We consider the system Applicative_first_order_05__#3.38.

  Alphabet:

    0 : [] --> c 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    rev : [d] --> d 
    rev1 : [c * d] --> c 
    rev2 : [c * d] --> d 
    s : [a] --> c 
    true : [] --> b 

  Rules:

    rev(nil) => nil 
    rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) 
    rev1(0, nil) => 0 
    rev1(s(x), nil) => s(x) 
    rev1(x, cons(y, z)) => rev1(y, z) 
    rev2(x, nil) => nil 
    rev2(x, cons(y, z)) => rev(cons(x, rev2(y, z))) 
    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) 

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:

  rev(nil) => nil 
  rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
  rev1(0, nil) => 0 
  rev1(s(X), nil) => s(X) 
  rev1(X, cons(Y, Z)) => rev1(Y, Z) 
  rev2(X, nil) => nil 
  rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 

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 mutermprover, this system is indeed terminating:

 || 
 || Problem 1: 
 || 
 || (VAR %X %Y %Z)
 || (RULES
 || rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 || rev(nil) -> nil
 || rev1(0,nil) -> 0
 || rev1(s(%X),nil) -> s(%X)
 || rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 || rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 || rev2(%X,nil) -> nil
 || )
 || 
 || Problem 1: 
 || 
 || Innermost Equivalent Processor:
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 ||  
 || 
 || Problem 1: 
 || 
 || Dependency Pairs Processor:
 || -> Pairs:
 ||  REV(cons(%X,%Y)) -> REV1(%X,%Y)
 ||  REV(cons(%X,%Y)) -> REV2(%X,%Y)
 ||  REV1(%X,cons(%Y,%Z)) -> REV1(%Y,%Z)
 ||  REV2(%X,cons(%Y,%Z)) -> REV(cons(%X,rev2(%Y,%Z)))
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  REV(cons(%X,%Y)) -> REV1(%X,%Y)
 ||  REV(cons(%X,%Y)) -> REV2(%X,%Y)
 ||  REV1(%X,cons(%Y,%Z)) -> REV1(%Y,%Z)
 ||  REV2(%X,cons(%Y,%Z)) -> REV(cons(%X,rev2(%Y,%Z)))
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  REV1(%X,cons(%Y,%Z)) -> REV1(%Y,%Z)
 || ->->-> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->->Cycle:
 || ->->-> Pairs:
 ||  REV(cons(%X,%Y)) -> REV2(%X,%Y)
 ||  REV2(%X,cons(%Y,%Z)) -> REV(cons(%X,rev2(%Y,%Z)))
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || ->->-> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || 
 || 
 || The problem is decomposed in 2 subproblems.
 || 
 || Problem 1.1: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  REV1(%X,cons(%Y,%Z)) -> REV1(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Projection:
 ||  pi(REV1) = 2
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.2: 
 || 
 || Reduction Pairs Processor:
 || -> Pairs:
 ||  REV(cons(%X,%Y)) -> REV2(%X,%Y)
 ||  REV2(%X,cons(%Y,%Z)) -> REV(cons(%X,rev2(%Y,%Z)))
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || -> Usable rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [rev](X) = X
 || [rev1](X1,X2) = X2 + 2
 || [rev2](X1,X2) = X2
 || [0] = 0
 || [cons](X1,X2) = 2.X2 + 2
 || [nil] = 1
 || [s](X) = 2
 || [REV](X) = 2.X + 1
 || [REV2](X1,X2) = 2.X2 + 2
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  REV2(%X,cons(%Y,%Z)) -> REV(cons(%X,rev2(%Y,%Z)))
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || ->->-> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || 
 || Problem 1.2: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  REV2(%X,cons(%Y,%Z)) -> REV2(%Y,%Z)
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Projection:
 ||  pi(REV2) = 2
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  rev(cons(%X,%Y)) -> cons(rev1(%X,%Y),rev2(%X,%Y))
 ||  rev(nil) -> nil
 ||  rev1(0,nil) -> 0
 ||  rev1(s(%X),nil) -> s(%X)
 ||  rev1(%X,cons(%Y,%Z)) -> rev1(%Y,%Z)
 ||  rev2(%X,cons(%Y,%Z)) -> rev(cons(%X,rev2(%Y,%Z)))
 ||  rev2(%X,nil) -> nil
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
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:

    rev(nil) => nil 
    rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
    rev1(0, nil) => 0 
    rev1(s(X), nil) => s(X) 
    rev1(X, cons(Y, Z)) => rev1(Y, Z) 
    rev2(X, nil) => nil 
    rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 
    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.