We consider the system uncurry.

  Alphabet:

    f : [] --> a -> b -> c 
    f1 : [a] --> b -> c 
    f2 : [a * b] --> c 

  Rules:

    f x => f1(x) 
    f1(x) y => f2(x, y) 

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:

  f1(X) Y => f2(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:
 || 
 ||    f1(%X, %Y) -> f2(%X, %Y)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(f1(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(f2(x_1, x_2)) = 1 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    f1(%X, %Y) -> f2(%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] f(X) Y =#> f1(X) Y   
    1] f#(X) =#> f1#(X)   

  Rules R_0:

    f(X) => f1(X) 
    f1(X) Y => f2(X, Y) 

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 :  

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.