We consider the system extrec.

  Alphabet:

    !plus : [nat * nat] --> nat 
    !times : [nat * nat] --> nat 
    0 : [] --> nat 
    rec : [nat * nat * nat -> nat -> nat] --> nat 
    s : [nat] --> nat 

  Rules:

    !plus(x, 0) => x 
    !plus(x, s(y)) => s(!plus(x, y)) 
    rec(0, x, f) => x 
    rec(s(x), y, f) => f x rec(x, y, f) 
    !times(x, y) => rec(y, 0, /\z./\u.!plus(x, u)) 

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:

  !plus(X, 0) => X 
  !plus(X, s(Y)) => s(!plus(X, 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 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:
 || 
 ||    !plus(%X, 0) -> %X
 ||    !plus(%X, s(%Y)) -> s(!plus(%X, %Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(!plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(0) = 2
 ||    POL(s(x_1)) = 1 + x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    !plus(%X, 0) -> %X
 ||    !plus(%X, s(%Y)) -> s(!plus(%X, %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] rec#(s(X), Y, F) =#> F(X, rec(X, Y, F))   
    1] rec#(s(X), Y, F) =#> rec#(X, Y, F)   
    2] !times#(X, Y) =#> rec#(Y, 0, /\x./\y.!plus(X, y))   
    3] !times#(X, Y) =#> !plus#(X, x)   

  Rules R_0:

    !plus(X, 0) => X 
    !plus(X, s(Y)) => s(!plus(X, Y)) 
    rec(0, X, F) => X 
    rec(s(X), Y, F) => F X rec(X, Y, F) 
    !times(X, Y) => rec(Y, 0, /\x./\y.!plus(X, 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 : 0, 1, 2, 3 
    * 1 : 0, 1 
    * 2 : 0, 1 
    * 3 :  

This graph has the following strongly connected components:

  P_1:

    rec#(s(X), Y, F) =#> F(X, rec(X, Y, F))   
    rec#(s(X), Y, F) =#> rec#(X, Y, F)   
    !times#(X, Y) =#> rec#(Y, 0, /\x./\y.!plus(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, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, 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:

  rec#(s(X), Y, F) >? F(X, rec(X, Y, F)) 
  rec#(s(X), Y, F) >? rec#(X, Y, F) 
  !times#(X, Y) >? rec#(Y, 0, /\x./\y.!plus-(X, y)) 
  !plus(X, 0) >= X 
  !plus(X, s(Y)) >= s(!plus(X, Y)) 
  rec(0, X, F) >= X 
  rec(s(X), Y, F) >= F X rec(X, Y, F) 
  !times(X, Y) >= rec(Y, 0, /\x./\y.!plus-(X, y)) 
  !plus-(X, Y) >= !plus(X, Y) 

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

  pi( !times(X, Y) ) = #argfun-!times#(rec(Y, 0, /\x./\y.!plus-(X, y))) 
  pi( !times#(X, Y) ) = #argfun-!times##(rec#(Y, 0, /\x./\y.!plus-(X, y))) 

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:

  [[#argfun-!times##(x_1)]] = x_1 
  [[0]] = _|_ 

We choose Lex = {} and Mul = {!plus, !plus-, !times, !times#, #argfun-!times#, @_{o -> o -> o}, @_{o -> o}, rec, rec#, s}, and the following precedence: !plus- > !plus > s > !times# > rec# > rec > @_{o -> o -> o} > @_{o -> o} > #argfun-!times# > !times

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

  rec#(s(X), Y, F) > @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) 
  rec#(s(X), Y, F) >= rec#(X, Y, F) 
  rec#(X, _|_, /\x./\y.!plus-(Y, y)) >= rec#(X, _|_, /\x./\y.!plus-(Y, y)) 
  !plus(X, _|_) >= X 
  !plus(X, s(Y)) >= s(!plus(X, Y)) 
  rec(_|_, X, F) >= X 
  rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) 
  #argfun-!times#(rec(X, _|_, /\x./\y.!plus-(Y, y))) >= rec(X, _|_, /\x./\y.!plus-(Y, y)) 
  !plus-(X, Y) >= !plus(X, Y) 

With these choices, we have:

