We consider the system 452.

  Alphabet:

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

  Rules:

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

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

  g(X) => h(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) QTRSRoofMatchBoundsTAProof [EQUIVALENT]
 || (2) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    g(%X) -> h(%X)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRoofMatchBoundsTAProof (EQUIVALENT)
 || The TRS R could be shown to be Match-Bounded [TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] by  1. Therefore it terminates.
 || 
 || The compatible tree automaton used to show the Match-Boundedness is represented by: 
 || final states : [0, 1]
 || transitions: 
 || g0(0) -> 0
 || g0(1) -> 0
 || h0(0) -> 1
 || h0(1) -> 1
 || h1(0) -> 0
 || h1(1) -> 0
 || 
 || ----------------------------------------
 || 
 || (2)
 || 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)) =#> X(X(a))   
    1] f#(/\x.X(x)) =#> X(a) {X : 1}  

  Rules R_0:

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

This combination (P_0, 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_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, 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(X(a)) 
  f#(/\x.X(x)) >? X(a) 

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

  pi( f#(F) ) = #argfun-f##(F (F a), F a) 

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

The following interpretation satisfies the requirements:

  #argfun-f## = \y0y1.3 + max(y0, y1) 
  a = 0 
  f# = \G0.0 

Using this interpretation, the requirements translate to:

  [[#argfun-f##((/\x._x0(x)) ((/\y._x0(y)) a), (/\z._x0(z)) a)]] = 3 + max(F0(0), F0(F0(0))) > F0(F0(0)) = [[_x0(_x0(a))]] 
  [[#argfun-f##((/\x._x0(x)) ((/\y._x0(y)) a), (/\z._x0(z)) a)]] = 3 + max(F0(0), F0(F0(0))) > F0(0) = [[_x0(a)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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.