In mathematics, Kruskal's tree theorem states that the set of finite trees over a well-quasi-ordered set of labels is itself well-quasi-ordered under homeomorphic embedding.
The theorem was conjectured by Andrew Vázsonyi and proved by ; a short proof was given by . It has since become a prominent example in reverse mathematics as a statement that cannot be proved in ATR0 (a second-order arithmetic theory with a form of arithmetical transfinite recursion).
In 2004, the result was generalized from trees to graphs as the Robertson–Seymour theorem, a result that has also proved important in reverse mathematics and leads to the even-faster-growing SSCG function, which dwarfs
TREE(3)
The version given here is that proven by Nash-Williams; Kruskal's formulation is somewhat stronger. All trees we consider are finite.
Given a tree with a root, and given vertices,, call a successor of if the unique path from the root to contains, and call an immediate successor of if additionally the path from to contains no other vertex.
Take to be a partially ordered set. If, are rooted trees with vertices labeled in, we say that is inf-embeddable in and write
T1\leqT2
F(v)
F(w)
F(v)
F(w1)
F(w2)
F(v)
Kruskal's tree theorem then states:
If is well-quasi-ordered, then the set of rooted trees with labels in is well-quasi-ordered under the inf-embeddable order defined above. (That is to say, given any infinite sequence of rooted trees labeled in, there is someso thati<j
.)Ti\leqTj
For a countable label set, Kruskal's tree theorem can be expressed and proven using second-order arithmetic. However, like Goodstein's theorem or the Paris–Harrington theorem, some special cases and variants of the theorem can be expressed in subsystems of second-order arithmetic much weaker than the subsystems where they can be proved. This was first observed by Harvey Friedman in the early 1980s, an early success of the field of reverse mathematics. In the case where the trees above are taken to be unlabeled (that is, in the case where has size one), Friedman found that the result was unprovable in ATR0,[1] thus giving the first example of a predicative result with a provably impredicative proof.[2] This case of the theorem is still provable by Π-CA0, but by adding a "gap condition"[3] to the definition of the order on trees above, he found a natural variation of the theorem unprovable in this system.[4] [5] Much later, the Robertson–Seymour theorem would give another theorem unprovable by Π-CA0.
Ordinal analysis confirms the strength of Kruskal's theorem, with the proof-theoretic ordinal of the theorem equaling the small Veblen ordinal (sometimes confused with the smaller Ackermann ordinal).
Suppose that
P(n)
There is some such that if is a finite sequence of unlabeled rooted trees where has
i+n
Ti\leqTj
i<j
All the statements
P(n)
P(n)
P(n)
P(n)
P(n)
Define
tree(n)
There is a sequence of unlabeled rooted trees, where each has at most
i+n
Ti\leqTj
i<j\leqm
It is known that
tree(1)=2
tree(2)=5
tree(3)\geq844,424,930,131,960
tree(4)\ggg64
g64
TREE(3)
| |||||||||||||||||||
tree |
(7).
To differentiate the two functions, "TREE" (with all caps) is the big TREE function, and "tree" (with all letters in lowercase) is the weak tree function.
By incorporating labels, Friedman defined a far faster-growing function. For a positive integer , take
TREE(n)
There is a sequence of rooted trees labelled from a set of labels, where each has at most vertices, such that
Ti\leqTj
i<j\leqm
The TREE sequence begins
TREE(1)=1
TREE(2)=3
TREE(3)
n(4)
nn(5)(5)
n(4)
TREE(3)
AA(187196)(1)
AA(187196)(1)
g | |
3\uparrow1871963 |
gx
Friedman originally denoted this function by TR[''n''].
n(k) is defined as the length of the longest possible sequence that can be constructed with a k-letter alphabet such that no block of letters xi,...,x2i is a subsequence of any later block xj,...,x2j.
n(1)=3,n(2)=11,rm{and}n(3)>2\uparrow7197158386
A(x) taking one argument, is defined as A(x, x), where A(k, n), taking two arguments, is a particular version of Ackermann's function defined as: A(1, n) = 2n, A(k+1, 1) = A(k, 1), A(k+1, n+1) = A(k, A(k+1, n)).
CitationsBibliography