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).

Symbol curry is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:

  Alphabet:

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

  Rules:

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

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.

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

  Dependency Pairs P_0:

    0] add X =#> plus X   
    1] add X Y =#> plus X Y   
    2] add# =#> plus#   

  Rules R_0:

    plus 0 X => X 
    plus s(X) Y => s(plus X Y) 
    add => 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 :  
    * 1 :  
    * 2 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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.