Logical Analysis and Verification of Cryptographic Protocols - Loria

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104 CHAPTER 4. PROTOCOLS WITH VULNERABLE SIGNATURE SCHEMES Let us now consider the case of the Apply rule, and let C ′ be the obtained modified I-constraint system. If the underlying intruder deduction rule is in LDSKS, the fact that t is not a variable implies that the variables of the right-hand side of the rule will be instantiated by the strict maximal subterms t1, . . . , tk of t. We will thus have: M(C ′ ) = M(C) ∪ {t1, . . . , tn} \ {t} and thus M(C ′ ) < M(C). It now suffices to prove the Lemma for the two rules in L”DSKS \ LDSKS: rule x, Sk(y) → sig(x, Sk(y)): The substitution � σ applied is the most general unifier of the unification system sig(x, Sk(y)) ? =∅ t, Sk(y) ? � =∅ u for some u ∈ E. Since this is syntactic unification and since we can assume neither u (by Lemma 9) nor t (by definition of the Apply rule) are variables, we must have u = Sk(u ′ ) and t = sig(t1, t2). The second equation thus yields y = u ′ , with u ′ ∈ Sub(C). Replacing in the first equation, σ is the most general unifier of the equation sig(x, Sk(u ′ )) ? =∅ � sig(t1, t2), which reduces into the set of equations x ? =∅ t1, Sk(u ′ ) ? � =∅ t2 . The first equation implies that x is instantiated by a strict subterm t1 of t. If the second equation is trivial we have M(C ′ ) = M(C) ∪ {t1} \ {t}, and thus M(C ′ ) < M(C). Otherwise, since V ar(Sk(u ′ )) ∪ V ar(t2) ⊆ V ar(C) we have nbv(C ′ ) < nbv(C). rule x, Sk ′ (P k(y), sig(x, Sk(y))) → sig(x, Sk(y)): The substitution σ applied � is the most general unifier of the sig(x, Sk(y)) unification system ? =∅ t, Sk ′ (P k(y), sig(x, Sk(y))) ? � =∅ u for some u ∈ E. Since this is syntactic unification and since we can assume neither u (by Lemma 9) nor t (by definition of the Apply rule) are variables, we must have u = Sk ′ (u ′ 1, u ′ 2) and t = sig(t1, t2). If σ is the identity on C, we are done, since in this case we have M(C ′ ) = M(C) ∪ {t1} \ {t} and thus M(C ′ ) < M(C). Otherwise let us examine how the unification system is solved. It is first transformed into: � x ? =∅ t1, Sk(y) ? =∅ t2, P k(y) ? =∅ u ′ 1, sig(x, Sk(y)) ? =∅ u ′ � 2 Resolving the first equation yields (note that x /∈ V ar(C)): � Sk(y) ? =∅ t2, P k(y) ? =∅ u ′ 1, sig(t1, Sk(y)) ? =∅ u ′ � 2 Let us consider two cases, depending on whether both u ′ 1 and t2 are variables:
4.4. DECIDABILITY RESULTS 105 � • If they are both variables, then solving the first equation removes t2 from V ar(C) but adds a variable y. The second equation will also remove u ′ 1, but since the variable y is already present, it will not add another variable. Since P k(y) and Sk(y) are not unifiable, we note that we must have t2 �= u ′ 1, and thus we have removed two variables and added one by solving the two first equations. The remaining equation contains only variables of the “intermediate” modified I-constraint system, and thus will not add any new variable. In conclusion, in this case, the number of variables of C decreases by at least 1. • If t2 is not a variable, and thus t2 = Sk(t ′ 2), with t ′ 2 ∈ sub(C). Resolving the first equation and injecting the solution in the remaining equations yields the unification system: � P k(t ′ 2) ? =∅ u ′ 1, sig(t1, Sk(t ′ 2)) ? =∅ u ′ � 2 Note that up to this point the substitution σ that we built does not affect any variable of C. If this remaining unification system is trivial, then the substitution applied on C is the identity, we are done (see above). Otherwise, since all the variables in this system are in V ar(C), it strictly reduces nbv(C). This terminates the proof of this case. Thus, if this rule is applied, either no substitution is applied on C and M(C) strictly decreases, or the number of variables in the resulting modified Iconstraint system C ′ is strictly smaller than the number of variables in C. Lemma 37 Let C = (E1 ⊲ t1, . . . , En ⊲ tn) be a modified I-constraint system. The application of transformation rules of the algorithm on C using L”DEO rules terminates. PROOF. Let C = (E1 ⊲t1, . . . , En⊲tn) be a modified I-constraint system not in solved form and let the complexity of C be a couple ordered lexicographically with the following components: • nbv(C), the number of distinct variables in C, • M(C) the multiset of the right-hand side of deduction constraints in C. We have to show that each rule reduces the complexity. The fact is obvious if the Unif rule is applied, since it amounts to the unification of two subterms of C. If is then well-known that if two subterms are not syntactically equal, then the number of variables in their most general unifier is strictly less than the union of their variables, which is included in V ar(C). If their are syntactically equal, then
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Logical Analysis and Verification o
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202 BIBLIOGRAPHY [12] M. Abadi and
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208 BIBLIOGRAPHY [83] H. Comon-Lund
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214 BIBLIOGRAPHY [160] J. C. Mitche
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216 BIBLIOGRAPHY [187] H. Seidl and
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Logical Analysis and Verification of Cryptographic Protocols - Loria

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