We consider the system 443.

  Alphabet:

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

  Rules:

    f(/\x.X(x), Y) => X(X(Y)) 
    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) RFCMatchBoundsTRSProof [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) 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) -> h(%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 h_1(0)* 3 to 3 labelled #_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), Y) =#> X(X(Y))   
    1] f#(/\x.X(x), Y) =#> X(Y) {X : 1}  

  Rules R_0:

    f(/\x.X(x), Y) => X(X(Y)) 
    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), Y) >? X(X(Y)) 
  f#(/\x.X(x), Y) >? X(Y) 

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

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

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) 
  f# = \G0y1.0 

Using this interpretation, the requirements translate to:

  [[#argfun-f##((/\x._x0(x)) ((/\y._x0(y)) _x1), (/\z._x0(z)) _x1)]] = 3 + max(x1, F0(x1), F0(max(x1, F0(x1)))) > F0(F0(x1)) = [[_x0(_x0(_x1))]] 
  [[#argfun-f##((/\x._x0(x)) ((/\y._x0(y)) _x1), (/\z._x0(z)) _x1)]] = 3 + max(x1, F0(x1), F0(max(x1, F0(x1)))) > F0(x1) = [[_x0(_x1)]] 

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.