We consider the system Applicative_05__Ex7_9.

  Alphabet:

    0 : [] --> a 
    cons : [b * c] --> c 
    d : [a * a] --> c 
    false : [] --> c 
    filter : [b -> c * c] --> c 
    gtr : [a * a] --> c 
    if : [c * c * c] --> c 
    len : [c] --> a 
    nil : [] --> c 
    s : [a] --> a 
    sub : [a * a] --> a 
    true : [] --> c 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    sub(x, 0) => x 
    sub(s(x), s(y)) => sub(x, y) 
    gtr(0, x) => false 
    gtr(s(x), 0) => true 
    gtr(s(x), s(y)) => gtr(x, y) 
    d(x, 0) => true 
    d(s(x), s(y)) => if(gtr(x, y), false, d(s(x), sub(y, x))) 
    len(nil) => 0 
    len(cons(x, y)) => s(len(y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, 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:

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  sub(X, 0) => X 
  sub(s(X), s(Y)) => sub(X, Y) 
  gtr(0, X) => false 
  gtr(s(X), 0) => true 
  gtr(s(X), s(Y)) => gtr(X, Y) 
  d(X, 0) => true 
  d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
  len(nil) => 0 
  len(cons(X, Y)) => s(len(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: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX
 ||     2: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY
 ||     3: sub(PeRCenTX,0()) -> PeRCenTX
 ||     4: sub(s(PeRCenTX),s(PeRCenTY)) -> sub(PeRCenTX,PeRCenTY)
 ||     5: gtr(0(),PeRCenTX) -> false()
 ||     6: gtr(s(PeRCenTX),0()) -> true()
 ||     7: gtr(s(PeRCenTX),s(PeRCenTY)) -> gtr(PeRCenTX,PeRCenTY)
 ||     8: d(PeRCenTX,0()) -> true()
 ||     9: d(s(PeRCenTX),s(PeRCenTY)) -> if(gtr(PeRCenTX,PeRCenTY),false(),d(s(PeRCenTX),sub(PeRCenTY,PeRCenTX)))
 ||     10: len(nil()) -> 0()
 ||     11: len(cons(PeRCenTX,PeRCenTY)) -> s(len(PeRCenTY))
 || Number of strict rules: 11
 || Direct POLO(bPol) ... failed.
 || Uncurrying ... failed.
 || Dependency Pairs:
 ||    #1: #d(s(PeRCenTX),s(PeRCenTY)) -> #if(gtr(PeRCenTX,PeRCenTY),false(),d(s(PeRCenTX),sub(PeRCenTY,PeRCenTX)))
 ||    #2: #d(s(PeRCenTX),s(PeRCenTY)) -> #gtr(PeRCenTX,PeRCenTY)
 ||    #3: #d(s(PeRCenTX),s(PeRCenTY)) -> #d(s(PeRCenTX),sub(PeRCenTY,PeRCenTX))
 ||    #4: #d(s(PeRCenTX),s(PeRCenTY)) -> #sub(PeRCenTY,PeRCenTX)
 ||    #5: #len(cons(PeRCenTX,PeRCenTY)) -> #len(PeRCenTY)
 ||    #6: #gtr(s(PeRCenTX),s(PeRCenTY)) -> #gtr(PeRCenTX,PeRCenTY)
 ||    #7: #sub(s(PeRCenTX),s(PeRCenTY)) -> #sub(PeRCenTX,PeRCenTY)
 || Number of SCCs: 4, DPs: 4
 ||   SCC { #5 }
 || POLO(Sum)... succeeded.
 ||       d 	w: 0
 ||       #len	w: x1
 ||       s 	w: 0
 ||       #gtr	w: 0
 ||       false	w: 0
 ||       gtr	w: 0
 ||       sub	w: 0
 ||       true	w: 0
 ||       #sub	w: 0
 ||       if	w: 0
 ||       0 	w: 0
 ||       nil	w: 0
 ||       #d 	w: 0
 ||       cons	w: x2 + 1
 ||       #if	w: 0
 ||       len	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #5
 || Number of SCCs: 3, DPs: 3
 ||   SCC { #7 }
 || POLO(Sum)... succeeded.
 ||       d 	w: 0
 ||       #len	w: 0
 ||       s 	w: x1 + 1
 ||       #gtr	w: 0
 ||       false	w: 0
 ||       gtr	w: 0
 ||       sub	w: 0
 ||       true	w: 0
 ||       #sub	w: x1
 ||       if	w: 0
 ||       0 	w: 0
 ||       nil	w: 0
 ||       #d 	w: 0
 ||       cons	w: 1
 ||       #if	w: 0
 ||       len	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #7
 || Number of SCCs: 2, DPs: 2
 ||   SCC { #6 }
 || POLO(Sum)... succeeded.
 ||       d 	w: 0
 ||       #len	w: 0
 ||       s 	w: x1 + 1
 ||       #gtr	w: x1
 ||       false	w: 0
 ||       gtr	w: 0
 ||       sub	w: 0
 ||       true	w: 0
 ||       #sub	w: 0
 ||       if	w: 0
 ||       0 	w: 0
 ||       nil	w: 0
 ||       #d 	w: 0
 ||       cons	w: 1
 ||       #if	w: 0
 ||       len	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #6
 || Number of SCCs: 1, DPs: 1
 ||   SCC { #3 }
 || POLO(Sum)... succeeded.
 ||       d 	w: 0
 ||       #len	w: 0
 ||       s 	w: x1 + 2
 ||       #gtr	w: 0
 ||       false	w: 0
 ||       gtr	w: 0
 ||       sub	w: x1 + 1
 ||       true	w: 0
 ||       #sub	w: 0
 ||       if	w: 0
 ||       0 	w: 1
 ||       nil	w: 0
 ||       #d 	w: x2
 ||       cons	w: 1
 ||       #if	w: 0
 ||       len	w: 0
 ||     USABLE RULES: { 3 4 }
 ||     Removed DPs: #3
 || 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] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    sub(X, 0) => X 
    sub(s(X), s(Y)) => sub(X, Y) 
    gtr(0, X) => false 
    gtr(s(X), 0) => true 
    gtr(s(X), s(Y)) => gtr(X, Y) 
    d(X, 0) => true 
    d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
    len(nil) => 0 
    len(cons(X, Y)) => s(len(Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, 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 :  
    * 1 : 0, 1, 2 
    * 2 : 0, 1, 2 

This graph has the following strongly connected components:

  P_1:

    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    filter#(F, cons(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).

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(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

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