We consider the system h00.

  Alphabet:

    0 : [] --> c 
    add : [] --> a -> c -> c 
    cons : [] --> a -> b -> b 
    fold : [] --> (a -> c -> c) -> c -> b -> c 
    mul : [] --> a -> c -> c 
    nil : [] --> b 
    plus : [] --> c -> c -> c 
    prod : [] --> b -> c 
    s : [] --> c -> c 
    sum : [] --> b -> c 
    times : [] --> c -> c -> c 

  Rules:

    fold (/\x./\y.f x y) z nil => z 
    fold (/\x./\y.f x y) z (cons u v) => f u (fold (/\w./\x'.f w x') z v) 
    plus 0 x => x 
    plus (s x) y => s (plus x y) 
    times 0 x => 0 
    times (s x) y => plus (times x y) y 
    sum x => fold (/\y./\z.add y z) 0 x 
    prod x => fold (/\y./\z.mul y z) (s 0) x 

Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.

We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:

  Alphabet:

    0 : [] --> c 
    add : [] --> a -> c -> c 
    cons : [a * b] --> b 
    fold : [a -> c -> c * c * b] --> c 
    mul : [] --> a -> c -> c 
    nil : [] --> b 
    plus : [c * c] --> c 
    prod : [b] --> c 
    s : [c] --> c 
    sum : [b] --> c 
    times : [c * c] --> c 
    ~AP1 : [a -> c -> c * a] --> c -> c 

  Rules:

    fold(/\x./\y.~AP1(F, x) y, X, nil) => X 
    fold(/\x./\y.~AP1(F, x) y, X, cons(Y, Z)) => ~AP1(F, Y) fold(/\z./\u.~AP1(F, z) u, X, Z) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(times(X, Y), Y) 
    sum(X) => fold(/\x./\y.~AP1(add, x) y, 0, X) 
    prod(X) => fold(/\x./\y.~AP1(mul, x) y, s(0), X) 
    ~AP1(F, X) => F X 

Symbol ~AP1 is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:

  Alphabet:

    0 : [] --> c 
    add : [a * c] --> c 
    cons : [a * b] --> b 
    fold : [a -> c -> c * c * b] --> c 
    mul : [a * c] --> c 
    nil : [] --> b 
    plus : [c * c] --> c 
    prod : [b] --> c 
    s : [c] --> c 
    sum : [b] --> c 
    times : [c * c] --> c 

  Rules:

    fold(/\x./\y.X(x, y), Y, nil) => Y 
    fold(/\x./\y.X(x, y), Y, cons(Z, U)) => X(Z, fold(/\z./\u.X(z, u), Y, U)) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(times(X, Y), Y) 
    sum(X) => fold(/\x./\y.add(x, y), 0, X) 
    prod(X) => fold(/\x./\y.mul(x, y), s(0), X) 

We observe that the rules contain a first-order subset:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  times(0, X) => 0 
  times(s(X), Y) => plus(times(X, Y), 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(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    times(0, %X) -> 0
 ||    times(s(%X), %Y) -> plus(times(%X, %Y), %Y)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || times_2 > plus_2 > s_1
 || times_2 > 0 > s_1
 || 
 || 
 || Status:
 || plus_2: multiset status
 || 0: multiset status
 || s_1: multiset status
 || times_2: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    times(0, %X) -> 0
 ||    times(s(%X), %Y) -> plus(times(%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.

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] fold#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(Z, fold(/\z./\u.X(z, u), Y, U))   
    1] fold#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> fold#(/\z./\u.X(z, u), Y, U) {X : 2}  
    2] fold#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u) {X : 2}  
    3] sum#(X) =#> fold#(/\x./\y.add(x, y), 0, X)   
    4] prod#(X) =#> fold#(/\x./\y.mul(x, y), s(0), X)   

  Rules R_0:

    fold(/\x./\y.X(x, y), Y, nil) => Y 
    fold(/\x./\y.X(x, y), Y, cons(Z, U)) => X(Z, fold(/\z./\u.X(z, u), Y, U)) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(times(X, Y), Y) 
    sum(X) => fold(/\x./\y.add(x, y), 0, X) 
    prod(X) => fold(/\x./\y.mul(x, y), s(0), X) 

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).

This combination (P_0, R_0) has no formative rules!  We will name the empty set of rules:R_1.

