We consider the system Applicative_first_order_05__perfect2.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    f : [a * a * a * a] --> b 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    if : [b * b * b] --> b 
    le : [a * a] --> b 
    map : [c -> c * d] --> d 
    minus : [a * a] --> a 
    nil : [] --> d 
    perfectp : [a] --> b 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    minus(0, x) => 0 
    minus(s(x), 0) => s(x) 
    minus(s(x), s(y)) => minus(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    if(true, x, y) => x 
    if(false, x, y) => y 
    perfectp(0) => false 
    perfectp(s(x)) => f(x, s(0), s(x), s(x)) 
    f(0, x, 0, y) => true 
    f(0, x, s(y), z) => false 
    f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) 
    f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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(0, X) => 0 
  minus(s(X), 0) => s(X) 
  minus(s(X), s(Y)) => minus(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  perfectp(0) => false 
  perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) => true 
  f(0, X, s(Y), Z) => false 
  f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 

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(0(),PeRCenTX) -> 0()
 ||     2: minus(s(PeRCenTX),0()) -> s(PeRCenTX)
 ||     3: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(PeRCenTX,PeRCenTY)
 ||     4: le(0(),PeRCenTX) -> true()
 ||     5: le(s(PeRCenTX),0()) -> false()
 ||     6: le(s(PeRCenTX),s(PeRCenTY)) -> le(PeRCenTX,PeRCenTY)
 ||     7: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX
 ||     8: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY
 ||     9: perfectp(0()) -> false()
 ||     10: perfectp(s(PeRCenTX)) -> f(PeRCenTX,s(0()),s(PeRCenTX),s(PeRCenTX))
 ||     11: f(0(),PeRCenTX,0(),PeRCenTY) -> true()
 ||     12: f(0(),PeRCenTX,s(PeRCenTY),PeRCenTZ) -> false()
 ||     13: f(s(PeRCenTX),0(),PeRCenTY,PeRCenTZ) -> f(PeRCenTX,PeRCenTZ,minus(PeRCenTY,s(PeRCenTX)),PeRCenTZ)
 ||     14: f(s(PeRCenTX),s(PeRCenTY),PeRCenTZ,PeRCenTU) -> if(le(PeRCenTX,PeRCenTY),f(s(PeRCenTX),minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU),f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU))
 || Number of strict rules: 14
 || Direct POLO(bPol) ... failed.
 || Uncurrying f
 || 1: minus(0(),PeRCenTX) -> 0()
 || 2: minus(s(PeRCenTX),0()) -> s(PeRCenTX)
 || 3: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(PeRCenTX,PeRCenTY)
 || 4: le(0(),PeRCenTX) -> true()
 || 5: le(s(PeRCenTX),0()) -> false()
 || 6: le(s(PeRCenTX),s(PeRCenTY)) -> le(PeRCenTX,PeRCenTY)
 || 7: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX
 || 8: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY
 || 9: perfectp(0()) -> false()
 || 10: perfectp(s(PeRCenTX)) -> f(PeRCenTX,s(0()),s(PeRCenTX),s(PeRCenTX))
 || 11: f^1_0(PeRCenTX,0(),PeRCenTY) -> true()
 || 12: f^1_0(PeRCenTX,s(PeRCenTY),PeRCenTZ) -> false()
 || 13: f^1_s(PeRCenTX,0(),PeRCenTY,PeRCenTZ) -> f(PeRCenTX,PeRCenTZ,minus(PeRCenTY,s(PeRCenTX)),PeRCenTZ)
 || 14: f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> if(le(PeRCenTX,PeRCenTY),f^1_s(PeRCenTX,minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU),f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU))
 || 15: f(0(),_3,_4,_5) ->= f^1_0(_3,_4,_5)
 || 16: f(s(_1),_4,_5,_6) ->= f^1_s(_1,_4,_5,_6)
 || Number of strict rules: 14
 || Direct POLO(bPol) ... failed.
