)) = 2 + 2*x_1 + x_2 + x_3
 ||    POL(true) = 2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
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] h#(F, G, s(X)) =#> h#(F, G, if(F X, X, 0))   
    1] h#(F, G, s(X)) =#> if#(F X, X, 0)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    h(F, G, 0) => 0 
    h(F, G, s(X)) => G h(F, G, if(F X, X, 0)) 

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 : 0, 1 
    * 1 :  

This graph has the following strongly connected components:

  P_1:

    h#(F, G, s(X)) =#> h#(F, G, if(F X, X, 0))   

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).

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).

The formative rules of (P_1, R_0) are R_1 ::=

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  h(F, G, s(X)) => G h(F, G, if(F X, X, 0)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, 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:

  h#(F, G, s(X)) >? h#(F, G, if(F X, X, 0)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  h(F, G, s(X)) >= G h(F, G, if(F X, X, 0)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

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

Argument functions:

  [[0]] = _|_ 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 

We choose Lex = {} and Mul = {@_{o -> o}, false, h, h#, if, s, true}, and the following precedence: h > @_{o -> o} > h# > s > if > true > false

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

  h#(F, G, s(X)) > h#(F, G, if(X, _|_)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  h(F, G, s(X)) >= @_{o -> o}(G, h(F, G, if(X, _|_))) 

With these choices, we have:

  1] h#(F, G, s(X)) > h#(F, G, if(X, _|_))  because [2], by definition 
  2] h#*(F, G, s(X)) >= h#(F, G, if(X, _|_))  because h# in Mul, [3], [4] and [5], by (Stat) 
  3] F >= F  by (Meta) 
  4] G >= G  by (Meta) 
  5] s(X) > if(X, _|_)  because [6], by definition 
  6] s*(X) >= if(X, _|_)  because s > if, [7] and [9], by (Copy) 
  7] s*(X) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 
  9] s*(X) >= _|_  by (Bot) 

  10] if(X, Y) >= X  because [11], by (Star) 
  11] if*(X, Y) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 

  13] if(X, Y) >= Y  because [14], by (Star) 
  14] if*(X, Y) >= Y  because [15], by (Select) 
  15] Y >= Y  by (Meta) 

  16] h(F, G, s(X)) >= @_{o -> o}(G, h(F, G, if(X, _|_)))  because [17], by (Star) 
  17] h*(F, G, s(X)) >= @_{o -> o}(G, h(F, G, if(X, _|_)))  because h > @_{o -> o}, [18] and [19], by (Copy) 
  18] h*(F, G, s(X)) >= G  because [4], by (Select) 
  19] h*(F, G, s(X)) >= h(F, G, if(X, _|_))  because h in Mul, [3], [4] and [5], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  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.