We consider the system fuhkop11frocos.

  Alphabet:

    append : [list * list] --> list 
    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    mirror : [list] --> list 
    nil : [] --> list 
    reverse : [list] --> list 
    shuffle : [list] --> list 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    reverse(nil) => nil 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    mirror(nil) => nil 
    mirror(cons(x, y)) => append(cons(x, mirror(y)), cons(x, nil)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 

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:

  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
  reverse(nil) => nil 
  shuffle(nil) => nil 
  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
  mirror(nil) => nil 
  mirror(cons(X, Y)) => append(cons(X, mirror(Y)), cons(X, nil)) 

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) QTRSRRRProof [EQUIVALENT]
 || (6) QTRS
 || (7) QTRSRRRProof [EQUIVALENT]
 || (8) QTRS
 || (9) RisEmptyProof [EQUIVALENT]
 || (10) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    shuffle(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 ||    mirror(nil) -> nil
 ||    mirror(cons(%X, %Y)) -> append(cons(%X, mirror(%Y)), cons(%X, nil))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(append(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(mirror(x_1)) = 2*x_1
 ||    POL(nil) = 0
 ||    POL(reverse(x_1)) = x_1
 ||    POL(shuffle(x_1)) = 1 + 2*x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    shuffle(nil) -> nil
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 ||    mirror(nil) -> nil
 ||    mirror(cons(%X, %Y)) -> append(cons(%X, mirror(%Y)), cons(%X, nil))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(append(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(mirror(x_1)) = 1 + 2*x_1
 ||    POL(nil) = 0
 ||    POL(reverse(x_1)) = x_1
 ||    POL(shuffle(x_1)) = 2*x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    mirror(nil) -> nil
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 ||    mirror(cons(%X, %Y)) -> append(cons(%X, mirror(%Y)), cons(%X, nil))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || append/2(YES,YES)
 || nil/0)
 || cons/2(YES,YES)
 || reverse/1)YES(
 || shuffle/1)YES(
 || mirror/1(YES)
 || 
 || Quasi precedence:
 || mirror_1 > append_2 > cons_2
 || mirror_1 > nil
 || 
 || 
 || Status:
 || append_2: [2,1]
 || nil: multiset status
 || cons_2: [2,1]
 || mirror_1: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    mirror(cons(%X, %Y)) -> append(cons(%X, mirror(%Y)), cons(%X, nil))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (6)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    reverse(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (7) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:reverse_1 > shuffle_1 > cons_2 > nil
 || 
 || and weight map:
 || 
 ||    nil=1
 ||    reverse_1=0
 ||    shuffle_1=1
 ||    cons_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:
 || 
 ||    reverse(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (8)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (9) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (10)
 || 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] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    reverse(nil) => nil 
    shuffle(nil) => nil 
    shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
    mirror(nil) => nil 
    mirror(cons(X, Y)) => append(cons(X, mirror(Y)), cons(X, nil)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, 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 apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

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

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.