We consider the system 726.

  Alphabet:

    0 : [] --> nat 
    dom : [nat * nat * nat] --> nat 
    eval : [nat * nat] --> nat 
    fun : [nat -> nat * nat * nat] --> nat 
    s : [nat] --> nat 

  Rules:

    dom(s(X), s(Y), s(Z)) => s(dom(X, Y, Z)) 
    dom(0, s(X), s(Y)) => s(dom(0, X, Y)) 
    dom(X, Y, 0) => X 
    dom(0, 0, X) => 0 
    eval(fun(/\x.X(x), Y, Z), U) => X(dom(Y, Z, U)) 

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

  dom(s(X), s(Y), s(Z)) => s(dom(X, Y, Z)) 
  dom(0, s(X), s(Y)) => s(dom(0, X, Y)) 
  dom(X, Y, 0) => X 
  dom(0, 0, X) => 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) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    dom(s(%X), s(%Y), s(%Z)) -> s(dom(%X, %Y, %Z))
 ||    dom(0, s(%X), s(%Y)) -> s(dom(0, %X, %Y))
 ||    dom(%X, %Y, 0) -> %X
 ||    dom(0, 0, %X) -> 0
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 2
 ||    POL(dom(x_1, x_2, x_3)) = 1 + 2*x_1 + 2*x_2 + x_3
 ||    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:
 || 
 ||    dom(s(%X), s(%Y), s(%Z)) -> s(dom(%X, %Y, %Z))
 ||    dom(0, s(%X), s(%Y)) -> s(dom(0, %X, %Y))
 ||    dom(%X, %Y, 0) -> %X
 ||    dom(0, 0, %X) -> 0
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%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 dynamic dependency pairs.

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

  Dependency Pairs P_0:

    0] eval#(fun(/\x.X(x), Y, Z), U) =#> X(dom(Y, Z, U))   
    1] eval#(fun(/\x.X(x), Y, Z), U) =#> dom#(Y, Z, U) {X : 1}  

  Rules R_0:

    dom(s(X), s(Y), s(Z)) => s(dom(X, Y, Z)) 
    dom(0, s(X), s(Y)) => s(dom(0, X, Y)) 
    dom(X, Y, 0) => X 
    dom(0, 0, X) => 0 
    eval(fun(/\x.X(x), Y, Z), U) => X(dom(Y, Z, U)) 

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:

    eval#(fun(/\x.X(x), Y, Z), U) =#> X(dom(Y, Z, U))   

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 ::=

  dom(X, Y, 0) => X 
  dom(0, 0, X) => 0 
  eval(fun(/\x.X(x), Y, Z), U) => X(dom(Y, Z, U)) 

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:

  eval#(fun(/\x.X(x), Y, Z), U) >? X(dom(Y, Z, U)) 
  dom(X, Y, 0) >= X 
  dom(0, 0, X) >= 0 
  eval(fun(/\x.X(x), Y, Z), U) >= X(dom(Y, Z, U)) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  dom = \y0y1y2.y0 
  eval = \y0y1.y0 
  eval# = \y0y1.3 + y0 
  fun = \G0y1y2.3 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[eval#(fun(/\x._x0(x), _x1, _x2), _x3)]] = 6 + F0(x1) > F0(x1) = [[_x0(dom(_x1, _x2, _x3))]] 
  [[dom(_x0, _x1, 0)]] = x0 >= x0 = [[_x0]] 
  [[dom(0, 0, _x0)]] = 0 >= 0 = [[0]] 
  [[eval(fun(/\x._x0(x), _x1, _x2), _x3)]] = 3 + F0(x1) >= F0(x1) = [[_x0(dom(_x1, _x2, _x3))]] 

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.