We consider the system 521.

  Alphabet:

    0 : [string] --> string 
    0s : [string] --> string 
    1 : [string] --> string 
    1s : [string] --> string 
    cmp : [string * string] --> bool 
    decide : [string] --> bool 
    false : [] --> bool 
    nil : [] --> string 
    true : [] --> bool 

  Rules:

    decide(X) => cmp(0s(X), 1s(X)) 
    0s(nil) => nil 
    0s(0(X)) => 0(0s(X)) 
    0s(1(X)) => 0s(X) 
    1s(nil) => nil 
    1s(1(X)) => 1(1s(X)) 
    1s(0(X)) => 1s(X) 
    cmp(0(X), 1(Y)) => cmp(X, Y) 
    cmp(0(X), nil) => false 
    cmp(nil, X) => true 

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

  decide(X) => cmp(0s(X), 1s(X)) 
  0s(nil) => nil 
  0s(0(X)) => 0(0s(X)) 
  0s(1(X)) => 0s(X) 
  1s(nil) => nil 
  1s(1(X)) => 1(1s(X)) 
  1s(0(X)) => 1s(X) 
  cmp(0(X), 1(Y)) => cmp(X, Y) 
  cmp(0(X), nil) => false 
  cmp(nil, X) => true 

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) QTRSRRRProof [EQUIVALENT]
 || (4) QTRS
 || (5) RisEmptyProof [EQUIVALENT]
 || (6) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    decide(%X) -> cmp(0s(%X), 1s(%X))
 ||    0s(nil) -> nil
 ||    0s(0(%X)) -> 0(0s(%X))
 ||    0s(1(%X)) -> 0s(%X)
 ||    1s(nil) -> nil
 ||    1s(1(%X)) -> 1(1s(%X))
 ||    1s(0(%X)) -> 1s(%X)
 ||    cmp(0(%X), 1(%Y)) -> cmp(%X, %Y)
 ||    cmp(0(%X), nil) -> false
 ||    cmp(nil, %X) -> true
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0(x_1)) = 1 + 2*x_1
 ||    POL(0s(x_1)) = x_1
 ||    POL(1(x_1)) = x_1
 ||    POL(1s(x_1)) = x_1
 ||    POL(cmp(x_1, x_2)) = x_1 + x_2
 ||    POL(decide(x_1)) = 1 + 2*x_1
 ||    POL(false) = 0
 ||    POL(nil) = 0
 ||    POL(true) = 0
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    decide(%X) -> cmp(0s(%X), 1s(%X))
 ||    1s(0(%X)) -> 1s(%X)
 ||    cmp(0(%X), 1(%Y)) -> cmp(%X, %Y)
 ||    cmp(0(%X), nil) -> false
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    0s(nil) -> nil
 ||    0s(0(%X)) -> 0(0s(%X))
 ||    0s(1(%X)) -> 0s(%X)
 ||    1s(nil) -> nil
 ||    1s(1(%X)) -> 1(1s(%X))
 ||    cmp(nil, %X) -> true
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:0s_1 > 0_1 > cmp_2 > true > 1s_1 > 1_1 > nil
 || 
 || and weight map:
 || 
 ||    nil=1
 ||    true=2
 ||    0s_1=2
 ||    0_1=1
 ||    1_1=1
 ||    1s_1=2
 ||    cmp_2=0
 || 
 || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    0s(nil) -> nil
 ||    0s(0(%X)) -> 0(0s(%X))
 ||    0s(1(%X)) -> 0s(%X)
 ||    1s(nil) -> nil
 ||    1s(1(%X)) -> 1(1s(%X))
 ||    cmp(nil, %X) -> true
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (6)
 || 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:


  Rules R_0:

    decide(X) => cmp(0s(X), 1s(X)) 
    0s(nil) => nil 
    0s(0(X)) => 0(0s(X)) 
    0s(1(X)) => 0s(X) 
    1s(nil) => nil 
    1s(1(X)) => 1(1s(X)) 
    1s(0(X)) => 1s(X) 
    cmp(0(X), 1(Y)) => cmp(X, Y) 
    cmp(0(X), nil) => false 
    cmp(nil, X) => true 

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:


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.