We consider the system AotoYamada_05__020.

  Alphabet:

    0 : [] --> a 
    comp : [b -> b * b -> b] --> b -> b 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    twice : [b -> b] --> b -> b 

  Rules:

    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    times(0, x) => 0 
    times(s(x), y) => plus(times(x, y), y) 
    comp(f, g) x => f (g x) 
    twice(f) => comp(f, f) 

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:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  times(0, X) => 0 
  times(s(X), Y) => plus(times(X, Y), Y) 

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) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    times(0, %X) -> 0
 ||    times(s(%X), %Y) -> plus(times(%X, %Y), %Y)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || times_2 > plus_2 > s_1
 || times_2 > 0 > s_1
 || 
 || 
 || Status:
 || plus_2: multiset status
 || 0: multiset status
 || s_1: multiset status
 || times_2: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    times(0, %X) -> 0
 ||    times(s(%X), %Y) -> plus(times(%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.

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] comp(F, G) X =#> F(G X)   
    1] comp(F, G) X =#> G(X)   
    2] twice(F) X =#> comp(F, F) X   
    3] twice#(F) =#> comp#(F, F)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(times(X, Y), Y) 
    comp(F, G) X => F (G X) 
    twice(F) => comp(F, F) 

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, 2, 3 
    * 1 : 0, 1, 2, 3 
    * 2 : 0, 1 
    * 3 :  

This graph has the following strongly connected components:

  P_1:

    comp(F, G) X =#> F(G X)   
    comp(F, G) X =#> G(X)   
    twice(F) X =#> comp(F, F) 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).

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:

  comp(F, G, X) >? F(G X) 
  comp(F, G, X) >? G(X) 
  twice(F, X) >? comp(F, F, X) 

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

  pi( comp(F, G, X) ) = #argfun-comp#(F (G X), G X) 
  pi( twice(F, X) ) = #argfun-twice#(#argfun-comp#(F (F X), F X)) 

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

The following interpretation satisfies the requirements:

  #argfun-comp# = \y0y1.3 + max(y0, y1) 
  #argfun-twice# = \y0.3 + y0 
  comp = \G0G1y2.0 
  twice = \G0y1.0 

Using this interpretation, the requirements translate to:

  [[#argfun-comp#(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F0(max(x2, F1(x2))) = [[_F0(_F1 _x2)]] 
  [[#argfun-comp#(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F1(x2) = [[_F1(_x2)]] 
  [[#argfun-twice#(#argfun-comp#(_F0 (_F0 _x1), _F0 _x1))]] = 6 + max(x1, F0(x1), F0(max(x1, F0(x1)))) > 3 + max(x1, F0(x1), F0(max(x1, F0(x1)))) = [[#argfun-comp#(_F0 (_F0 _x1), _F0 _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.