Euclid 300 BC 1. Is the side-angle-side (SAS) theorem: "If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent." true? 1. Is the angle-side-side (SAS) theorem: "If two sides and a non-included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent." true? How would Euclid approach these questions? Axiomatic System consists: 1. Axioms 2. Rules to derive new statements from old statements. Euclid's 5 axioms of Geometry: 1. A straight line may be drawn between any two points. 2. Any terminated straight line may be extended indefinitely. 3. A circle may be drawn with any given point as center and any given radius. 4. All right angles are equal. 5'. For any given point not on a given line, there is exactly one line through the point that does not meet the given line. Models of first 4 axioms: Normal Plane, Sphere where "line" is a great circle and "point" is the intersection of two great circles See https://en.wikipedia.org/wiki/Foundations_of_geometry for info about axiomization of geometry Example: Axioms of Group Theory 1. Closure. For all a, b in G, the result of the operation a + b is also in G. 2. Associativity.For all a, b and c in G, the equation (a + b) + c = a + (b + c) holds. 3. Identity element. There exists an element e in G, such that for all elements a in G , the equation e + a = a + e = a holds. 4. Inverse element. For each a in G, there exists an element b in G such that a + b = b + a = e, where e is the identity element. Models of groups: (Z, +) (R^+, *) Example proof rule: Modus ponens from X and X->Y one can conclude Y Abstract example: Axioms: 000, 0000000 Proof rule: If you have O^i and 0^k and a j such that i < j < k then you can conclude 0^j Possible models: 000, 0000, 00000, 000000, 0000000 00, 000, 0000, 00000, 000000, 0000000, 00000000 We write M |= S to mean that S is true in model M We write A |= S to mean that S is true in every model of the axioms A. For convenience we generally just say that S is true Definition: PROOF of statement S = sequence (p_1, ... p_n =S) of sentences such that each formula is an AXIOM or each formula follows from sentences by a PROOF RULE. We write A |- S to mean that one can prove S using axioms A. Question: Which options are possible for a particular S? M |= S M |= not S 1 0 0 1 0 0 1 1 A |= S A |= not S 1 0 0 1 0 0 1 1 A |- S A |- not S 1 0 0 1 0 0 1 1 Back to abstract example with axioms 000 and 0000000 Give two different proofs of 00000 Build the tree of all proofs where the number of proofs with k statements is equal to the number of nodes of depth k Definition: A set is recursively enumerable iff there is a program that outputs only members of that set, and every member of the that set is eventually output Warmup: Positive even integers are recursively enumerate. Proof: Program i= 1 repeat write(2i) i++ Warmup: Even integers are recursively enumerate. Wrong Proof: Program i= 0 repeat write(2i) i++ i= 1 repeat write(-2i) i++ Correct Proof: Program i= 0 repeat write(2i) write(-2i) i++ Note that it is ok for an element (in this case zero) of the enumerated set to be listed more than once Lemma: Theoremhood = {S | Axioms |- S} is recursively enumerable if the axiomization is finite. Proof1: If there only only finitely many consequence of any proof rule then just do a breadth first search of the proof tree of possible proves Proof2: What if you have a proof rule like Forall x phi(x) yields phi(1) phi (2) .....? Answer: homework Definition: An axiomization A of a model M is SOUND if A |- phi then M |= phi That is, one can only derive true statements from phi Definition: An axiomization A is COMPLETE if M |= phi then A |- phi That is, one can derive every true statement using phi Warm-up Lemma 2: If there is a sound and complete finite axiomization of a model M then {phi | M |= phi} is recursive/decidable Proof: Search for all proofs until you reach phi or not phi 1900: Hilbert's 23 problems for the next century http://en.wikipedia.org/wiki/Hilbert%27s_problems 2nd Problem: Prove that the axioms of arithmetic are consistent. Godel's Incompleteness Theorem 1930's (along with Godel's second incompleteness proof kind of kill's Hilbert's second problem): There is no sound and complete axiomization of the standard model of arithmetic Proof: By reduction from the Halting problem. Let P and I be the input to the halting problem. We want to create a first order sentence of the form THEREEXISTS w P(w), where P(w) verifies that w is a valid computation history of P on I. Now since {phi | Number theory |= phi} is recursive, we can to determine whether THEREEXISTS w P(w) or NOT THEREEXISTS w P(w) is true. This allows us to decide whether P halts on I. How find P(w)? Configurations are encoded as strings as proof of term rewriting. If there are b symbols then these strings are then read as base b integers. Yields(C, D) where C are D are integers that encode configurations can be expressed in first order logic. The div and mod operators allow you to cut the string There exist two places to cut C and D such that C and D are equal to the left of the left cut, and to the right of the right cut, and are related by the Table relation in the fixed number of places between the cuts. The fact that a sequence of configurations (put a new symbol between each configuration) encodes a computation can be expressed in first order logic in the following way: Each configuration yields the next, the first configuration is the starting configuration, and the last configuration is the accepting configuration. Proof by example: Consider the Turing Machine states: p=start, q, h=halt input symbols:0,1,space (p,0)->(q,0,right) (p,1)->(p,1,right) (q,0)->(p,0,left) (q,1)->(p,1,right) (p,space)->(h,space,stay) (q,space)->(h,space,stay) Input: 0110 w = #p0110#0q110#01p10#011p0#0110q_#0110h_#