Logical Analysis and Verification of Cryptographic Protocols - Loria

6.5. A DECIDABILITY RESULT 171
A → ∗
R g
S
A ′ and by minimality of A ′ in MB the first case implies A ′ → R g
S B.
By transitivity we thus have A → ∗
R g
S
Thus for every atom A ∈ MB we have A →∗ R g B, which contradicts the assump-
S
tion B is maximal in M, and therefore M = ∅. �
Lemma 66 Let S be a set of clauses and C be a ground clause, C = A1, . . . , An → B.
Let π be a ground proof of S |= C and let A be an atom in π maximal with respect to
atoms of π for the ordering ≻π. There exists a clause C ′ ∈ S ∪ � n
i=1 Ai ∪ ¬B and an
atom A ′ ∈ µ(C ′ ) such that A ′ in maximal with respect to atoms of C ′ for the ordering
≻a and A ′ σ = A.
PROOF.
Let π be a ground proof of S |= C. By definition, leaves of π are labelled
by ground instances of clauses in S, positive unit clauses ∅ → Ai for 1 ≤ i ≤
n and negative unit clause B → ∅. Let A ∈ µ(π) be such that A is maximal
with respect to atoms of π for the ordering ≻π. As atoms in π are ground, we
have either A ∈ {A1, . . . , An, B} or there is a ground instance C1 of a clause
in S, such that A ∈ µ(C1). If A ∈ {A1, . . . , An, B} and as each atom in the
set {A1, . . . , An, B} is a unit clause then we conclude the lemma directly. Now,
suppose that A �∈ {A1, . . . , An, B}. Then there exists a ground instance of a
clause in S, C1 such that A ∈ µ(C1). As C1 is a ground instance of a clause in
S, let C s 1 be that clause in S and let A s be an atom in C s 1 such that A is a ground
instance of A s . We have σ is the applied ground substition. Now, let us prove
that A s is maximal with respect to atoms in C s 1 for the ordering ≻a. Let the set
of atoms M = {D such that D is an atom in C s 1 and D is maximal in C s 1 for ≻a}
and let us prove that A s ∈ M. By contradiction, suppose that A s �∈ M, then
there is an atom E s ∈ M such that E s ≻aA s and then E s σ ≻π A s σ, by definition
of the ordering ≻π. We have that E s σ ∈ C1 and then E s σ ≻π A s σ is an atom
of the proof π and that contradicts the maximality of A = A s σ with respect to
atoms of π for the ordering ≻π. We conclude that A s is maximal with respect to
atoms of C s 1 for the ordering ≻a which concludes the proof. �
Let π be a proof of S |= C where S is a set of clauses and C is a ground clause.
By definition of (refutational) proof, we remark that every atom appearing in π
appears in a clause labelling a leave of π.
Lemma 67 Let S be a finite saturated set of clauses, C be a ground clause and Π be the
non-empty set of refutational ground proofs of S |= C. Given π ∈ Π let:
B.
δS(π, C) = (µ(π) ↓S) \ (µ(C) ↓S)
If π ∈ Π is such that δS(π, C) is minimal then δS(π, C) = ∅.