We consider the system Applicative_first_order_05__#3.2.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    minus : [a * a] --> a 
    nil : [] --> d 
    pred : [a] --> a 
    quot : [a * a] --> a 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    pred(s(x)) => x 
    minus(x, 0) => x 
    minus(x, s(y)) => pred(minus(x, y)) 
    quot(0, s(x)) => 0 
    quot(s(x), s(y)) => s(quot(minus(x, y), s(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) 

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:

  pred(s(X)) => X 
  minus(X, 0) => X 
  minus(X, s(Y)) => pred(minus(X, Y)) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(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 nattprover, this system is indeed terminating:

 || Input TRS:
 ||     1: pred(s(PeRCenTX)) -> PeRCenTX
 ||     2: minus(PeRCenTX,0()) -> PeRCenTX
 ||     3: minus(PeRCenTX,s(PeRCenTY)) -> pred(minus(PeRCenTX,PeRCenTY))
 ||     4: quot(0(),s(PeRCenTX)) -> 0()
 ||     5: quot(s(PeRCenTX),s(PeRCenTY)) -> s(quot(minus(PeRCenTX,PeRCenTY),s(PeRCenTY)))
 || Number of strict rules: 5
 || Direct POLO(bPol) ... failed.
 || Uncurrying ... failed.
 || Dependency Pairs:
 ||    #1: #quot(s(PeRCenTX),s(PeRCenTY)) -> #quot(minus(PeRCenTX,PeRCenTY),s(PeRCenTY))
 ||    #2: #quot(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY)
 ||    #3: #minus(PeRCenTX,s(PeRCenTY)) -> #pred(minus(PeRCenTX,PeRCenTY))
 ||    #4: #minus(PeRCenTX,s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY)
 || Number of SCCs: 2, DPs: 2
 ||   SCC { #4 }
 || POLO(Sum)... succeeded.
 ||       s 	w: x1 + 1
 ||       minus	w: 0
 ||       pred	w: 0
 ||       0 	w: 0
 ||       quot	w: 0
 ||       #minus	w: x2
 ||       #pred	w: 0
 ||       #quot	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #4
 || Number of SCCs: 1, DPs: 1
 ||   SCC { #1 }
 || POLO(Sum)... succeeded.
 ||       s 	w: x1 + 2
 ||       minus	w: x1 + 1
 ||       pred	w: x1
 ||       0 	w: 1
 ||       quot	w: 0
 ||       #minus	w: 0
 ||       #pred	w: 0
 ||       #quot	w: x1
 ||     USABLE RULES: { 1..3 }
 ||     Removed DPs: #1
 || Number of SCCs: 0, DPs: 0
 || 
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:

    pred(s(X)) => X 
    minus(X, 0) => X 
    minus(X, s(Y)) => pred(minus(X, Y)) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(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.