We consider the system sort.

  Alphabet:

    0 : [] --> nat 
    ascending!6220sort : [list] --> list 
    cons : [nat * list] --> list 
    descending!6220sort : [list] --> list 
    insert : [nat * list * nat -> nat -> nat * nat -> nat -> nat] --> list 
    max : [nat * nat] --> nat 
    min : [nat * nat] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 
    sort : [list * nat -> nat -> nat * nat -> nat -> nat] --> list 

  Rules:

    max(0, x) => x 
    max(x, 0) => x 
    max(s(x), s(y)) => s(max(x, y)) 
    min(0, x) => 0 
    min(x, 0) => 0 
    min(s(x), s(y)) => s(min(x, y)) 
    insert(x, nil, f, g) => cons(x, nil) 
    insert(x, cons(y, z), f, g) => cons(f x y, insert(g x y, z, f, g)) 
    sort(nil, f, g) => nil 
    sort(cons(x, y), f, g) => insert(x, sort(y, f, g), f, g) 
    ascending!6220sort(x) => sort(x, /\y./\z.min(y, z), /\u./\v.max(u, v)) 
    descending!6220sort(x) => sort(x, /\y./\z.max(y, z), /\u./\v.min(u, v)) 

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:

  max(0, X) => X 
  max(X, 0) => X 
  max(s(X), s(Y)) => s(max(X, Y)) 
  min(0, X) => 0 
  min(X, 0) => 0 
  min(s(X), s(Y)) => s(min(X, Y)) 

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:
 || 
 ||    max(0, %X) -> %X
 ||    max(%X, 0) -> %X
 ||    max(s(%X), s(%Y)) -> s(max(%X, %Y))
 ||    min(0, %X) -> 0
 ||    min(%X, 0) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:~PAIR_2 > 0 > min_2 > max_2 > s_1
 || 
 || and weight map:
 || 
 ||    0=1
 ||    s_1=1
 ||    max_2=0
 ||    min_2=0
 ||    ~PAIR_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:
 || 
 ||    max(0, %X) -> %X
 ||    max(%X, 0) -> %X
 ||    max(s(%X), s(%Y)) -> s(max(%X, %Y))
 ||    min(0, %X) -> 0
 ||    min(%X, 0) -> 0
 ||    min(s(%X), s(%Y)) -> s(min(%X, %Y))
 ||    ~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] insert#(X, cons(Y, Z), F, G) =#> insert#(G X Y, Z, F, G)   
    1] sort#(cons(X, Y), F, G) =#> insert#(X, sort(Y, F, G), F, G)   
    2] sort#(cons(X, Y), F, G) =#> sort#(Y, F, G)   
    3] ascending!6220sort#(X) =#> sort#(X, /\x./\y.min(x, y), /\z./\u.max(z, u))   
    4] ascending!6220sort#(X) =#> min#(Y, Z)   
    5] ascending!6220sort#(X) =#> max#(Y, Z)   
    6] descending!6220sort#(X) =#> sort#(X, /\x./\y.max(x, y), /\z./\u.min(z, u))   
    7] descending!6220sort#(X) =#> max#(Y, Z)   
    8] descending!6220sort#(X) =#> min#(Y, Z)   

  Rules R_0:

    max(0, X) => X 
    max(X, 0) => X 
    max(s(X), s(Y)) => s(max(X, Y)) 
    min(0, X) => 0 
    min(X, 0) => 0 
    min(s(X), s(Y)) => s(min(X, Y)) 
    insert(X, nil, F, G) => cons(X, nil) 
    insert(X, cons(Y, Z), F, G) => cons(F X Y, insert(G X Y, Z, F, G)) 
    sort(nil, F, G) => nil 
    sort(cons(X, Y), F, G) => insert(X, sort(Y, F, G), F, G) 
    ascending!6220sort(X) => sort(X, /\x./\y.min(x, y), /\z./\u.max(z, u)) 
    descending!6220sort(X) => sort(X, /\x./\y.max(x, y), /\z./\u.min(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 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 : 1, 2 
    * 3 : 1, 2 
    * 4 :  
    * 5 :  
    * 6 : 1, 2 
    * 7 :  
    * 8 :  

This graph has the following strongly connected components:

  P_1:

    insert#(X, cons(Y, Z), F, G) =#> insert#(G X Y, Z, F, G)   

  P_2:

    sort#(cons(X, Y), F, G) =#> sort#(Y, F, G)   

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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_2, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(sort#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sort#(cons(X, Y), F, G)) = cons(X, Y) |> Y = nu(sort#(Y, F, G)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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(insert#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(insert#(X, cons(Y, Z), F, G)) = cons(Y, Z) |> Z = nu(insert#(G X Y, Z, F, G)) 

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.