  1] rec#(s(X), Y, F) > @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because [2], by definition 
  2] rec#*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because rec# > @_{o -> o}, [3] and [10], by (Copy) 
  3] rec#*(s(X), Y, F) >= @_{o -> o -> o}(F, X)  because rec# > @_{o -> o -> o}, [4] and [6], by (Copy) 
  4] rec#*(s(X), Y, F) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] rec#*(s(X), Y, F) >= X  because [7], by (Select) 
  7] s(X) >= X  because [8], by (Star) 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] rec#*(s(X), Y, F) >= rec(X, Y, F)  because rec# > rec, [6], [11] and [4], by (Copy) 
  11] rec#*(s(X), Y, F) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] rec#(s(X), Y, F) >= rec#(X, Y, F)  because rec# in Mul, [14], [15] and [16], by (Fun) 
  14] s(X) >= X  because [8], by (Star) 
  15] Y >= Y  by (Meta) 
  16] F >= F  by (Meta) 

  17] rec#(X, _|_, /\x./\y.!plus-(Y, y)) >= rec#(X, _|_, /\x./\y.!plus-(Y, y))  because rec# in Mul, [18], [19] and [20], by (Fun) 
  18] X >= X  by (Meta) 
  19] _|_ >= _|_  by (Bot) 
  20] /\x./\y.!plus-(Y, y) >= /\x./\y.!plus-(Y, y)  because [21], by (Abs) 
  21] /\y.!plus-(Y, y) >= /\y.!plus-(Y, y)  because [22], by (Abs) 
  22] !plus-(Y, x) >= !plus-(Y, x)  because !plus- in Mul, [23] and [24], by (Fun) 
  23] Y >= Y  by (Meta) 
  24] x >= x  by (Var) 

  25] !plus(X, _|_) >= X  because [26], by (Star) 
  26] !plus*(X, _|_) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 

  28] !plus(X, s(Y)) >= s(!plus(X, Y))  because [29], by (Star) 
  29] !plus*(X, s(Y)) >= s(!plus(X, Y))  because !plus > s and [30], by (Copy) 
  30] !plus*(X, s(Y)) >= !plus(X, Y)  because !plus in Mul, [31] and [32], by (Stat) 
  31] X >= X  by (Meta) 
  32] s(Y) > Y  because [33], by definition 
  33] s*(Y) >= Y  because [34], by (Select) 
  34] Y >= Y  by (Meta) 

  35] rec(_|_, X, F) >= X  because [36], by (Star) 
  36] rec*(_|_, X, F) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 

  38] rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because [39], by (Star) 
  39] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because rec > @_{o -> o}, [40] and [43], by (Copy) 
  40] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X)  because rec > @_{o -> o -> o}, [41] and [42], by (Copy) 
  41] rec*(s(X), Y, F) >= F  because [16], by (Select) 
  42] rec*(s(X), Y, F) >= X  because [14], by (Select) 
  43] rec*(s(X), Y, F) >= rec(X, Y, F)  because rec in Mul, [44], [15] and [16], by (Stat) 
  44] s(X) > X  because [45], by definition 
  45] s*(X) >= X  because [9], by (Select) 

  46] #argfun-!times#(rec(X, _|_, /\x./\y.!plus-(Y, y))) >= rec(X, _|_, /\x./\y.!plus-(Y, y))  because [47], by (Star) 
  47] #argfun-!times#*(rec(X, _|_, /\x./\y.!plus-(Y, y))) >= rec(X, _|_, /\x./\y.!plus-(Y, y))  because [48], by (Select) 
  48] rec(X, _|_, /\x./\y.!plus-(Y, y)) >= rec(X, _|_, /\x./\y.!plus-(Y, y))  because rec in Mul, [18], [19] and [20], by (Fun) 

  49] !plus-(X, Y) >= !plus(X, Y)  because [50], by (Star) 
  50] !plus-*(X, Y) >= !plus(X, Y)  because !plus- > !plus, [51] and [53], by (Copy) 
  51] !plus-*(X, Y) >= X  because [52], by (Select) 
  52] X >= X  by (Meta) 
  53] !plus-*(X, Y) >= Y  because [54], by (Select) 
  54] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_2, R_0, minimal, formative), where P_2 consists of:

  rec#(s(X), Y, F) =#> rec#(X, Y, F)   
  !times#(X, Y) =#> rec#(Y, 0, /\x./\y.!plus(X, y))   

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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 : 0 
    * 1 : 0 

This graph has the following strongly connected components:

  P_3:

    rec#(s(X), Y, F) =#> rec#(X, Y, F)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_0, m, f) by (P_3, R_0, m, f).

Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(rec#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(rec#(s(X), Y, F)) = s(X) |> X = nu(rec#(X, Y, F)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, 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.