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


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YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> nat 
    apply : [nat * nat] --> nat 
    avg : [nat * nat] --> nat 
    check : [nat] --> nat 
    fun : [nat -> nat] --> nat 
    s : [nat] --> nat 

  Rules:

    avg(s(x), y) => avg(x, s(y)) 
    avg(x, s(s(s(y)))) => s(avg(s(x), y)) 
    avg(0, 0) => 0 
    avg(0, s(0)) => 0 
    avg(0, s(s(0))) => s(0) 
    apply(fun(f), x) => f check(x) 
    check(s(x)) => s(check(x)) 
    check(0) => 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:

  avg(s(X), Y) => avg(X, s(Y)) 
  avg(X, s(s(s(Y)))) => s(avg(s(X), Y)) 
  avg(0, 0) => 0 
  avg(0, s(0)) => 0 
  avg(0, s(s(0))) => s(0) 
  check(s(X)) => s(check(X)) 
  check(0) => 0 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-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) QTRSRRRProof [EQUIVALENT]
 || (4) QTRS
 || (5) RisEmptyProof [EQUIVALENT]
 || (6) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    avg(s(%X), %Y) -> avg(%X, s(%Y))
 ||    avg(%X, s(s(s(%Y)))) -> s(avg(s(%X), %Y))
 ||    avg(0, 0) -> 0
 ||    avg(0, s(0)) -> 0
 ||    avg(0, s(s(0))) -> s(0)
 ||    check(s(%X)) -> s(check(%X))
 ||    check(0) -> 0
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 1
 ||    POL(avg(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(check(x_1)) = 2 + 2*x_1
 ||    POL(s(x_1)) = 1 + x_1
 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    avg(s(%X), %Y) -> avg(%X, s(%Y))
 ||    avg(0, 0) -> 0
 ||    avg(0, s(0)) -> 0
 ||    avg(0, s(s(0))) -> s(0)
 ||    check(s(%X)) -> s(check(%X))
 ||    check(0) -> 0
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    avg(%X, s(s(s(%Y)))) -> s(avg(s(%X), %Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || avg_2 > s_1
 || 
 || 
 || Status:
 || avg_2: [2,1]
 || s_1: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    avg(%X, s(s(s(%Y)))) -> s(avg(s(%X), %Y))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (6)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] apply#(fun(F), X) =#> F(check(X))   
    1] apply#(fun(F), X) =#> check#(X)   

  Rules R_0:

    avg(s(X), Y) => avg(X, s(Y)) 
    avg(X, s(s(s(Y)))) => s(avg(s(X), Y)) 
    avg(0, 0) => 0 
    avg(0, s(0)) => 0 
    avg(0, s(s(0))) => s(0) 
    apply(fun(F), X) => F check(X) 
    check(s(X)) => s(check(X)) 
    check(0) => 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:

    apply#(fun(F), X) =#> F(check(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).

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 ::=

  apply(fun(F), X) => F check(X) 

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].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  apply#(fun(F), X) >? F(check(X)) 
  apply(fun(F), X) >= F check(X) 

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

The following interpretation satisfies the requirements:

  apply = \y0y1.3 + 3y0 
  apply# = \y0y1.3 + y0 
  check = \y0.0 
  fun = \G0.3 + G0(0) 

Using this interpretation, the requirements translate to:

  [[apply#(fun(_F0), _x1)]] = 6 + F0(0) > F0(0) = [[_F0(check(_x1))]] 
  [[apply(fun(_F0), _x1)]] = 12 + 3F0(0) >= F0(0) = [[_F0 check(_x1)]] 

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.