We consider the system 519.

  Alphabet:

    0 : [string] --> string 
    1 : [string] --> string 
    false : [] --> bool 
    fold : [bool -> bool * bool -> bool * bool * string] --> bool 
    nil : [] --> string 
    not : [bool] --> bool 
    true : [] --> bool 

  Rules:

    not(true) => false 
    not(false) => true 
    fold(/\x.X(x), /\y.Y(y), Z, nil) => Z 
    fold(/\x.X(x), /\y.Y(y), Z, 0(U)) => fold(/\z.X(z), /\u.Y(u), X(Z), U) 
    fold(/\x.X(x), /\y.Y(y), Z, 1(U)) => fold(/\z.X(z), /\u.Y(u), Y(Z), U) 

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

  not(true) => false 
  not(false) => 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) QTRS Reverse [EQUIVALENT]
 || (2) QTRS
 || (3) RFCMatchBoundsTRSProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    not(true) -> false
 ||    not(false) -> true
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRS Reverse (EQUIVALENT)
 || We applied the QTRS Reverse Processor [REVERSE].
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    true'(not(x)) -> false'(x)
 ||    false'(not(x)) -> true'(x)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RFCMatchBoundsTRSProof (EQUIVALENT)
 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R.
 || The following rules were used to construct the certificate:
 || 
 ||    true'(not(x)) -> false'(x)
 ||    false'(not(x)) -> true'(x)
 || 
 || The certificate found is represented by the following graph.
 || The certificate consists of the following enumerated nodes:
 || 2, 4
 || 
 || Node 2 is start node and node 4 is final node.
 || 
 || Those nodes are connected through the following edges:
 || 
 || * 2 to 4 labelled false'_1(0), true'_1(0), true'_1(1), false'_1(1)* 4 to 4 labelled #_1(0)
 || 
 || 
 || ----------------------------------------
 || 
 || (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] fold#(/\x.X(x), /\y.Y(y), Z, 0(U)) =#> fold#(/\z.X(z), /\u.Y(u), X(Z), U)   
    1] fold#(/\x.X(x), /\y.Y(y), Z, 1(U)) =#> fold#(/\z.X(z), /\u.Y(u), Y(Z), U)   

  Rules R_0:

    not(true) => false 
    not(false) => true 
    fold(/\x.X(x), /\y.Y(y), Z, nil) => Z 
    fold(/\x.X(x), /\y.Y(y), Z, 0(U)) => fold(/\z.X(z), /\u.Y(u), X(Z), U) 
    fold(/\x.X(x), /\y.Y(y), Z, 1(U)) => fold(/\z.X(z), /\u.Y(u), Y(Z), U) 

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 apply the subterm criterion with the following projection function:

  nu(fold#) = 4 

Thus, we can orient the dependency pairs as follows:

  nu(fold#(/\x.X(x), /\y.Y(y), Z, 0(U))) = 0(U) |> U = nu(fold#(/\z.X(z), /\u.Y(u), X(Z), U)) 
  nu(fold#(/\x.X(x), /\y.Y(y), Z, 1(U))) = 1(U) |> U = nu(fold#(/\z.X(z), /\u.Y(u), Y(Z), U)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_0, 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.