PROOF-OF-WELL-FOUNDEDNESS

a proof that o< is well-founded on o-ps
Major Section:  MISCELLANEOUS

The soundness of ACL2 rests in part on the well-foundedness of o< on o-ps. This can be taken as obvious if one is willing to grant that those concepts are simply encodings of the standard mathematical notions of the ordinals below epsilon-0 and its natural ordering relation. But it is possible to prove that o< is well-founded on o-ps without having to assert any connection to the ordinals and that is what we do here. The book books/ordinals/proof-of-well-foundedness carries out the proof outlined below in ACL2, using only that the natural numbers are well-founded.

Before outlining the above mentioned proof, we note that in the analogous documentation page of ACL2 Version_2.7, there is a proof of the well-foundedness of e0-ord-< on e0-ordinalps, the less-than relation and recognizer for the old ordinals (that is, for the ordinals appearing in ACL2 up through that version). Manolios and Vroon have given a proof in ACL2 Version_2.7 that the current ordinals (based on o< and o-p) are order-isomorphic to the old ordinals (based on e0-ord-< and e0-ordinalp). Their proof establishes that switching from the old ordinals to the current ordinals preserves the soundness of ACL2. For details see their paper:

Manolios, Panagiotis & Vroon, Daron.
Ordinal arithmetic in ACL2.
Kaufmann, Matt, & Moore, J Strother (eds).
Fourth International Workshop on the ACL2 Theorem
Prover and Its Applications (ACL2-2003),
July, 2003.
See http://www.cs.utexas.edu/users/moore/acl2/workshop-2003/.

We now give an outline of the above mentioned proof of well-foundedness. We first observe three facts about o< on ordinals that have been proved by ACL2 using only structural induction on lists. These theorems can be proved by hand.

(defthm transitivity-of-o<
  (implies (and (o< x y)
                (o< y z))
           (o< x z))
  :rule-classes nil)

(defthm non-circularity-of-o<
  (implies (o< x y)
           (not (o< y x)))
  :rule-classes nil)

(defthm trichotomy-of-o<
  (implies (and (o-p x)
                (o-p y))
           (or (equal x y)
               (o< x y)
               (o< y x)))
  :rule-classes nil)
These three properties establish that o< orders the o-ps. To put such a statement in the most standard mathematical nomenclature, we can define the macro:
(defmacro o<= (x y)
  `(not (o< ,y ,x)))
and then establish that o<= is a relation that is a simple, complete (i.e., total) order on ordinals by the following three lemmas, which have been proved:
(defthm antisymmetry-of-o<=
  (implies (and (o-p x)
                (o-p y)
                (o<= x y)
                (o<= y x))
           (equal x y))
  :rule-classes nil
  :hints (("Goal" :use non-circularity-of-o<)))

(defthm transitivity-of-o<=
  (implies (and (o-p x)
                (o-p y)
                (o<= x y)
                (o<= y z))
           (o<= x z))
  :rule-classes nil
  :hints (("Goal" :use transitivity-of-o<)))

(defthm trichotomy-of-o<=
  (implies (and (o-p x)
                (o-p y))
           (or (o<= x y)
               (o<= y x)))
  :rule-classes nil
  :hints (("Goal" :use trichotomy-of-o<)))
Crucially important to the proof of the well-foundedness of o< on o-ps is the concept of ordinal-depth, abbreviated od:
(defun od (l)
  (if (o-finp l)
      0
    (1+ (od (o-first-expt l)))))
If the od of an o-p x is smaller than that of an o-p y, then x is o< y:
(defun od-1 (x y)
  (if (o-finp x)
      (list x y)
    (od-1 (o-first-expt x) (o-first-expt y))))

(defthm od-implies-ordlessp
  (implies (and (o-p x)
                (< (od x) (od y)))
           (o< x y))
  :hints (("Goal"
           :induct (od-1 x y))))
Remark. A consequence of this lemma is the fact that if s = s(1), s(2), ... is an infinite, o< descending sequence of o-ps, then od(s(1)), od(s(2)), ... is a ``weakly'' descending sequence of non-negative integers: od(s(i)) is greater than or equal to od(s(i+1)).

Lemma Main. For each non-negative integer n, o< well-orders the set of o-ps with od less than or equal to n .

 Base Case.  n = 0.  The o-ps with 0 od are the non-negative
 integers.  On the non-negative integers, o< is the same as <.

 Induction Step.  n > 0.  We assume that o< well-orders the
 o-ps with od less than n.

   If o< does not well-order the o-ps with od less than or equal to n,
   consider, D, the set of infinite, o< descending sequences of o-ps of od
   less than or equal to n.  The first element of a sequence in D has od n.
   Therefore, the o-first-expt of the first element of a sequence in D has od
   n-1.  Since o<, by IH, well-orders the o-ps with od less than n, the set
   of o-first-expts of first elements of the sequences in D has a minimal
   element, which we denote by B and which has od of n-1.

   Let k be the minimum integer such that for some infinite, o< descending
   sequence s of o-ps with od less than or equal to n, the first element of s
   has an o-first-expt of B and an o-first-coeff of k.  Notice that k is
   positive.

   Having fixed B and k, let s = s(1), s(2), ... be an infinite, o<
   descending sequence of o-ps with od less than or equal to n such that s(1)
   has a o-first-expt of B and an o-first-coeff of k.

   We show that each s(i) has a o-first-expt of B and an o-first-coeff of
   k. For suppose that s(j) is the first member of s either with o-first-expt
   B and o-first-coeff m (m neq k) or with o-first-expt B' and o-first-coeff
   B' (B' neq B). If (o-first-expt s(j)) = B', then B' has od n-1 (otherwise,
   by IH, s would not be infinite) and B' is o< B, contradicting the
   minimality of B. If 0 < m < k, then the fact that the sequence beginning
   at s(j) is infinitely descending contradicts the minimality of k. If m >
   k, then s(j) is greater than its predecessor; but this contradicts the
   fact that s is descending.

   Thus, by the definition of o<, for s to be a decreasing sequence of o-ps,
   (o-rst s(1)), (o-rst s(2)), ... must be a decreasing sequence. We end by
   showing this cannot be the case. Let t = t(1), t(2), ... be an infinite
   sequence of o-ps such that t(i) = (o-rst s(i)). Then t is infinitely
   descending. Furthermore, t(1) begins with an o-p B' that is o< B. Since t
   is in D, t(1) has od n, therefore, B' has od n-1. But this contradicts the
   minimality of B. Q.E.D.
Theorem. o< well-orders the o-ps. Proof. Every infinite, o< descending sequence of o-ps has the property that each member has od less than or equal to the od, n, of the first member of the sequence. This contradicts Lemma Main. Q.E.D.