We consider the system 472.

  Alphabet:

    f : [o -> o] --> o 
    g : [o] --> o 
    h : [] --> o 
    i : [o] --> o 

  Rules:

    f(/\x.X(x)) => g(X(h)) 
    g(X) => i(X) 

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

  g(X) => i(X) 

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) QTRS Reverse [EQUIVALENT]
 || (2) QTRS
 || (3) RFCMatchBoundsTRSProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    g(%X) -> i(%X)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRS Reverse (EQUIVALENT)
 || We applied the QTRS Reverse Processor [REVERSE].
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    g(x) -> i(x)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RFCMatchBoundsTRSProof (EQUIVALENT)
 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R.
 || The following rules were used to construct the certificate:
 || 
 ||    g(x) -> i(x)
 || 
 || The certificate found is represented by the following graph.
 || The certificate consists of the following enumerated nodes:
 || 1, 3
 || 
 || Node 1 is start node and node 3 is final node.
 || 
 || Those nodes are connected through the following edges:
 || 
 || * 1 to 3 labelled i_1(0)* 3 to 3 labelled #_1(0)
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] f#(/\x.X(x)) =#> g#(X(h))   
    1] f#(/\x.X(x)) =#> X(h)   

  Rules R_0:

    f(/\x.X(x)) => g(X(h)) 
    g(X) => i(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 :  
    * 1 : 0, 1 

This graph has the following strongly connected components:

  P_1:

    f#(/\x.X(x)) =#> X(h)   

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).

This combination (P_1, R_0) has no formative rules!  We will name the empty set of rules:R_1.

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:

  f#(/\x.X(x)) >? X(h) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( f#(F) ) = #argfun-f##(F h) 

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

The following interpretation satisfies the requirements:

  #argfun-f## = \y0.1 + y0 
  f# = \G0.0 
  h = 0 

Using this interpretation, the requirements translate to:

  [[#argfun-f##((/\x._x0(x)) h)]] = 1 + F0(0) > F0(0) = [[_x0(h)]] 

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.