We consider the system h55.

  Alphabet:

    cons : [] --> a -> b -> b 
    foldr : [] --> (a -> b -> b) -> b -> b -> b 
    nil : [] --> b 

  Rules:

    foldr (/\x./\y.f x y) z nil => z 
    foldr (/\x./\y.f x y) z (cons u v) => f u (foldr (/\w./\x'.f w x') z v) 

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:

    cons : [a * b] --> b 
    foldr : [a -> b -> b * b * b] --> b 
    nil : [] --> b 
    ~AP1 : [a -> b -> b * a] --> b -> b 

  Rules:

    foldr(/\x./\y.~AP1(F, x) y, X, nil) => X 
    foldr(/\x./\y.~AP1(F, x) y, X, cons(Y, Z)) => ~AP1(F, Y) foldr(/\z./\u.~AP1(F, z) u, X, Z) 
    foldr(/\x./\y.cons(x, y), X, nil) => X 
    foldr(/\x./\y.cons(x, y), X, cons(Y, Z)) => cons(Y, foldr(/\z./\u.cons(z, u), X, Z)) 
    ~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.  Additionally, we can remove some (now-)redundant rules.  This gives:

  Alphabet:

    cons : [a * b] --> b 
    foldr : [a -> b -> b * b * b] --> b 
    nil : [] --> b 

  Rules:

    foldr(/\x./\y.X(x, y), Y, nil) => Y 
    foldr(/\x./\y.X(x, y), Y, cons(Z, U)) => X(Z, foldr(/\z./\u.X(z, u), Y, U)) 

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] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(Z, foldr(/\z./\u.X(z, u), Y, U))   
    1] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> foldr#(/\z./\u.X(z, u), Y, U) {X : 2}  
    2] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u) {X : 2}  

  Rules R_0:

    foldr(/\x./\y.X(x, y), Y, nil) => Y 
    foldr(/\x./\y.X(x, y), Y, cons(Z, U)) => X(Z, foldr(/\z./\u.X(z, u), Y, U)) 

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 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:

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(Z, foldr(/\z./\u.X(z, u), Y, U)) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldr#(/\z./\u.X(z, u), Y, U) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\z./\u.X(z, u), Y, U)) 

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:

  [[~c0]] = _|_ 
  [[~c1]] = _|_ 

We choose Lex = {} and Mul = {cons, foldr, foldr#, nil}, and the following precedence: foldr = foldr# > nil > cons

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

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(Z, foldr(/\x./\y.X(x, y), Y, U)) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr#(/\x./\y.X(x, y), Y, U) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U)) 

With these choices, we have:

  1] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [2], by definition 
  2] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [3], by (Select) 
  3] X(foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr#*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [4] and [8], by (Meta) 
  4] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [5], by (Select) 
  5] cons(Z, U) >= Z  because [6], by (Star) 
  6] cons*(Z, U) >= Z  because [7], by (Select) 
  7] Z >= Z  by (Meta) 
  8] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr(/\x./\y.X(x, y), Y, U)  because foldr# = foldr, foldr# in Mul, [9], [14] and [15], by (Stat) 
  9] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [10], by (Abs) 
  10] /\z.X(y, z) >= /\z.X(y, z)  because [11], by (Abs) 
  11] X(y, x) >= X(y, x)  because [12] and [13], by (Meta) 
  12] y >= y  by (Var) 
  13] x >= x  by (Var) 
  14] Y >= Y  by (Meta) 
  15] cons(Z, U) > U  because [16], by definition 
  16] cons*(Z, U) >= U  because [17], by (Select) 
  17] U >= U  by (Meta) 

  18] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr#(/\x./\y.X(x, y), Y, U)  because foldr# in Mul, [9], [14] and [19], by (Fun) 
  19] cons(Z, U) >= U  because [16], by (Star) 

  20] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [21], by (Star) 
  21] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [22], by (Select) 
  22] X(foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr#*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(_|_, _|_)  because [23] and [24], by (Meta) 
  23] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 
  24] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 

  25] foldr(/\x./\y.X(x, y), Y, nil) >= Y  because [26], by (Star) 
  26] foldr*(/\x./\y.X(x, y), Y, nil) >= Y  because [27], by (Select) 
  27] Y >= Y  by (Meta) 

