We consider the system 431.

  Alphabet:

    0 : [] --> nat 
    rec : [nat -> A -> A * A * nat] --> A 
    s : [nat] --> nat 

  Rules:

    rec(/\x./\y.X(x, y), Y, 0) => Y 
    rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  rec(/\x./\y.X(x, y), Y, 0) >? Y 
  rec(/\x./\y.X(x, y), Y, s(Z)) >? X(Z, rec(/\z./\u.X(z, u), Y, Z)) 

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

We choose Lex = {} and Mul = {0, rec, s}, and the following precedence: 0 > rec > s

With these choices, we have:

  1] rec(/\x./\y.X(x, y), Y, 0) >= Y  because [2], by (Star) 
  2] rec*(/\x./\y.X(x, y), Y, 0) >= Y  because [3], by (Select) 
  3] Y >= Y  by (Meta) 

  4] rec(/\x./\y.X(x, y), Y, s(Z)) > X(Z, rec(/\x./\y.X(x, y), Y, Z))  because [5], by definition 
  5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(Z, rec(/\x./\y.X(x, y), Y, Z))  because [6], by (Select) 
  6] X(rec*(/\x./\y.X(x, y), Y, s(Z)), rec*(/\z./\u.X(z, u), Y, s(Z))) >= X(Z, rec(/\x./\y.X(x, y), Y, Z))  because [7] and [11], by (Meta) 
  7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= Z  because [8], by (Select) 
  8] s(Z) >= Z  because [9], by (Star) 
  9] s*(Z) >= Z  because [10], by (Select) 
  10] Z >= Z  by (Meta) 
  11] rec*(/\x./\y.X(x, y), Y, s(Z)) >= rec(/\x./\y.X(x, y), Y, Z)  because rec in Mul, [12], [17] and [18], by (Stat) 
  12] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [13], by (Abs) 
  13] /\z.X(y, z) >= /\z.X(y, z)  because [14], by (Abs) 
  14] X(y, x) >= X(y, x)  because [15] and [16], by (Meta) 
  15] y >= y  by (Var) 
  16] x >= x  by (Var) 
  17] Y >= Y  by (Meta) 
  18] s(Z) > Z  because [19], by definition 
  19] s*(Z) >= Z  because [10], by (Select) 

We can thus remove the following rules:

  rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  rec(/\x./\y.X(x, y), Y, 0) >? Y 

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

We choose Lex = {} and Mul = {0, rec}, and the following precedence: 0 > rec

With these choices, we have:

  1] rec(/\x./\y.X(x, y), Y, 0) > Y  because [2], by definition 
  2] rec*(/\x./\y.X(x, y), Y, 0) >= Y  because [3], by (Select) 
  3] Y >= Y  by (Meta) 

We can thus remove the following rules:

  rec(/\x./\y.X(x, y), Y, 0) => Y 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.