We consider the system extrec.

  Alphabet:

    !plus : [nat * nat] --> nat 
    !times : [nat * nat] --> nat 
    0 : [] --> nat 
    rec : [nat * nat * nat -> nat -> nat] --> nat 
    s : [nat] --> nat 

  Rules:

    !plus(x, 0) => x 
    !plus(x, s(y)) => s(!plus(x, y)) 
    rec(0, x, f) => x 
    rec(s(x), y, f) => f x rec(x, y, f) 
    !times(x, y) => rec(y, 0, /\z./\u.!plus(x, u)) 

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(X, 0) => X 
  !plus(X, s(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(%X, 0) -> %X
 ||    !plus(%X, s(%Y)) -> s(!plus(%X, %Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(!plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(0) = 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(%X, 0) -> %X
 ||    !plus(%X, s(%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 static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]).

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

  Dependency Pairs P_0:

    0] rec#(s(X), Y, F) =#> rec#(X, Y, F)   
    1] !times#(X, Y) =#> rec#(Y, 0, /\x./\y.!plus(X, y))   
    2] !times#(X, Y) =#> !plus#(X, Z)   

  Rules R_0:

    !plus(X, 0) => X 
    !plus(X, s(Y)) => s(!plus(X, Y)) 
    rec(0, X, F) => X 
    rec(s(X), Y, F) => F X rec(X, Y, F) 
    !times(X, Y) => rec(Y, 0, /\x./\y.!plus(X, y)) 

Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_0, R_0, computable, 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 : 0 
    * 2 :  

This graph has the following strongly connected components:

  P_1:

    rec#(s(X), Y, F) =#> rec#(X, Y, F)   

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, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(rec#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(rec#(s(X), Y, F)) = s(X) |> X = nu(rec#(X, Y, F)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, f).  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 +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.