{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

FiniteRevMath101901

# FiniteRevMath101901 - 1 FINITE REVERSE MATHEMATICS Harvey M...

This preview shows pages 1–4. Sign up to view the full content.

1 FINITE REVERSE MATHEMATICS Harvey M. Friedman* Department of Mathematics Ohio State University [email protected] http://www.math.ohio-state.edu/~friedman/ \ December 22, 1999 October 19, 2001 Abstract. We present some formal systems in the language of linearly ordered rings with finite sets whose nonlogical axioms are strictly mathematical, which correspond to polynomially bounded arithmetic. With an additional strictly mathematical axiom, the systems correspond to exponentially bounded arithmetic. 1. T 0 and I S 0 . In this section, we introduce the system T 0 , and show that it corresponds to the system I S 0 of polynomially bounded arithmetic (presented below). Let T 0 be the following system in the two sorted language with variables over integers and variables over finite sets of integers. For the integer sort, we use the language 0,1,+,- ,•,<,= of linearly ordered rings. We use between integers and sets. Equality is used only between integers. The official integer variables are x 0 ,x 1 ,..., and the official set variables are A 0 ,A 1 ,... . The nonlogical axioms of T 0 are as follows. 1. Linearly ordered ring axioms. 2. Finite interval. ( \$ A)( " x)(x A (y < x x < z)). 3. Boolean difference. ( \$ C)( " x)(x C (x A ÿ (x B))). 4. Set addition. ( \$ C)( " x)(x C ( \$ y)( \$ z)(y A z B x = y+z)). 5. Set multiplication. ( \$ C)( " x)(x C ( \$ y)( \$ z)(y A z B x = y•z)). 6. Least element. ( \$ x)(x A) Æ ( \$ x)(x A ÿ ( \$ y)(y A y < x)). The linearly ordered ring axioms are as follows.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 a. x+0 = x. b. x+y = y+x. c. x+(y+z) = (x+y)+z. d. x+(-x) = 0. e. x•1 = x. f. x•y = y•x. g. x•(y•z) = (x•y)•z. h. x•(y+z) = (x•y)+(x•z). i. (x < y y < z) Æ x < z. j. ÿ (x < x). k. x = y x < y y < x. l. 0 < 1. m. x < y Æ x+z < y+z. n. 0 < x Æ (y < z Æ x•y < x•z). We write x-y for x+(-y), x > y for y < x, x £ y for (x < y x = y), x y for y £ x, and x y for ÿ (x = y). Note that these ordered ring axioms suffice to prove the following. o. x < y 0 < y-x. p. x < y -y < -x. q. x•y = 0 (x = 0 y = 0). r. 0 < x•y ((0 < x 0 < y) (x < 0 y < 0)). s. 0 < x Æ (y < z x•y < x•z). t. x < 0 Æ (y < z x•z < x•y). With regard to the interval axiom, we use the notation (a,b) for {x: a < x x < b}. With regard to set difference, we write A\B for {x: x A x B}. With regard to set addition and set multiplication, we write A+B = {x+y: x A x B}, A•B = {x•y: x A x B}. We will often use scalar addition and scalar multiplication. We write A+x = x+A = A+{x}, and A•x = x•A = A•{x}. In Lemmas 1.1-1.27, it is understood that we are asserting provability within T 0 . LEMMA 1.1. Let n 0. i) ÿ (x < y y < x+1); ii) [a,b],[a,b),(a,b] exist; iii) ,{x} exists;
3 iv) x•A = {x•y: y A} exists; v) every nonempty set has a greatest element; vi) every set is included in some interval [a,b]; vii) sets are closed under pairwise union and pairwise intersection; viii) {x 1 ,...,x n } exists; ix) the set of all positive (negative, nonnegative, nonpositive) elements of any set exists. Proof: For i), assume 0 < x < 1. By axiom n, 0 = 0•x < x•x < 1•x = x. Hence there is no least y such that 0 < y < 1. By finite interval, (0,1) exists. By least element, there is a least y such that 0 < y < 1. This is a contradiction. So ÿ (0 < x < 1). Now suppose x < y < x+1. Then 0 < y-x < 1, which is a contradiction.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}