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University of Twente The Netherlands Centre for Telematics and Information Technology Verification of Security Protocols Sandro Etalle etalle@cs.utwente.nl
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University of Twente The Netherlands Centre for Telematics and Information Technology Outline Day 1: Practice Using the tool we developed in Twente Day 2: Theory the constraint-solving algorithm
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University of Twente The Netherlands Centre for Telematics and Information Technology Schema of Day 1 A couple of words on Security Protocols How to specify a protocol How to specify a particular session How to find security and authentication flaws Interpreting the result of the tool
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University of Twente The Netherlands Centre for Telematics and Information Technology Part 0 What are security protocols anyway?
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University of Twente The Netherlands Centre for Telematics and Information Technology Security Protocols Use symmetric or public key cryptography. May achieve Confidentiality. Authentication. For assigning responsibility. For giving credit. Non-Repudiation. …. Difficult to do right!
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University of Twente The Netherlands Centre for Telematics and Information Technology The same, old example: Needham-Schroeder The same, old example: Needham-Schroeder A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) Notation msg*k: asymmetric encryption Na, Nb: nonces A, B: Agents (Alice and Bob) pk(A): public key of A
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University of Twente The Netherlands Centre for Telematics and Information Technology Goals of NS A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) Exchange two secrets can be used to form a key Authentication
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University of Twente The Netherlands Centre for Telematics and Information Technology What can go wrong A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) A -> I: [A,Na]*pk(I) I(A) -> B:[A,Na]*pk(B) B-> A:[Na,Nb]*pk(A) A->I : [Nb]*pk(I) I(A)->B : [Nb]*pk(B) Secrecy (Na and Nb are disclosed) Authentication B “thinks” he is talking to A, while he is talking to I I is the intruder.
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University of Twente The Netherlands Centre for Telematics and Information Technology Finding Flaws is not Easy Protocol dates: 1978. Flaw found in 1995. Difficult to do “by hand” One can use Belief logics (e.g. BAN logic). Theorem Proving. Model Checking and alike.
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University of Twente The Netherlands Centre for Telematics and Information Technology The Model Checking Approach Basic idea: Model the protocol (finite!). Model the intruder (Dolev-Yao intruder). Can intercept & learn messages. Can forge new messages using his knowledge. Problems (source of infiniteness) Forging new messages (can be fixed). Number of parallel sessions.
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University of Twente The Netherlands Centre for Telematics and Information Technology Two Aspects of Authentication Authentication can be used: To assign responsibility. a message is supported by someone To assign credit.
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University of Twente The Netherlands Centre for Telematics and Information Technology Example A -> B: [A,B,[K,A,B,T]*sk(A)] A -> B: [A,B,[M]*K -1 ] (2) T: timestamp K, K -1 : asymmetric key pair sk(A): secret key of A When A sends (2) we can assume he takes responsibility for it Should we give her credit as well?
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University of Twente The Netherlands Centre for Telematics and Information Technology Not suitable for giving Credit A -> C: [A,B,[K,A,B,T]*sk(A)] C -> B: [C,B,[K,C,B,T]*sk(C)] A -> C: [A,B,[M]*K -1 ] (2) C -> B: [C,B,[M]*K -1 ] (2) C could fake himself in between. and get the credit It is a man-in-the-middle attack.
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University of Twente The Netherlands Centre for Telematics and Information Technology Another Example A -> B: [A,K]*pk(B) A -> B: [M1]*K (2) B -> A: [M2]*K (3) Can we assign responsibility/credit to A for M1? Can we assign responsibility/credit to B for M2?
