We consider the system 520.

  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) 
    0(1(X)) => 1(0(X)) 

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

  not(true) => false 
  not(false) => true 
  0(1(X)) => 1(0(X)) 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-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:
 || 
 ||    not(true) -> false
 ||    not(false) -> true
 ||    0(1(%X)) -> 1(0(%X))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0(x_1)) = 2 + 2*x_1
 ||    POL(1(x_1)) = 1 + x_1
 ||    POL(false) = 2
 ||    POL(not(x_1)) = 2 + 2*x_1
 ||    POL(true) = 1
 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    not(true) -> false
 ||    not(false) -> true
 ||    0(1(%X)) -> 1(0(%X))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %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] 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) 
    0(1(X)) => 1(0(X)) 

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.