We consider the system AotoYamada_05__005.

  Alphabet:

    0 : [] --> a 
    add : [] --> a -> a -> a 
    curry : [a -> a -> a] --> a -> a -> a 
    plus : [] --> a -> a -> a 
    s : [a] --> a 

  Rules:

    plus 0 x => x 
    plus s(x) y => s(plus x y) 
    curry(f) x y => f x y 
    add => curry(plus) 

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) 

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))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 2
 ||    POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(s(x_1)) = 1 + x_1
 || 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))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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] curry(F) X Y =#> F(X, Y)   
    1] add X =#> curry(plus) X   
    2] add X Y =#> curry(plus) X Y   
    3] add# =#> curry#(plus)   
    4] add# =#> plus#   

  Rules R_0:

    plus 0 X => X 
    plus s(X) Y => s(plus X Y) 
    curry(F) X Y => F X Y 
    add => curry(plus) 

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, 4 
    * 1 :  
    * 2 : 0 
    * 3 :  
    * 4 :  

This graph has the following strongly connected components:

  P_1:

    curry(F) X Y =#> F(X, Y)   
    add X Y =#> curry(plus) X Y   

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:

  curry(F, X, Y) >? F(X, Y) 
  add(X, Y) >? curry(plus, X, Y) 

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

  pi( curry(F, X, Y) ) = #argfun-curry#(F X Y) 
  pi( add(X, Y) ) = #argfun-add#(#argfun-curry#(plus X Y)) 

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

The following interpretation satisfies the requirements:

  #argfun-add# = \y0.3 + y0 
  #argfun-curry# = \y0.3 + y0 
  add = \y0y1.0 
  curry = \G0y1y2.0 
  plus = \y0y1.0 

Using this interpretation, the requirements translate to:

  [[#argfun-curry#(_F0 _x1 _x2)]] = 3 + max(x1, x2, F0(x1,x2)) > F0(x1,x2) = [[_F0(_x1, _x2)]] 
  [[#argfun-add#(#argfun-curry#(plus _x0 _x1))]] = 6 + max(x0, x1) > 3 + max(x0, x1) = [[#argfun-curry#(plus _x0 _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.