# Summary of Last Lecture. Automated Reasoning. Outline of the Lecture. Definition. Example. Lemma non-redundant superposition inferences are liftable

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1 Summary Automated Reasoning Georg Moser Institute of Computer UIBK Summary of Last Lecture C A D B (C D)σ C s = t D A[s ] (C D A[t])σ ORe OPm(L) C s = t D u[s ] v SpL (C D u[t] v)σ C s t Cσ ERR C A B (C A)σ OFc C s = t D A[s ] (C D A[t])σ OPm(R) C s = t D u[s ] = v SpR (C D u[t] = v)σ C u = v s = t (C v t u = t)σ EFc Winter 2013 ORe and OFc are ordered resolution and ordered factoring OPm(L), OPm(R), SpL, SpR stands for ordered paramodulation and superpostion (left or right) ERR means equality resolution and EFc means equality factoring GM (Institute of Computer UIBK) Automated Reasoning 320/1 Summary Example re-consider C = {c d, b = d, a d a = c, a = b a = d} together with the term order: a b c d; without equality factoring only the following clause is derivable: a d b = c a = d here the atom order is the multiset extension of : a = b {a, b} {a, d} a = d and the literal order L is the multiset extenion of the atom order: a = c L a d non-redundant superposition inferences are liftable Theorem superposition is sound and complete; let F be a sentence and C its clause form; then F is unsatisfiable iff Res SP (C) GM (Institute of Computer UIBK) Automated Reasoning 321/1 Summary Outline of the Lecture Early Approaches in Automated Reasoning short recollection of Herbrand s theorem, Gilmore s prover, method of Davis and Putnam Starting Points resolution, tableau provers, Skolemisation, ordered resolution, redundancy and deletion Automated Reasoning with Equality paramodulation, ordered completion and proof orders, superposition Applications of Automated Reasoning Neuman-Stubblebine Key Exchange Protocol, group theory Robbin s problem GM (Institute of Computer UIBK) Automated Reasoning 322/1

2 Summary Application ➀: Neuman-Stubblebine Key Exchange Protocol Description Neuman-Stubblebine key exchange protocol aims to establish a secure key between two agents that already share secure keys with a trusted third party principals: Alice, Bob, Server Conventions A, B, T identifiers of Alice, Bob, Server K at key between A and T N a, N b nonce created by Alice, Bob K bt key between B and T Time time span of key K ab K ab key between A and B E key (message) encryption of message using key we write A B: M Alice sends Bob message M GM (Institute of Computer UIBK) Automated Reasoning 323/1 GM (Institute of Computer UIBK) Automated Reasoning 324/1 The Protocol 1 A B: A, N a Alice sends to Bob her identifier a freshly generated nonce 2 B T: B, E Kbt (A, N a, Time), N b Bob encrypts the triple (A, N a, Time) and sends to Server his identity encryption of (A, N a, Time) new nonce 3 T A: E Kat (B, N a, K ab, Time), E Kbt (A, K ab, Time), N b Server generates K ab and sends to Alice encryption of K ab with key for Alice encryption of K ab with key for Bob N b 4 A B: E Kbt (A, K ab, Time), E Kab (N b ) Alice encrypts Bob s nonce with K ab and forwards part of message GM (Institute of Computer UIBK) Automated Reasoning 325/1 The Attack Assumptions 1 intruder can intercept and record all sent messages 2 intruder can send messages and can forge the sender of a message 3 intruder can encrypt messages, when he finds out a key 4 intruder has no access to information private to Alice, Bob, or Server the server. 5 intruder cannot break any secure key still Intruder (denoted I) can break the protocol 1 I(A) B: A, N a 2 B I(T): B, E Kbt (A, N a, Time), N b. 3 I(A) B: E Kbt (A, N a, Time), E Na (N b ). the problem is that keys and nonces can be confused E Kbt (A, K ab, Time) and E Kbt (A, N a, Time) GM (Institute of Computer UIBK) Automated Reasoning 326/1