  28] foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [29], by (Star) 
  29] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [30], by (Select) 
  30] X(foldr*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [31] and [32], by (Meta) 
  31] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [5], by (Select) 
  32] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr(/\x./\y.X(x, y), Y, U)  because foldr in Mul, [9], [14] and [15], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of:

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> foldr#(/\z./\u.X(z, u), Y, U)   
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u)   

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:

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldr#(/\z./\u.X(z, u), Y, U) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\z./\u.X(z, u), Y, U)) 

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:

  [[~c0]] = _|_ 
  [[~c1]] = _|_ 

We choose Lex = {} and Mul = {cons, foldr, foldr#, nil}, and the following precedence: cons > foldr > foldr# > nil

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

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > foldr#(/\x./\y.X(x, y), Y, U) 
  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U)) 

With these choices, we have:

  1] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > foldr#(/\x./\y.X(x, y), Y, U)  because [2], by definition 
  2] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr#(/\x./\y.X(x, y), Y, U)  because foldr# in Mul, [3], [8] and [9], by (Stat) 
  3] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [4], by (Abs) 
  4] /\z.X(y, z) >= /\z.X(y, z)  because [5], by (Abs) 
  5] X(y, x) >= X(y, x)  because [6] and [7], by (Meta) 
  6] y >= y  by (Var) 
  7] x >= x  by (Var) 
  8] Y >= Y  by (Meta) 
  9] cons(Z, U) > U  because [10], by definition 
  10] cons*(Z, U) >= U  because [11], by (Select) 
  11] U >= U  by (Meta) 

  12] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [13], by (Star) 
  13] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [14], by (Select) 
  14] X(foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr#*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(_|_, _|_)  because [15] and [16], by (Meta) 
  15] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 
  16] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 

  17] foldr(/\x./\y.X(x, y), Y, nil) >= Y  because [18], by (Star) 
  18] foldr*(/\x./\y.X(x, y), Y, nil) >= Y  because [19], by (Select) 
  19] Y >= Y  by (Meta) 

  20] foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [21], by (Star) 
  21] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [22], by (Select) 
  22] X(foldr*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [23] and [27], by (Meta) 
  23] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [24], by (Select) 
  24] cons(Z, U) >= Z  because [25], by (Star) 
  25] cons*(Z, U) >= Z  because [26], by (Select) 
  26] Z >= Z  by (Meta) 
  27] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr(/\x./\y.X(x, y), Y, U)  because foldr in Mul, [3], [8] and [9], by (Stat) 

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:

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u)   

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 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:

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\z./\u.X(z, u), Y, U)) 

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:

  [[~c0]] = _|_ 
  [[~c1]] = _|_ 

We choose Lex = {} and Mul = {cons, foldr, foldr#, nil}, and the following precedence: cons > foldr > foldr# > nil

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

  foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(_|_, _|_) 
  foldr(/\x./\y.X(x, y), Y, nil) >= Y 
  foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U)) 

With these choices, we have:

  1] foldr#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(_|_, _|_)  because [2], by definition 
  2] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [3], by (Select) 
  3] X(foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr#*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(_|_, _|_)  because [4] and [5], by (Meta) 
  4] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 
  5] foldr#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 

  6] foldr(/\x./\y.X(x, y), Y, nil) >= Y  because [7], by (Star) 
  7] foldr*(/\x./\y.X(x, y), Y, nil) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] foldr(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [10], by (Star) 
  10] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [11], by (Select) 
  11] X(foldr*(/\x./\y.X(x, y), Y, cons(Z, U)), foldr*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(Z, foldr(/\x./\y.X(x, y), Y, U))  because [12] and [16], by (Meta) 
  12] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [13], by (Select) 
  13] cons(Z, U) >= Z  because [14], by (Star) 
  14] cons*(Z, U) >= Z  because [15], by (Select) 
  15] Z >= Z  by (Meta) 
  16] foldr*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldr(/\x./\y.X(x, y), Y, U)  because foldr in Mul, [17], [22] and [23], by (Stat) 
  17] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [18], by (Abs) 
  18] /\z.X(y, z) >= /\z.X(y, z)  because [19], by (Abs) 
  19] X(y, x) >= X(y, x)  because [20] and [21], by (Meta) 
  20] y >= y  by (Var) 
  21] x >= x  by (Var) 
  22] Y >= Y  by (Meta) 
  23] cons(Z, U) > U  because [24], by definition 
  24] cons*(Z, U) >= U  because [25], by (Select) 
  25] U >= U  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  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.