Goodstein's theorem
In mathematical logic, Goodstein's theorem is a statement about the natural numbers, proved by Reuben Goodstein in 1944, which states that every "Goodstein sequence" eventually terminates at 0. Kirby and Paris showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such as second-order arithmetic). This was the third example of a true statement that is unprovable in Peano arithmetic, after the examples provided by Gödel's incompleteness theorem and Gerhard Gentzen's 1943 direct proof of the unprovability of ε-induction in Peano arithmetic. The Paris–Harrington theorem  gave another example. Laurence Kirby and Jeff Paris introduced a graph-theoretic hydra game with behavior similar to that of Goodstein sequences: the "Hydra" (named for the mythological multi-headed Hydra of Lerna) is a rooted tree, and a move consists of cutting off one of its "heads" (a branch of the tree), to which the hydra responds by growing a finite number of new heads according to certain rules. Kirby and Paris proved that the Hydra will eventually be killed, regardless of the strategy that Hercules uses to chop off its heads, though this may take a very long  time. Just like for Goodstein sequences, Kirby and Paris showed that it cannot be proven in Peano arithmetic alone. Hereditary base-"n" notation. Goodstein sequences are defined in terms of a concept called "hereditary base-"n" notation". This notation is very similar to usual base-"n" positional notation, but the usual notation does not suffice for the purposes of Goodstein's theorem. In ordinary base-"n" notation, where "n" is a natural number greater than 1, an arbitrary natural number "m" is written as a sum of multiples of powers of "n": where each coefficient "a" satisfies , and . For example, in base 2 , Thus the base-2 representation of 35 is 100011, which means . Similarly, 100 represented in base-3 is 10201: Note that the exponents themselves are not written in base-"n" notation. For example, the expressions above include 2 and 3, and 5 > 2, 4 > 3. To convert a base-"n" representation to hereditary base-"n" notation, first rewrite all of the exponents in base-"n" notation. Then rewrite any exponents inside the exponents, and continue in this way until every number appearing in the expression (except the bases themselves) has been converted to base-"n" notation. For example, while 35 in ordinary base-2 notation is  , it is written in hereditary base-2 notation as using the fact that Similarly, 100 in hereditary base-3 notation is Goodstein sequences. The Goodstein sequence "G"("m") of a number "m" is a sequence of natural numbers. The first element in the sequence "G"("m") is "m" itself. To get the second, "G"("m")(2), write "m" in hereditary base-2 notation, change all the 2s to 3s, and then subtract 1 from the result. In general, the term, , of the Goodstein sequence of "m" is as follows: Early Goodstein sequences terminate quickly. For example, "G"(3) terminates at the 6th step: Later Goodstein sequences increase for  a very large number of steps. For example, "G"(4) starts as follows: Elements of "G"(4) continue to increase for a while, but at base formula_6, they reach the maximum of formula_7, stay there for the next formula_6 steps, and then begin their descent. The value 0 is reached at base formula_9. (Curiously, this is a Woodall number: formula_10. This is also the case with all other final bases for starting values greater than 4.) However, even "G"(4) doesn't give a good idea of just "how" quickly the elements of a Goodstein sequence can increase. "G"(19) increases much more rapidly  and starts as follows: In spite of this rapid growth, Goodstein's theorem states that every Goodstein sequence eventually terminates at 0, no matter what the starting value is. Proof of Goodstein's theorem. Goodstein's theorem can be proved (using techniques outside Peano arithmetic, see below) as follows: Given a Goodstein sequence "G"("m"), we construct a parallel sequence "P"("m") of ordinal numbers which is strictly decreasing and terminates. Then "G"("m") must terminate too, and it can terminate only when it goes to 0. A common misunderstanding of this proof is to believe that "G"("m") goes to 0 "because" it  is dominated by "P"("m"). Actually, the fact that "P"("m") dominates "G"("m") plays no role at all. The important point is: "G"("m")("k") exists if and only if "P"("m")("k") exists (parallelism). Then if "P"("m") terminates, so does "G"("m"). And "G"("m") can terminate only when it comes to 0. We define a function formula_11 which computes the hereditary base representation of and then replaces each occurrence of the base with the first infinite ordinal number ω. For example, formula_12. Each term "P"("m")("n") of the sequence "P"("m") is then defined as "f"("G"("m")("n"),"n+1"). For example, and . Addition, multiplication and exponentiation of ordinal numbers are well  defined. We claim that formula_13: Let formula_14 be "G"("m")("n") after applying the first, "base-changing" operation in generating the next element of the Goodstein sequence, but before the second "minus 1" operation in this generation. Observe that formula_15. Then clearly, formula_16. Now we apply the "minus 1" operation, and formula_17, as formula_18. For example, formula_19 and formula_20, so formula_21 and formula_22, which is strictly smaller. Note that in order to calculate "f(G(m)(n),n+1)", we first need to write "G"("m")("n") in hereditary base notation, as for instance the expression formula_23 either makes no sense or is equal to formula_24. Thus the sequence  P("m") is strictly decreasing. As the standard order < on ordinals is well-founded, an infinite strictly decreasing sequence cannot exist, or equivalently, every strictly decreasing sequence of ordinals terminates (and cannot be infinite). But "P"("m")("n") is calculated directly from "G"("m")("n"). Hence the sequence "G"("m") must terminate as well, meaning that it must reach 0. While this proof of Goodstein's theorem is fairly easy, the "Kirby–Paris theorem", which shows that Goodstein's theorem is not a theorem of Peano arithmetic, is technical and considerably more difficult. It makes use of countable nonstandard models of Peano arithmetic . Extended Goodstein's theorem. Suppose the definition of the Goodstein sequence is changed so that instead of replacing each occurrence of the base "b" with it replaces it with . Would the sequence still terminate? More generally, let "b", "b", "b", … be any sequences of integers. Then let the term of the extended Goodstein sequence of "m" be as follows: take the hereditary base "b" representation of "G"("m")("n") and replace each occurrence of the base "b" with and then subtract one. The claim is that this sequence still terminates. The extended proof defines as follows: take the hereditary base "b" representation  of "G"("m")("n"), and replace each occurrence of the base "b" with the first infinite ordinal number ω. The "base-changing" operation of the Goodstein sequence when going from "G"("m")("n") to still does not change the value of "f". For example, if and if , then formula_25, hence the ordinal formula_26 is strictly greater than the ordinal formula_27 Sequence length as a function of the starting value. The Goodstein function, formula_28, is defined such that formula_29 is the length of the Goodstein sequence that starts with "n". (This is a total function since every Goodstein sequence terminates.) The extreme growth-rate of  formula_30 can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions formula_31 in the Hardy hierarchy, and the functions formula_32 in the fast-growing hierarchy of Löb and Wainer: Some examples: Application to computable functions. Goodstein's theorem can be used to construct a total computable function that Peano arithmetic cannot prove to be total. The Goodstein sequence of a number can be effectively enumerated by a Turing machine; thus the function which maps "n" to the number of steps required for the Goodstein sequence of "n" to terminate is computable  by a particular Turing machine. This machine merely enumerates the Goodstein sequence of "n" and, when the sequence reaches "0", returns the length of the sequence. Because every Goodstein sequence eventually terminates, this function is total. But because Peano arithmetic does not prove that every Goodstein sequence terminates, Peano arithmetic does not prove that this Turing machine computes a total function.