INF3170 / INF4171 Models and axioms Exercises Autumn 2015 Exercise 1 In this exercise we will use the axiom of set existence (axiom 0) 9x(x = x); the axiom of extentionality (axiom 1) 8x8y[8z(z 2 x $ z 2 y) ! x = y]; and the comprehension scheme (axiom 3) 8y9z(x 2 z $ x 2 y ^ φ(x)): Part 1a Explain why any model satisfying the above axioms must also satisfy 9y8x(x 62 y): Hint: φ can be a contradiction. Part 1b Explain how we can use comprehension to show that 9!y8x(x 62 y); where 9!y (y) is equivalent to 9y( (y) ^ 8x( (x) ! x = y)). Hint: If y is empty then x 62 y for any x. 1 Part 1c Since the there is only one empty set, we can introduce the symbol ; de- noting the unique set with no elements. That is, φ(;) is equivalent to φ(z) ^ 8x(x 62 z). Rewrite 9z(; 2 z ^ 8x(x 2 z ! S(x) 2 z)) to a formula that does not contain the symbol ;. Make sure not to reuse variables. Exercise 2 Part 2a (challenging) The axiom of pairing (axiom 4) 8x8y9z(x 2 z ^ y 2 z) sais that as long as we have two sets x and y, we have at least one set large enough that it contains both x and y. We now express a formula PAIR(x; y; z) that states that z is a set containing only x and y. PAIR(x; y; z) $ x 2 z ^ y 2 z ^ 8w(w 2 z ! (w = x _ w = y)) Use the axioms stated so far to prove that 8x8y9!zPAIR(x; y; z). Hint: For existence, you can divide the formula into two conjuncts, one equal to the axioms of pairing, and one similar to the axioms of comprehen- sion with φ(w) $ (w = x _ w = y). For uniqueness, assume there are two solutions and prove that they are equal using extensionality. Part 2b Using the result from part a, we are justified in defining a function pair(x; y) so that pair(x; y) = z , PAIR(x; y; z): Instead of pair(x; y), we often write fx; yg. If we do not want to use this new notation, we can rewrite any formula φ(fx; yg) to φ(z) ^ PAIR(x; y; z). Write w = fx; fy; zgg without the new notation (you can use the short- hand PAIR(x; y; z) for the long formula). 2 Exercise 3 The axiom of union (axiom 5) 8F9A8Y 8x [(x 2 Y ^ Y 2 F) ! x 2 A] ; says that given a set of sets, we can form the union of the inner sets. Com- bining this with comprehension, we can jusify the notation [ [ F = Y Y 2F = fx j 9Y (x 2 Y ^ Y 2 F)g: Using the union operator, we can now build sets with more than two elemets. Given sets a; b; c; d; e; f, we can build the set containing exactly these as elements as follows: [ [ pair pair(a; b); pair [pair(c; d); pair(e; f)] [ [ = pair fa; bg; pair [fc; dg; fe; fg] [ [ = pair fa; bg; ffc; dg; fe; fgg [ = pair (fa; bg; fc; d; e; fg) [ = ffa; bg; fc; d; e; fgg = fa; b; c; d; e; fg: We can build singletons fxg using pair(x; x). Use only [ and pair to build the set fa; b; c; d; e; f; gg: Exercise 4 (challenging) Using union, and pairing, we define S(x) = x [ fxg, or [ S(x) = pair(x; pair(x; x)): We use this definition to create sets to represent the natural numbers. 0 = ; 1 = S(0) = f;g = f0g 2 = S(1) = f;; f;gg = f0; 1g 3 = S(2) = f;; f;g; ff;; f;gggg = f0; 1; 2g . 3 Since for every set x we can prove that there is a set S(x), any model satis- fying axioms 0, 1, 3, 4 and 5 must have an infinite domain. The axiom of infinity (axiom 7) 9x [; 2 x ^ 8y(y 2 x ! S(y) 2 x)] states that our domain must also contain an infinite set. That is, there is some set d where ci 2 d for an infinite number of ci. Using comprehension with a carefully selected formula, we can form the set ! = f0; 1; 2;:::g: That is, the set containing exactly the successor sequence starting at ;. Show that 2 is a strict total order on !, that is, it behaves as the less-than relation on numbers. A strict total order R must satisfy the following: •8 x8y8z(xRy ^ yRz ! xRz (transitivity) •8 x8y(xRy !:yRx) (asymmetry) •8 x8y(xRy _ yRx _ x = y) (totality) 4.