We consider the system from.

  Alphabet:

    chain : [N -> N * list] --> list 
    cons : [N * list] --> list 
    false : [] --> B 
    from : [N * list] --> list 
    if : [B * list * list] --> list 
    incch : [list] --> list 
    lteq : [N * N] --> B 
    nil : [] --> list 
    o : [] --> N 
    s : [N] --> N 
    true : [] --> B 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    lteq(s(x), o) => false 
    lteq(o, x) => true 
    lteq(s(x), s(y)) => lteq(x, y) 
    from(x, nil) => nil 
    from(x, cons(y, z)) => if(lteq(x, y), cons(y, z), from(x, z)) 
    chain(f, nil) => nil 
    chain(f, cons(x, y)) => cons(f x, chain(f, from(f x, y))) 
    incch(x) => chain(/\y.s(y), x) 

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 
  lteq(s(X), o) => false 
  lteq(o, X) => true 
  lteq(s(X), s(Y)) => lteq(X, Y) 
  from(X, nil) => nil 
  from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 

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 the external first-order termination prover, this system is indeed 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:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    lteq(s(%X), o) -> false
 ||    lteq(o, %X) -> true
 ||    lteq(s(%X), s(%Y)) -> lteq(%X, %Y)
 ||    from(%X, nil) -> nil
 ||    from(%X, cons(%Y, %Z)) -> if(lteq(%X, %Y), cons(%Y, %Z), from(%X, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || from_2 > [if_3, cons_2]
 || from_2 > lteq_2 > [true, false, o]
 || from_2 > nil
 || 
 || 
 || Status:
 || if_3: multiset status
 || true: multiset status
 || false: multiset status
 || lteq_2: [2,1]
 || s_1: [1]
 || o: multiset status
 || from_2: multiset status
 || nil: multiset status
 || cons_2: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    lteq(s(%X), o) -> false
 ||    lteq(o, %X) -> true
 ||    lteq(s(%X), s(%Y)) -> lteq(%X, %Y)
 ||    from(%X, nil) -> nil
 ||    from(%X, cons(%Y, %Z)) -> if(lteq(%X, %Y), cons(%Y, %Z), from(%X, %Z))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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 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] chain#(F, cons(X, Y)) =#> chain#(F, from(F X, Y))   
    1] chain#(F, cons(X, Y)) =#> from#(F X, Y)   
    2] incch#(X) =#> chain#(/\x.s(x), X)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    lteq(s(X), o) => false 
    lteq(o, X) => true 
    lteq(s(X), s(Y)) => lteq(X, Y) 
    from(X, nil) => nil 
    from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
    chain(F, nil) => nil 
    chain(F, cons(X, Y)) => cons(F X, chain(F, from(F X, Y))) 
    incch(X) => chain(/\x.s(x), X) 

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 
    * 1 :  
    * 2 : 0, 1 

This graph has the following strongly connected components:

  P_1:

    chain#(F, cons(X, Y)) =#> chain#(F, from(F X, 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).

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

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  lteq(s(X), o) => false 
  lteq(o, X) => true 
  lteq(s(X), s(Y)) => lteq(X, Y) 
  from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, cons(X, Y)) => cons(F X, chain(F, from(F X, Y))) 
  incch(X) => chain(/\x.s(x), X) 

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

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

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  chain#(F, cons(X, Y)) >? chain#(F, from(F X, Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  lteq(s(X), o) >= false 
  lteq(o, X) >= true 
  lteq(s(X), s(Y)) >= lteq(X, Y) 
  from(X, cons(Y, Z)) >= if(lteq(X, Y), cons(Y, Z), from(X, Z)) 
  chain(F, cons(X, Y)) >= cons(F X, chain(F, from(F X, Y))) 
  incch(X) >= chain(/\x.s(x), X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( incch(X) ) = #argfun-incch#(chain(/\x.s(x), X)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[chain#(x_1, x_2)]] = chain#(x_2, x_1) 
  [[from(x_1, x_2)]] = from(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[incch(x_1)]] = x_1 
  [[lteq(x_1, x_2)]] = x_1 
  [[true]] = _|_ 

We choose Lex = {chain#} and Mul = {#argfun-incch#, @_{o -> o}, chain, cons, false, from, if, o, s}, and the following precedence: chain# > #argfun-incch# > chain > s > cons > @_{o -> o} > from > if > o > false