 || Dependency Pairs:
 ||    #1: #le(s(PeRCenTX),s(PeRCenTY)) -> #le(PeRCenTX,PeRCenTY)
 ||    #2: #f^1_s(PeRCenTX,0(),PeRCenTY,PeRCenTZ) -> #f(PeRCenTX,PeRCenTZ,minus(PeRCenTY,s(PeRCenTX)),PeRCenTZ)
 ||    #3: #f^1_s(PeRCenTX,0(),PeRCenTY,PeRCenTZ) -> #minus(PeRCenTY,s(PeRCenTX))
 ||    #4: #f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> #if(le(PeRCenTX,PeRCenTY),f^1_s(PeRCenTX,minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU),f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU))
 ||    #5: #f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> #le(PeRCenTX,PeRCenTY)
 ||    #6: #f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> #f^1_s(PeRCenTX,minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU)
 ||    #7: #f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> #minus(PeRCenTY,PeRCenTX)
 ||    #8: #f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> #f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU)
 ||    #9: #perfectp(s(PeRCenTX)) -> #f(PeRCenTX,s(0()),s(PeRCenTX),s(PeRCenTX))
 ||    #10: #f(s(_1),_4,_5,_6) ->? #f^1_s(_1,_4,_5,_6)
 ||    #11: #minus(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY)
 ||    #12: #f(0(),_3,_4,_5) ->? #f^1_0(_3,_4,_5)
 || Number of SCCs: 3, DPs: 6
 ||   SCC { #1 }
 || POLO(Sum)... succeeded.
 ||       le	w: 0
 ||       s 	w: x1 + 1
 ||       #le	w: x1
 ||       minus	w: 0
 ||       #f^1_s	w: 0
 ||       #perfectp	w: 0
 ||       false	w: 0
 ||       true	w: 0
 ||       f 	w: 0
 ||       0 	w: 0
 ||       if	w: 0
 ||       #f 	w: 0
 ||       #minus	w: 0
 ||       #f^1_0	w: 0
 ||       #if	w: 0
 ||       perfectp	w: 0
 ||       f^1_0	w: 0
 ||       f^1_s	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #1
 || Number of SCCs: 2, DPs: 5
 ||   SCC { #11 }
 || POLO(Sum)... succeeded.
 ||       le	w: 0
 ||       s 	w: x1 + 1
 ||       #le	w: 0
 ||       minus	w: 0
 ||       #f^1_s	w: 0
 ||       #perfectp	w: 0
 ||       false	w: 0
 ||       true	w: 0
 ||       f 	w: 0
 ||       0 	w: 0
 ||       if	w: 0
 ||       #f 	w: 0
 ||       #minus	w: x1
 ||       #f^1_0	w: 0
 ||       #if	w: 0
 ||       perfectp	w: 0
 ||       f^1_0	w: 0
 ||       f^1_s	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #11
 || Number of SCCs: 1, DPs: 4
 ||   SCC { #2 #6 #8 #10 }
 || POLO(Sum)... succeeded.
 ||       le	w: 0
 ||       s 	w: x1 + 2
 ||       #le	w: 0
 ||       minus	w: x1 + x2 + 1
 ||       #f^1_s	w: x1 + x4 + 1
 ||       #perfectp	w: 0
 ||       false	w: 0
 ||       true	w: 0
 ||       f 	w: 0
 ||       0 	w: 1
 ||       if	w: 0
 ||       #f 	w: x1 + x4
 ||       #minus	w: 0
 ||       #f^1_0	w: 0
 ||       #if	w: 0
 ||       perfectp	w: 0
 ||       f^1_0	w: 0
 ||       f^1_s	w: 0
 ||     USABLE RULES: { }
 ||     Removed DPs: #2 #8 #10
 || Number of SCCs: 1, DPs: 1
 ||   SCC { #6 }
 || POLO(Sum)... succeeded.
 ||       le	w: 0
 ||       s 	w: x1 + 2
 ||       #le	w: 0
 ||       minus	w: x1 + 1
 ||       #f^1_s	w: x2 + 1
 ||       #perfectp	w: 0
 ||       false	w: 0
 ||       true	w: 0
 ||       f 	w: 0
 ||       0 	w: 1
 ||       if	w: 0
 ||       #f 	w: 0
 ||       #minus	w: 0
 ||       #f^1_0	w: 0
 ||       #if	w: 0
 ||       perfectp	w: 0
 ||       f^1_0	w: 0
 ||       f^1_s	w: 0
 ||     USABLE RULES: { 1..3 }
 ||     Removed DPs: #6
 || 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:

    minus(0, X) => 0 
    minus(s(X), 0) => s(X) 
    minus(s(X), s(Y)) => minus(X, Y) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    perfectp(0) => false 
    perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
    f(0, X, 0, Y) => true 
    f(0, X, s(Y), Z) => false 
    f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
    f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
    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.