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University of Twente The Netherlands Centre for Telematics and Information Technology Another Example A -> B: [A,K]*pk(B) A -> B: [M1]*K (2) B -> B: [M2]*K (3) B can assign credit to A for M1. because A included his name in (1). No responsibility, though: anyone could have generated message 1. A can hold B responsible for M2. But A could have faked (3) so it would be difficult for A to prove that B “did it”. Not really suitable for credit…
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University of Twente The Netherlands Centre for Telematics and Information Technology Part 1 Using CoProVe
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University of Twente The Netherlands Centre for Telematics and Information Technology Preliminaries: Prolog’s notation variables: begin with uppercase or with _ Na,Nb,A,B, _a are variables a,na,nb,b are non-variable terms variable are terms Complex terms can be built using predicate (function) symbols: pk(b) is a non-variable term ( pk is a function symbol) pk(B) Nb*pk(B) is the same as *(Nb,pk(B)) : * is an infix- operator. send(Nb*pk(B))
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University of Twente The Netherlands Centre for Telematics and Information Technology Learning by example: the Needham-Schroeder A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) Notation [t1,t2]: pairing (these are lists in PROLOG) msg*k: asymmetric encryption Conventions Na, Nb: nonces A, B: Agents (Alice and Bob) pk(A): public key of A
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University of Twente The Netherlands Centre for Telematics and Information Technology Roles A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) Here we have 2 ROLES one INITIATOR (A) one RESPONDER (B) A role is specified as a sequence of EVENTS
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University of Twente The Netherlands Centre for Telematics and Information Technology Events events are actions, two kind: send(t) recv(t) t is a term (a message) the crucial part of a role is a list of his actions: [recv([B]), %forget about this one for a moment send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B))] [t1,…,tn]: is a list in Prolog
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University of Twente The Netherlands Centre for Telematics and Information Technology Specifying a Role Fixed (abstract) notation: name(Variables) = [Actions]. E.g. initiator(A,B,Na,Nb) = [ send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B))]. The tool notation is different! compiler notation vs abstract notation (this one)
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University of Twente The Netherlands Centre for Telematics and Information Technology The Responder How does the responder look like? Just exchange “send” and “recv” responder(A,B,Na,Nb) = [ recv([A,Na]*pk(B)), send([Na,Nb]*pk(A)), recv(Nb*pk(B))]). Any name is good (not only “responder) Notice ALL THESE VARIABLES! names & nonces are not fixed roles are parametric
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University of Twente The Netherlands Centre for Telematics and Information Technology Summarizing: We specified the roles of NS: initiator(A,B,Na, Nb), responder(A,B,Na,Nb) We still have to specify how our session looks like how many initiators & how many responders NB: a recent result by Comon-Lundh & Cortier states that 2 agents are sufficient (but give no limit on the number of sessions) The names of the agents are there agents playing both as initiator and responders? We need to define a scenario
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University of Twente The Netherlands Centre for Telematics and Information Technology Part 2 How to specify a particular session
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University of Twente The Netherlands Centre for Telematics and Information Technology System Scenarios Protocol roles provide ‘templates’ Set up a finite scenario for verification choose roles participating in the session instantiate the variables of the roles Instantiation: used for: Say who is playing which role Introduce fresh nonces
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University of Twente The Netherlands Centre for Telematics and Information Technology System Scenarios cont’d A->B : [A,Na]*pk(B) B->A : [Na,Nb]*pk(A) A->B : [Nb]*pk(B) A possible scenario: s1 = {initiator(a,B,na,Nb), responder(A,b,Na,nb)} one INITIATOR A played by agent a one RESPONDER B played by agent b
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University of Twente The Netherlands Centre for Telematics and Information Technology Variables & non-variables Consider the scenario {initiator(a,B,na,Nb), responder(A,b,Na,nb)} Variables indicate parameters that may assume any value (their value is not known at the start). For instance, the initiator here does not know in advance the name of the responder. Fresh nonces = new terms (ground terms that don’t occur elsewhere ).
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University of Twente The Netherlands Centre for Telematics and Information Technology More System Scenarios for NS {initiator(a,b,na,nb), responder(a,b,na,nb)} –the ‘honest’ scenario (but unrealistic) {initiator(a,B,na,Nb), responder(A,b,Na,nb)} –may not communicate with each other {initiator(a,b,na,nb), responder(A,B,Na,Nb)} –a may also play the responder role {initiator(a,b,na,nb), responder(c,d,nc,nd)} –no communication!
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University of Twente The Netherlands Centre for Telematics and Information Technology The network model Network/Intruder Scenario Agent Role Network - intruder: Dolev-Yao. send(t) recv(t)
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University of Twente The Netherlands Centre for Telematics and Information Technology Constraint Store knowledge (K) the intruder’s knowledge: the set of intercepted messages constraint store: {msg_1:K_1, …, msg_n:K_n} msg_1, …, msg_n: messages (terms) K_1, …, K_n: knowledges (set of messages) Is satisfiable: each msg_i is generable using K_i.