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  chain#(F, cons(X, Y)) > chain#(F, from(Y)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  s(X) >= false 
  o >= _|_ 
  s(X) >= X 
  from(cons(X, Y)) >= if(cons(X, Y), from(Y)) 
  chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y))) 
  #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X) 

With these choices, we have:

  1] chain#(F, cons(X, Y)) > chain#(F, from(Y))  because [2], by definition 
  2] chain#*(F, cons(X, Y)) >= chain#(F, from(Y))  because [3], [7] and [9], by (Stat) 
  3] cons(X, Y) > from(Y)  because [4], by definition 
  4] cons*(X, Y) >= from(Y)  because cons > from and [5], by (Copy) 
  5] cons*(X, Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 
  7] chain#*(F, cons(X, Y)) >= F  because [8], by (Select) 
  8] F >= F  by (Meta) 
  9] chain#*(F, cons(X, Y)) >= from(Y)  because chain# > from and [10], by (Copy) 
  10] chain#*(F, cons(X, Y)) >= Y  because [11], by (Select) 
  11] cons(X, Y) >= Y  because [5], by (Star) 

  12] if(X, Y) >= X  because [13], by (Star) 
  13] if*(X, Y) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 

  15] if(X, Y) >= Y  because [16], by (Star) 
  16] if*(X, Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 

  18] s(X) >= false  because [19], by (Star) 
  19] s*(X) >= false  because s > false, by (Copy) 

  20] o >= _|_  by (Bot) 

  21] s(X) >= X  because [22], by (Star) 
  22] s*(X) >= X  because [23], by (Select) 
  23] X >= X  by (Meta) 

  24] from(cons(X, Y)) >= if(cons(X, Y), from(Y))  because [25], by (Star) 
  25] from*(cons(X, Y)) >= if(cons(X, Y), from(Y))  because from > if, [26] and [30], by (Copy) 
  26] from*(cons(X, Y)) >= cons(X, Y)  because [27], by (Select) 
  27] cons(X, Y) >= cons(X, Y)  because cons in Mul, [28] and [29], by (Fun) 
  28] X >= X  by (Meta) 
  29] Y >= Y  by (Meta) 
  30] from*(cons(X, Y)) >= from(Y)  because [31], by (Select) 
  31] cons(X, Y) >= from(Y)  because [32], by (Star) 
  32] cons*(X, Y) >= from(Y)  because cons > from and [33], by (Copy) 
  33] cons*(X, Y) >= Y  because [29], by (Select) 

  34] chain(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because [35], by (Star) 
  35] chain*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), chain(F, from(Y)))  because chain > cons, [36] and [42], by (Copy) 
  36] chain*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because chain > @_{o -> o}, [37] and [38], by (Copy) 
  37] chain*(F, cons(X, Y)) >= F  because [8], by (Select) 
  38] chain*(F, cons(X, Y)) >= X  because [39], by (Select) 
  39] cons(X, Y) >= X  because [40], by (Star) 
  40] cons*(X, Y) >= X  because [41], by (Select) 
  41] X >= X  by (Meta) 
  42] chain*(F, cons(X, Y)) >= chain(F, from(Y))  because chain in Mul, [43] and [3], by (Stat) 
  43] F >= F  by (Meta) 

  44] #argfun-incch#(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because [45], by (Star) 
  45] #argfun-incch#*(chain(/\x.s(x), X)) >= chain(/\x.s(x), X)  because #argfun-incch# > chain, [46] and [50], by (Copy) 
  46] #argfun-incch#*(chain(/\x.s(x), X)) >= /\x.s(x)  because [47], by (F-Abs) 
  47] #argfun-incch#*(chain(/\x.s(x), X), y) >= s(y)  because #argfun-incch# > s and [48], by (Copy) 
  48] #argfun-incch#*(chain(/\x.s(x), X), y) >= y  because [49], by (Select) 
  49] y >= y  by (Var) 
  50] #argfun-incch#*(chain(/\x.s(x), X)) >= X  because [51], by (Select) 
  51] chain(/\x.s(x), X) >= X  because [52], by (Star) 
  52] chain*(/\x.s(x), X) >= X  because [53], by (Select) 
  53] X >= X  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  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.