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, 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:

  fold#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(Z, fold(/\z./\u.X(z, u), Y, U)) 
  fold#(/\x./\y.X(x, y), Y, cons(Z, U)) >? fold#(/\z./\u.X(z, u), Y, U) 
  fold#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  sum#(X) >? fold#(/\x./\y.add-(x, y), 0, X) 
  prod#(X) >? fold#(/\x./\y.mul-(x, y), s(0), X) 
  add-(X, Y) >= add(X, Y) 
  mul-(X, Y) >= mul(X, Y) 

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

  pi( prod#(X) ) = #argfun-prod##(fold#(/\x./\y.mul-(x, y), s(0), X)) 
  pi( sum#(X) ) = #argfun-sum##(fold#(/\x./\y.add-(x, y), 0, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-prod## = \y0.3 + y0 
  #argfun-sum## = \y0.3 + y0 
  0 = 0 
  add = \y0y1.0 
  add- = \y0y1.3 + 3y0 + 3y1 
  cons = \y0y1.3 + y0 + 3y1 
  fold = \G0y1y2.0 
  fold# = \G0y1y2.3 + y2 + 2y2y2G0(y2,y2) + 2G0(y2,y2) + 3y1y2G0(y1,y2) + 3y1y2G0(y2,y1) 
  mul = \y0y1.0 
  mul- = \y0y1.3 + 3y0 + 3y1 
  prod# = \y0.0 
  s = \y0.0 
  sum# = \y0.0 
  ~c0 = 0 
  ~c1 = 0 

Using this interpretation, the requirements translate to:

  [[fold#(/\x./\y._x0(x, y), _x1, cons(_x2, _x3))]] = 6 + x2 + 3x3 + 2x2x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 3x1x2F0(x1,3 + x2 + 3x3) + 3x1x2F0(3 + x2 + 3x3,x1) + 9x1x3F0(x1,3 + x2 + 3x3) + 9x1x3F0(3 + x2 + 3x3,x1) + 9x1F0(x1,3 + x2 + 3x3) + 9x1F0(3 + x2 + 3x3,x1) + 12x2x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 12x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 18x3x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 20F0(3 + x2 + 3x3,3 + x2 + 3x3) + 36x3F0(3 + x2 + 3x3,3 + x2 + 3x3) > F0(x2,0) = [[_x0(_x2, fold(/\x./\y._x0(x, y), _x1, _x3))]] 
  [[fold#(/\x./\y._x0(x, y), _x1, cons(_x2, _x3))]] = 6 + x2 + 3x3 + 2x2x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 3x1x2F0(x1,3 + x2 + 3x3) + 3x1x2F0(3 + x2 + 3x3,x1) + 9x1x3F0(x1,3 + x2 + 3x3) + 9x1x3F0(3 + x2 + 3x3,x1) + 9x1F0(x1,3 + x2 + 3x3) + 9x1F0(3 + x2 + 3x3,x1) + 12x2x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 12x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 18x3x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 20F0(3 + x2 + 3x3,3 + x2 + 3x3) + 36x3F0(3 + x2 + 3x3,3 + x2 + 3x3) > 3 + x3 + 2x3x3F0(x3,x3) + 2F0(x3,x3) + 3x1x3F0(x1,x3) + 3x1x3F0(x3,x1) = [[fold#(/\x./\y._x0(x, y), _x1, _x3)]] 
  [[fold#(/\x./\y._x0(x, y), _x1, cons(_x2, _x3))]] = 6 + x2 + 3x3 + 2x2x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 3x1x2F0(x1,3 + x2 + 3x3) + 3x1x2F0(3 + x2 + 3x3,x1) + 9x1x3F0(x1,3 + x2 + 3x3) + 9x1x3F0(3 + x2 + 3x3,x1) + 9x1F0(x1,3 + x2 + 3x3) + 9x1F0(3 + x2 + 3x3,x1) + 12x2x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 12x2F0(3 + x2 + 3x3,3 + x2 + 3x3) + 18x3x3F0(3 + x2 + 3x3,3 + x2 + 3x3) + 20F0(3 + x2 + 3x3,3 + x2 + 3x3) + 36x3F0(3 + x2 + 3x3,3 + x2 + 3x3) > F0(0,0) = [[_x0(~c0, ~c1)]] 
  [[#argfun-sum##(fold#(/\x./\y.add-(x, y), 0, _x0))]] = 12 + 6x0x0 + 12x0x0x0 + 13x0 > 9 + 6x0x0 + 12x0x0x0 + 13x0 = [[fold#(/\x./\y.add-(x, y), 0, _x0)]] 
  [[#argfun-prod##(fold#(/\x./\y.mul-(x, y), s(0), _x0))]] = 12 + 6x0x0 + 12x0x0x0 + 13x0 > 9 + 6x0x0 + 12x0x0x0 + 13x0 = [[fold#(/\x./\y.mul-(x, y), s(0), _x0)]] 
  [[add-(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 0 = [[add(_x0, _x1)]] 
  [[mul-(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 0 = [[mul(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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 +++

[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.