We consider the system 02Ackermann.

  Alphabet:

    ack : [N * N] --> N 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    ack(z, x) => s(x) 
    ack(s(x), z) => ack(x, s(z)) 
    ack(s(x), s(y)) => ack(x, ack(s(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:

  ack(z, X) => s(X) 
  ack(s(X), z) => ack(X, s(z)) 
  ack(s(X), s(Y)) => ack(X, ack(s(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:
 || 
 ||    ack(z, %X) -> s(%X)
 ||    ack(s(%X), z) -> ack(%X, s(z))
 ||    ack(s(%X), s(%Y)) -> ack(%X, ack(s(%X), %Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || [ack_2, z] > s_1
 || 
 || 
 || Status:
 || ack_2: [1,2]
 || z: multiset status
 || s_1: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    ack(z, %X) -> s(%X)
 ||    ack(s(%X), z) -> ack(%X, s(z))
 ||    ack(s(%X), s(%Y)) -> ack(%X, ack(s(%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:


  Rules R_0:

    ack(z, X) => s(X) 
    ack(s(X), z) => ack(X, s(z)) 
    ack(s(X), s(Y)) => ack(X, ack(s(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:


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.