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University of Twente The Netherlands Centre for Telematics and Information Technology Overview of the Verification Algorithm A step of the verification algorithm: choose an event e from a role of S Two cases: e = send(t) –t is added to the intruder’s knowledge e = recv(t) –add constraint t:K to the constraint store –if constraint store is solvable, proceed –otherwise, stop
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University of Twente The Netherlands Centre for Telematics and Information Technology Part 3 Using the tool in practice How to find security and authentication flaws
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University of Twente The Netherlands Centre for Telematics and Information Technology Finding Secrecy flaws What is a secrecy flaw? To check if na remains secret, one just has to add to the scenario the singleton role [recv(na)] na remains secret the intruder cannot output it! in practice we define a special role secrecy(X) = [recv(X)].
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University of Twente The Netherlands Centre for Telematics and Information Technology Finding Authentication Flaws More complex than checking secrecy. What is an authentication flaw? Various definitions. Basically: an input event recv(t) without corresponding preceding output event send(t). Can be checked by e.g., running the responder strand without an initiator role. We are working on it.
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University of Twente The Netherlands Centre for Telematics and Information Technology From abstract notation to implementation notation Abstract notation role_name(Var1,…,VarN) = [Events]. Concrete notation role_name(Var1,...,VarN,[Events]). Abstract Notation initiator(A,B,Na,Nb) = [ send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B)) ]). % Implementation Notation initiator(A,B,Na,Nb,[ send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B)) ]).
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University of Twente The Netherlands Centre for Telematics and Information Technology Specification of NS % Initiator role initiator(A,B,Na,Nb,[ send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B)) ]). % Responder role responder(A,B,Na,Nb,[ recv([A,Na]*pk(B)), send([Na,Nb]*pk(A)), recv(Nb*pk(B)) ]). % Standard secrecy-checking role secrecy(X,[recv(X)]).
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University of Twente The Netherlands Centre for Telematics and Information Technology Scenarios in Practice scenario([ [name_1,Var_1],..., [name_n,Var_n] ] ) :- role_1(...,Var_1),... role_n(...,Var_n).
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University of Twente The Netherlands Centre for Telematics and Information Technology For Instance What do we achieve with this scenario? scenario([ [alice,Init1], [bob,Resp1], [sec,Secr1] ] ) :- initiator(a,B,na,Nb,Init1), responder(a,b,Na,nb,Resp1), secrecy(nb, Secr1).
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University of Twente The Netherlands Centre for Telematics and Information Technology Last Details (1): Initial intruder knowledge & has_to_finish % Set up the initial intruder knowledge initial_intruder_knowledge([a,b,e]). % specify which roles we want to force to % finish (only sec in this example) has_to_finish([sec]).
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University of Twente The Netherlands Centre for Telematics and Information Technology The Origination assumption Roles are ‘parametric’, i.e. may contain variables We have to avoid sending out uninstantiated variables (only the intruder may do so). We must satisfy the origination assumption: Any variable should appear for the first time in a recv event So, we add events of the form recv(X), where appropriate
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University of Twente The Netherlands Centre for Telematics and Information Technology Specification of NS with O.A. % Initiator role initiator(A,B,Na,Nb,[ recv(B), send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B)) ]). % Responder role responder(A,B,Na,Nb,[ recv([A,Na]*pk(B)), send([Na,Nb]*pk(A)), recv(Nb*pk(B)) ]). scenario([[alice,Init1], [bob,Resp1], [sec,Secr1]]) :- initiator(a,B,na,Nb,Init1), responder(a,b,Na,nb,Resp1), secrecy(nb, Secr1).
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University of Twente The Netherlands Centre for Telematics and Information Technology Last steps before execution… Decide whether we want Prolog stop at first solution it finds, or iterate and show them all. Click on Verify
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University of Twente The Netherlands Centre for Telematics and Information Technology The Results For each run, the tool visualizes: which events of a role could not be completed (nb: the tools tries to complete each role, but this is sometimes impossible) the complete run.