3 Formalisation in First-Order definition of the language L of the formalisation 1 individual constants: a, b, t, na, at, bt a, b, t are to be interpreted as the identifiers A, B, and T constant na refers to Alics s nonce at (bt) represents the key K at (K bt ) 2 function constants: nb, tb, kt, key, sent, pair, triple, encr, quadr nb, tb, kt are unary; key, pair, encr are binary; sent, triple are ternary, and quadr is 4-ary nb, tb compute Bob s fresh nonce and the time-stamp Time kt computes of the new key the other constants act as containers as the formalisation is based on unary predictes ( (cont d)) 4 predicate constants: Ak, Bk, Tk, P, M, Fresh, Nonce, Store a, Store b Ak, Bk, Tk assert together with key existence of keys P represents principals M represents messages using the function sent Fresh asserts that Bob is only interested in fresh nonces Nonce denotes that its argument is a nonce Store a, Store b denote information that is in the store of Alice or Bob Notation we indicate the type of a bound variable in its name as subscript the bound variable x na indicates that this variable plays the role of the nonce N a GM (Institute of Computer UIBK) Automated Reasoning 327/1 GM (Institute of Computer UIBK) Automated Reasoning 328/1 Formalisation of Protocol A B: A, N a 1: Ak(key(at, t)) 2: P(a) 3: M(sent(a, b, pair(a, na))) Store a (pair(b, na)) B T: B, E Kbt (A, N a, Time), N b 4: Bk(key(bt, t)) 5: P(b) 6: Fresh(na) 7: x a x na (M(sent(x a, b, pair(x a, x na ))) Fresh(x na ) Store b (pair(x a, x na )) M(sent(b, t, triple(b, nb(x na ), encr(triple(x a, x na, tb(x na )), bt))))) GM (Institute of Computer UIBK) Automated Reasoning 329/1 T A: E Kat (B, N a, K ab, Time), E Kbt (A, K ab, Time), N b 8: Tk(key(at, a)) Tk(key(bt, b)) 9: P(t) 10: x b x nb x a x na x time x bt x at (M(sent(x b, t, triple(x b, x nb, encr(triple(x a, x na, x time ), x bt )))) Tk(key(x at, x a )) Tk(key(x bt, x b )) Nonce(x na ) M(sent(t, x a, 11: Nonce(na) triple(encr(quadr(x b, x na, kt(x na ), x time ), x at ), encr(triple(x a, kt(x na ), x time ), x bt ), x nb )))) 12: x Nonce(kt(x)) 13: x (Nonce(tb(x)) Nonce(nb(x))) Remark formulas are not part of the protocol, but prevents that the intruder can generate arbitrarily many keys GM (Institute of Computer UIBK) Automated Reasoning 330/1

4 A B: E Kbt (A, K ab, Time), E Kab (N b ) 14: x nb x k x m x b x na x time ((M(sent(t, a, triple(encr(quadr(x b, x na, x k, x time ), at), x m, x nb ))) Store a (pair(x b, x na ))) M(sent(a, x b, pair(x m, encr(x nb, x k )))) Ak(key(x k, x b ))) 15: x k x a x na ((M(sent(x a, b, pair(encr(triple(x a, x k, tb(x na )), bt), encr(nb(x na ), x k )))) Store b (pair(x a, x na ))) Bk(key(x k, x a ))) Fact SPASS verifies that the protocol terminates in less than a millisecond G = x(ak(key(x, a)) Bk(key(x, b))) Formalisation of the Intruder extend L by predicate constants Ik and Im Behaviour of Intruder 16: x a x b x m (M(sent(x a, x b, x m )) Im(x m )) 17: u v (Im(pair(u, v)) Im(u) Im(v)). 20: u v (Im(u) Im(v) Im(pair(u, v))). 23: x y u ((P(x) P(y) Im(u)) M(sent(x, y, u))) 24: u v ((Im(u) P(v)) Ik(key(u, v))) 25: u v w ((Im(u) Ik(key(v, w) P(w)) Im(encr(u, v))) Fact H extends G by SPASS shows that the protocol insecure in less than a millisecond H = x(ik(key(x, b)) Bk(key(x, a))) GM (Institute of Computer UIBK) Automated Reasoning 331/1 GM (Institute of Computer UIBK) Automated Reasoning 332/1 Application ➁: Robbin s Problem Huntington s Basis B = B; +,,, 0, 1 is a Boolean algebra if 1 B; +, 0 and B;, 1 are commutative monoids 2 a, b, c B: a (b + c) = (a b) + (a c) a + (b c) = (a + b) (a + c) 3 a B: a + a = 1 and a a = 0 a is called complement (or negation) of a consider the following axioms: x + y = y + x n(n(x) + y) + n(n(x) + n(y)) = x the operation n( ) is just complement commutativity (x + y) + z = x + (y + z) associativity Huntington equation GM (Institute of Computer UIBK) Automated Reasoning 333/1 GM (Institute of Computer UIBK) Automated Reasoning 334/1

