We consider the system Applicative_05__mapDivMinusHard.

  Alphabet:

    0 : [] --> c 
    cons : [a * b] --> b 
    div : [c * c] --> c 
    map : [a -> a * b] --> b 
    minus : [c * c] --> c 
    nil : [] --> b 
    p : [c] --> c 
    s : [c] --> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(p(s(x)), p(s(y))) 
    p(s(x)) => x 
    div(0, s(x)) => 0 
    div(s(x), s(y)) => s(div(minus(x, y), s(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:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(p(s(X)), p(s(Y))) 
  p(s(X)) => X 
  div(0, s(X)) => 0 
  div(s(X), s(Y)) => s(div(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: minus(PeRCenTX,0()) -> PeRCenTX
 ||     2: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(p(s(PeRCenTX)),p(s(PeRCenTY)))
 ||     3: p(s(PeRCenTX)) -> PeRCenTX
 ||     4: div(0(),s(PeRCenTX)) -> 0()
 ||     5: div(s(PeRCenTX),s(PeRCenTY)) -> s(div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY)))
 || Number of strict rules: 5
 || Direct POLO(bPol) ... failed.
 || Uncurrying p
 || 1: minus(PeRCenTX,0()) -> PeRCenTX
 || 2: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(p^1_s(PeRCenTX),p^1_s(PeRCenTY))
 || 3: p^1_s(PeRCenTX) -> PeRCenTX
 || 4: div(0(),s(PeRCenTX)) -> 0()
 || 5: div(s(PeRCenTX),s(PeRCenTY)) -> s(div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY)))
 || 6: p(s(_1)) ->= p^1_s(_1)
 || Number of strict rules: 5
 || Direct POLO(bPol) ... failed.
 || Dependency Pairs:
 ||    #1: #minus(s(PeRCenTX),s(PeRCenTY)) -> #minus(p^1_s(PeRCenTX),p^1_s(PeRCenTY))
 ||    #2: #minus(s(PeRCenTX),s(PeRCenTY)) -> #p^1_s(PeRCenTX)
 ||    #3: #minus(s(PeRCenTX),s(PeRCenTY)) -> #p^1_s(PeRCenTY)
 ||    #4: #p(s(_1)) ->? #p^1_s(_1)
 ||    #5: #div(s(PeRCenTX),s(PeRCenTY)) -> #div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY))
 ||    #6: #div(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY)
 || Number of SCCs: 2, DPs: 2
 ||   SCC { #1 }
 || POLO(Sum)... succeeded.
 ||       #div	w: 0
 ||       s 	w: x1 + 2
 ||       #p^1_s	w: 0
 ||       minus	w: 0
 ||       div	w: 0
 ||       p^1_s	w: x1 + 1
 ||       #p 	w: 0
 ||       p 	w: 0
 ||       0 	w: 0
 ||       #minus	w: x1 + x2
 ||     USABLE RULES: { 3 }
 ||     Removed DPs: #1
 || Number of SCCs: 1, DPs: 1
 ||   SCC { #5 }
 || POLO(Sum)... succeeded.
 ||       #div	w: x1
 ||       s 	w: x1 + 2
 ||       #p^1_s	w: 0
 ||       minus	w: x1 + 1
 ||       div	w: 0
 ||       p^1_s	w: x1 + 1
 ||       #p 	w: 0
 ||       p 	w: 0
 ||       0 	w: 1
 ||       #minus	w: x1 + x2
 ||     USABLE RULES: { 1..3 }
 ||     Removed DPs: #5
 || 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)   

  Rules R_0:

    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(p(s(X)), p(s(Y))) 
    p(s(X)) => X 
    div(0, s(X)) => 0 
    div(s(X), s(Y)) => s(div(minus(X, Y), s(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.