We consider the system Applicative_first_order_05__#3.45.

  Alphabet:

    0 : [] --> d 
    cons : [d * d] --> d 
    f : [d] --> a 
    false : [] --> c 
    filter : [d -> c * d] --> d 
    filter2 : [c * d -> c * d * d] --> d 
    g : [d] --> d 
    h : [d] --> b 
    map : [d -> d * d] --> d 
    nil : [] --> d 
    s : [d] --> d 
    true : [] --> c 

  Rules:

    f(s(x)) => f(x) 
    g(cons(0, x)) => g(x) 
    g(cons(s(x), y)) => s(x) 
    h(cons(x, y)) => h(g(cons(x, y))) 
    map(i, nil) => nil 
    map(i, cons(x, y)) => cons(i x, map(i, y)) 
    filter(i, nil) => nil 
    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
    filter2(true, i, x, y) => cons(x, filter(i, y)) 
    filter2(false, i, x, y) => filter(i, 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:

  f(s(X)) => f(X) 
  g(cons(0, X)) => g(X) 
  g(cons(s(X), Y)) => s(X) 
  h(cons(X, Y)) => h(g(cons(X, Y))) 

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) AAECC Innermost [EQUIVALENT]
 || (6) QTRS
 || (7) DependencyPairsProof [EQUIVALENT]
 || (8) QDP
 || (9) DependencyGraphProof [EQUIVALENT]
 || (10) TRUE
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    f(s(%X)) -> f(%X)
 ||    g(cons(0, %X)) -> g(%X)
 ||    g(cons(s(%X), %Y)) -> s(%X)
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 1
 ||    POL(cons(x_1, x_2)) = x_1 + 2*x_2
 ||    POL(f(x_1)) = x_1
 ||    POL(g(x_1)) = x_1
 ||    POL(h(x_1)) = 2*x_1
 ||    POL(s(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:
 || 
 ||    g(cons(0, %X)) -> g(%X)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    f(s(%X)) -> f(%X)
 ||    g(cons(s(%X), %Y)) -> s(%X)
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(cons(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(f(x_1)) = x_1
 ||    POL(g(x_1)) = x_1
 ||    POL(h(x_1)) = x_1
 ||    POL(s(x_1)) = 1 + x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    f(s(%X)) -> f(%X)
 ||    g(cons(s(%X), %Y)) -> s(%X)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) AAECC Innermost (EQUIVALENT)
 || We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
 || 
 || The TRS R 2 is 
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || The signature Sigma is {h_1}
 || ----------------------------------------
 || 
 || (6)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    h(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (7) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (8)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    H(cons(%X, %Y)) -> H(g(cons(%X, %Y)))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    h(cons(%X, %Y)) -> h(g(cons(%X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    h(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (9) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (10)
 || TRUE
 || 
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)   
    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    f(s(X)) => f(X) 
    g(cons(0, X)) => g(X) 
    g(cons(s(X), Y)) => s(X) 
    h(cons(X, Y)) => h(g(cons(X, Y))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(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 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 : 2, 3 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, Y)   

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(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by (P_3, R_0, computable, f), where P_3 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_3, 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 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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(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_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.