5 Huntington s Basis Theorem the provided axioms form a minimal axiomatisation of Boolean algebras, that is all axioms are independent from each other Example recall x y = x + y, thus x + y + x + y = x y + x y = x (y + y) = x Question ➀ Does Huntington s equation follow from (i) commutativity (ii) associativity and (iii) Robbins equation? Answer McCune (or better EQP) says yes Robbins equation: Example x + y + x + y = x x + y + x + y = (x + y) (x + y) = x + (yy) = x (R) a Robbins algebra is an algrebra satisfying (i) commutativity (ii) associativity and (iii) Robbins equation Question ➁ Is any Robbins algebra a Boolean algebra? GM (Institute of Computer UIBK) Automated Reasoning 335/1 GM (Institute of Computer UIBK) Automated Reasoning 336/1 Auxiliary s a Robbins algebra satisfying x(x + x = x) is a Boolean algebra automatically provable by EQP in about 5 seconds a Robbins algebra satisfying x y(x + y = x) is a Boolean algebra 1 originally the was manually proven by Steve Winker 2 based on the above lemma EQP can find a proof in about 40 minutes a Robbins algebra satisfying x y(x + y = x) is a Boolean algebra originally the was manually proven by Steve Winker all Robbin algebras satisfy x y(x + y = x) by EQP, dedicated (incomplete) heuristics are essential Theorem commutativity, associativity, and Robinns equation minimally axiomatise Boolean algebra GM (Institute of Computer UIBK) Automated Reasoning 337/1 GM (Institute of Computer UIBK) Automated Reasoning 338/1

6 Proof (of last lemma). n(n(n(x) + y) + n(x + y)) = y 7, (R) n(n(n(x + y) + n(x) + y) + y) = n(x + y) 10, [7 7] n(n(n(n(x) + y) + x + y) + y) = n(n(x) + y) 11, [7 7] n(n(n(n(x) + y) + x + 2y) + n(n(x) + y)) = y 29, [11 7] n(n(n(n(n(x) + y) + x + 2y) + n(n(x) + y) + z) + + n(y + z)) = z 54, [29 7] n(n(n(n(n(x) + y) + x + 2y) + n(n(x) + y) + + n(y + z) + z) + z) = n(y + z) 217, [54 7] n(n(n(n(n(n(x) + y) + x + 2y) + n(n(x) + y) + + n(y + z) + z) + z + u) + n(n(y + z) + u)) = u 674, [217 7] n(n(n(n(3x) + x) + n(3x)) + n(n(n(3x) + x) + 5x)) = = n(n(3x) + x) 6736, [10 674] Proof. n(n(n(3x) + x) + 5x) = n(3x) 8855, [6736 7] n(n(n(n(3x) + x) + n(3x) + 2x)) = n(n(3x) + x) + 2x 8865, [8855 7] n(n(n(3x) + x) + n(3x)) = x 8866, [8855 7] n(n(n(n(3x) + x) + n(3x) + y) + n(x + y)) = y 8870, [8866 7] n(n(3x) + x) + 2x = 2x 8871, [8865] last line asserts: x y(x + y = x) also derived: x y(x + y = x) Remarks SPASS could not find proof in 12 hours mkbtt cannot parse the problem GM (Institute of Computer UIBK) Automated Reasoning 339/1 GM (Institute of Computer UIBK) Automated Reasoning 340/1 EQP is restricted to equational logic and performs AC unification and matching based on basic superposition, that is, paramodulation into substitution parts of terms are forbidded incomplete heuristics AC unifiers are found by finding a basis of a linear Diophantine equation the complete set of unifiers is given as linear combinations of (members of) the basis a subset yields potential unifier if all unification conditions except unification of subterms are fulfilled the super-0 strategy restricts the number of AC unifiers by ignoring supersets if a potential unifier is found NB: the super-0 strategy yields incompleteness for AC matching a dedicated algorithm based on backtracking is used s the weight of a pair of equations be the sum of the size of its members the age of a pair is the sum of the ages of its members GM (Institute of Computer UIBK) Automated Reasoning 341/1 GM (Institute of Computer UIBK) Automated Reasoning 342/1

7 a pairing algorithm used to select the next equation: 1 either the lightest or the oldest pair (not yet selected) is chosen 2 pair selection ratio specifies the ratio lightest oldest 3 default ratio is 1 0 Use of EQP successful attack took place over the course of five weeks the following search parameters were varied 1 limit on the size of retained equations 2 with or without super-0 heuristics 3 with or without basic restriction Thank You for Your Attention! 4 pair selection ratio 1 0 or 1 1 subsequent experiments searched for shorter proofs yielded direct proof without the use of Winker s lemmas GM (Institute of Computer UIBK) Automated Reasoning 343/1 GM (Institute of Computer UIBK) Automated Reasoning 344/1

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