We consider the system filter.

  Alphabet:

    0 : [] --> nat 
    bool : [nat] --> boolean 
    cons : [nat * list] --> list 
    consif : [boolean * nat * list] --> list 
    false : [] --> boolean 
    filter : [nat -> boolean * list] --> list 
    nil : [] --> list 
    rand : [nat] --> nat 
    s : [nat] --> nat 
    true : [] --> boolean 

  Rules:

    rand(x) => x 
    rand(s(x)) => rand(x) 
    bool(0) => false 
    bool(s(0)) => true 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) 
    consif(true, x, y) => cons(x, y) 
    consif(false, x, y) => 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:

  rand(X) => X 
  rand(s(X)) => rand(X) 
  bool(0) => false 
  bool(s(0)) => true 
  consif(true, X, Y) => cons(X, Y) 
  consif(false, X, Y) => Y 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    rand(%X) -> %X
 ||    rand(s(%X)) -> rand(%X)
 ||    bool(0) -> false
 ||    bool(s(0)) -> true
 ||    consif(true, %X, %Y) -> cons(%X, %Y)
 ||    consif(false, %X, %Y) -> %Y
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 2
 ||    POL(bool(x_1)) = 1 + 2*x_1
 ||    POL(cons(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(consif(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + 2*x_3
 ||    POL(false) = 1
 ||    POL(rand(x_1)) = 2 + 2*x_1
 ||    POL(s(x_1)) = 2 + 2*x_1
 ||    POL(true) = 2
 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    rand(%X) -> %X
 ||    rand(s(%X)) -> rand(%X)
 ||    bool(0) -> false
 ||    bool(s(0)) -> true
 ||    consif(true, %X, %Y) -> cons(%X, %Y)
 ||    consif(false, %X, %Y) -> %Y
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] filter#(F, cons(X, Y)) =#> consif#(F X, X, filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> F(X)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    rand(X) => X 
    rand(s(X)) => rand(X) 
    bool(0) => false 
    bool(s(0)) => true 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 
    consif(true, X, Y) => cons(X, Y) 
    consif(false, X, Y) => Y 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, 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)) =#> F(X)   
    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, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  rand(X) => X 
  rand(s(X)) => rand(X) 
  bool(0) => false 
  bool(s(0)) => true 
  filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 
  consif(true, X, Y) => cons(X, Y) 
  consif(false, X, Y) => Y 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  filter#(F, cons(X, Y)) >? F(X) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  rand(X) >= X 
  rand(s(X)) >= rand(X) 
  bool(0) >= false 
  bool(s(0)) >= true 
  filter(F, cons(X, Y)) >= consif(F X, X, filter(F, Y)) 
  consif(true, X, Y) >= cons(X, Y) 
  consif(false, X, Y) >= Y 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 3 
  bool = \y0.3 
  cons = \y0y1.y0 + y1 
  consif = \y0y1y2.y1 + y2 
  false = 0 
  filter = \G0y1.y1 
  filter# = \G0y1.3 + G0(y1) 
  rand = \y0.y0 
  s = \y0.3 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 3 + F0(x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 + F0(x1 + x2) >= 3 + F0(x2) = [[filter#(_F0, _x2)]] 
  [[rand(_x0)]] = x0 >= x0 = [[_x0]] 
  [[rand(s(_x0))]] = 3 + x0 >= x0 = [[rand(_x0)]] 
  [[bool(0)]] = 3 >= 0 = [[false]] 
  [[bool(s(0))]] = 3 >= 0 = [[true]] 
  [[filter(_F0, cons(_x1, _x2))]] = x1 + x2 >= x1 + x2 = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] 
  [[consif(true, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[consif(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

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

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_1, minimal, 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)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, minimal, 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.