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University of Twente The Netherlands Centre for Telematics and Information Technology Examples of Results (1) SOLUTION FOUND State: [[alice,[]],[bob,[recv(nb * pk(b))]],[sec,[]]]. alice finishedsec finished! bob did not finish
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University of Twente The Netherlands Centre for Telematics and Information Technology Examples of Results (2) SOLUTION FOUND State: [[a,[]],[b,[recv(nb * pk(b))]],[sec,[]]] Trace: [a,send([a,na] * pk(e))] [b,recv([a,na] * pk(b))] [b,send([na,nb] * pk(a))] [a,recv([na,nb] * pk(a))] [a,send(nb * pk(e))] [sec,recv(nb)]
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University of Twente The Netherlands Centre for Telematics and Information Technology What if we try another scenario? scenario([ [alice1,Init1], [alice2,Init2], [bob,Resp1], [sec,Secr1] ] ) :- initiator(a,b,na,Nb,Init1), initiator(b,A,na,Nb,Init1), responder(a,b,Na,nb,Resp1), secrecy(nb, Secr1). TRY THIS!
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University of Twente The Netherlands Centre for Telematics and Information Technology Exercise 1: Modify NS as Lowe proposed A->B : [A,Na]*pk(B) B->A : [Na,Nb,B]*pk(A) A->B : [Nb]*pk(B) To do implement the roles Try bigger scenarios, with at least two parallel sessions Find Millen’s type flaw attack
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University of Twente The Netherlands Centre for Telematics and Information Technology Millen’s type flaw attack a1. E(A) -> B : {A, E}pk(B) (E is the intruder name, should be a nonce!) a2. B -> E(A): {E, Nb, B}pk(A) b1. E -> A : {E, Nb, B}pk(A) (here is the field confusion) b2. A -> E : {Nb,B, Nb2, A}pk(E)
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University of Twente The Netherlands Centre for Telematics and Information Technology Looking for authentication flaws in Needham-Schroeder Consider (again) the scenario: No secrecy check this time. But, if B is not b, and the responder role finishes, we have an authentication attack! {initiator(a,B,na,Nb), responder(a,b,Na,nb)}
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University of Twente The Netherlands Centre for Telematics and Information Technology Looking for authentication flaws in Needham-Schroeder cont’d We only have to specify has_to_finish to b: has_to_finish([b]). And verify again…
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University of Twente The Netherlands Centre for Telematics and Information Technology Results: the first reported trace SOLUTION FOUND State: [[a,[]],[b,[]]] Trace: [a,send([a,na] * pk(b))] [b,recv([a,na] * pk(b))] [b,send([na,nb] * pk(a))] [a,recv([na,nb] * pk(a))] [a,send(nb * pk(b))] [b,recv(nb * pk(b))] This is a normal run This is a correct trace. To find a flaw we must look for B not instantiated to b!
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University of Twente The Netherlands Centre for Telematics and Information Technology Results: the right trace SOLUTION FOUND State: [[a,[]],[b,[]]] Trace: [a,send([a,na] * pk(e))] [b,recv([a,na] * pk(b))] [b,send([na,nb] * pk(a))] [a,recv([na,nb] * pk(a))] [a,send(nb * pk(e))] [b,recv(nb * pk(b))]
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University of Twente The Netherlands Centre for Telematics and Information Technology Another protocol: Yahalom A->B : A,Na B->S : [A, Na,Nb]+Kbs S->A : [B, Kab, Na, Nb]+Kas, [A,Kab]+Kbs A->B : [A, Kab]+Kbs, [Nb]+Kab [t]+k: symmetric encryption Kxs: shared key between x and s Na, Nb: nonces Goal: establish a secret session key Kab Incorrect (see Clark and Jacob library)
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University of Twente The Netherlands Centre for Telematics and Information Technology Exercise for home For the yahalom protocol: Encode the protocol Verify the protocol: try many scenarios Could you find any flaw? Model leakage of Nb (i.e., B sends Nb in plain at some point) Verify again the protocol: could you find any flaw? Compare this attack to the one described by Clark & Jacob 2. Try the other protocols listed in the online tool. http://130.89.144.15/cgi-bin/show.cgi www.cs.utwente.nl/~etalle
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