We consider the system 1037.

  Alphabet:

    0 : [] --> nat 
    rec : [nat * nat * nat -> nat -> nat] --> nat 
    s : [nat] --> nat 
    v : [var] --> nat 
    xplus : [nat * nat] --> nat 
    xtimes : [nat * nat] --> nat 

  Rules:

    xplus(X, 0) => X 
    xplus(X, s(Y)) => s(xplus(X, Y)) 
    rec(0, X, /\x./\y.Y(x, y)) => X 
    rec(s(v(X)), Y, /\x./\y.Z(x, y)) => Z(v(X), rec(v(X), Y, /\z./\u.Z(z, u))) 
    xtimes(X, Y) => rec(Y, 0, /\x./\y.xplus(X, y)) 

We observe that the rules contain a first-order subset:

  xplus(X, 0) => X 
  xplus(X, s(Y)) => s(xplus(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:
 || 
 ||    xplus(%X, 0) -> %X
 ||    xplus(%X, s(%Y)) -> s(xplus(%X, %Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 2
 ||    POL(s(x_1)) = 1 + x_1
 ||    POL(xplus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    xplus(%X, 0) -> %X
 ||    xplus(%X, s(%Y)) -> s(xplus(%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(v(X)), Y, /\x./\y.Z(x, y)) =#> rec#(v(X), Y, /\z./\u.Z(z, u)) {Z : 2}  
    1] xtimes#(X, Y) =#> rec#(Y, 0, /\x./\y.xplus(X, y))   
    2] xtimes#(X, Y) =#> xplus#(X, Z)   

  Rules R_0:

    xplus(X, 0) => X 
    xplus(X, s(Y)) => s(xplus(X, Y)) 
    rec(0, X, /\x./\y.Y(x, y)) => X 
    rec(s(v(X)), Y, /\x./\y.Z(x, y)) => Z(v(X), rec(v(X), Y, /\z./\u.Z(z, u))) 
    xtimes(X, Y) => rec(Y, 0, /\x./\y.xplus(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 :  
    * 1 : 0 
    * 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 +++

[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.