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The aim of this section is to describe how the ElementaryDPLL(T ) procedure can be transposed into a proof-search process for sequents of theLK^p(T ) calculus.

Again, a complete and successful run of DPLL(T ) is a sequence of transitions ∅kφ ⇒^∗UNSAT, which ensures that the set of clauses φ is inconsistent modulo the theory. Hence, we are devising a proof-search process aiming at building anLK^p(T ) proof-tree for sequents of the form φ⁰`, where φ⁰ represents the set of clauses φ as a sequent calculus structure, in the following sense:

Definition 47 (Representation of clauses as formulae) An LK^p(T ) formula C⁰ represents a DPLL(T ) clause {lj}j=1...p if C⁰ = l1∨⁻. . . ∨⁻lp.

A set of formulae φ⁰ represents a set of clauses φ if there is a bijection f from φ to φ⁰ such that

for all clauses C in φ, f(φ) represents C. ※

Remark 46 If C⁰ represents C, then ](C⁰) ≤ 2](C) (there are fewer symbols ∨⁻ than there are literals in C).

Note, here that we carefully use the negative disjunction connective to translate DPLL(T ) clauses. This is crucial not only to mimic DPLL(T ) without duplicating formulae but more generally to control the search space. Again coming back to the DPLL(T ) transition sequence

∅kφ ⇒^∗∆kφ and its intuitive counterpart in sequent calculus, we have to formalise the notion of incomplete proof-tree together with the notion of “filling its holes”:

Definition 48 (Incomplete proof-tree, extension of an incomplete proof-tree) An incomplete proof-tree in LK^p(T ) is a tree labelled with sequents,

• whose leaves are tagged as either open or closed;

• whose open leaves are labelled with sequents without focus or right-hand side;

• and such that every node that is not an open leaf, together with its children, forms an instance of the LK^p(T ) rules.

The size of a incomplete proof-tree is its number of nodes.

An incomplete proof-tree π⁰ is an extension of π, if there is a tree (edge and nodes preserving) homomorphism from π to π⁰. It is an n-extension of π, if moreover the difference of size between π⁰

and π is less than or equal to n. ※

Remark 47 A incomplete proof-tree that has no open leaf is (isomorphic to) a well-formed com-pleteLK^p(T ) proof of the sequent labelling its root. In that case, we say the proof-tree is complete.

The intuition that an intermediate DPLL(T ) state describes an “interface” between a incom-plete proof-tree and the comincom-plete proof-trees that should be plugged into its holes, is formalised as follows:

Definition 49 (Correspondence)

An incomplete proof-tree π corresponds to a DPLL(T ) state ∆kφ if:

• the length of ∆::J∆K is the number of open leaves of π;

• if ∆i is the i^th element of ∆::J∆K, then the sequent labelling the i

th open leaf of π (taken left-to-right) is of the form ∆⁰, φ⁰ `^∆i , where:

– φ⁰ represents φ (in the sense of Definition47);

– for all l ∈ ∆⁰, l ∈ ∆i

– for all l ∈ ∆i, ∆⁰ |=_T l.¹

A incomplete proof-tree π corresponds to the state UNSAT if it has no open leaf. ※ Remark 48 In the general case, different incomplete proof-trees might correspond to a same DPLL(T ) state (just like different DPLL(T ) runs may reach that state from the initial one).

Note that we do not require anything from the conclusion of a incomplete proof-tree corres-ponding to ∆kφ: just as our correspondence says nothing about the DPLL(T ) transitions taking place after ∆kφ (nor about the trees to be plugged into the open leaves), it says nothing about the transitions taking place before ∆kφ (nor about the incomplete proof-tree, except for its open leaves).

If an incomplete proof-tree π corresponds to a DPLL(T ) state ∆kφ where there is no decision literals in ∆, then there is exactly one open leaf in π, and it is labelled by a sequent of the form

∆⁰, φ⁰ `^∆ , where φ⁰ represents φ and Cloφ(∆) =Cloφ(∆⁰).

To the initial state ∅kφ of a run of theDPLL(T ) procedure corresponds the incomplete proof-tree consisting of one node (both root and open leaf) labelled with the sequent φ⁰ ` , where φ⁰ represents φ.

The simulation theorem below provides a systematic way of interpreting any DPLL(T ) trans-ition as a completion of incomplete proof-trees that preserves the correspondence given in Defini-tion49and controls the growth of the proof trees.

Theorem 49 (Simulation of DPLL(T ) in LK^p(T )) If ∆kφ ⇒ S2is a valid DPLL(T ) transition, and π1 is an incomplete proof tree in LK^p(T ) corresponding to ∆kφ, then there exists a (2](φ) + 3)-extension π2 of π1that corresponds to S2.

Proof: By case analysis on the nature of the transition, completing the leftmost open leaf of π1:

• Decide: ∆kφ ⇒ ∆, l^dkφ where l 6∈ ∆, l^⊥ 6∈ ∆, l ∈lit(φ).

Let π1 be an incomplete proof-tree corresponding to ∆kφ. The leftmost leaf (corresponding to ∆) is of the form ∆⁰, φ⁰ `^∆ where φ⁰ represents φ and Cloφ(∆) =Cloφ(∆⁰).

We extend π₁into π₂by replacing the leftmost leaf by the following (incomplete) proof-tree:

∆⁰, l, φ⁰ `^∆;l

∆⁰, φ⁰ `^∆ l^⊥

∆⁰, l^⊥, φ⁰ `^∆;l⊥

∆⁰, φ⁰ `^∆l

∆⁰, φ⁰ `^∆

1The last two conditions entail in particular that Cloφ(∆⁰) = Cloφ(∆i).

Note that we use here the analytic cut rule of LK^p(T ). π2 is a 3-extension of π1 that cor-responds to ∆, l^dkφ. Indeed, we have ∆, l^d::J∆, l

d

K = (∆, l) :: (∆, l

⊥) ::J∆K and Cloφ(∆, l) = Cloφ(∆⁰, l) and Cloφ(∆, l^⊥) =Cloφ(∆⁰, l^⊥). The two new leaves are tagged as open.

• Propagate: ∆kφ, C ∨ l ⇒ ∆, lkφ, C ∨ l where ∆ |= ¬C, l 6∈ ∆, l^⊥6∈ ∆.

Let π1be an incomplete proof-tree corresponding to ∆kφ, C ∨ l. The open leaf corresponding to ∆ is of the form ∆⁰, φ⁰, C⁰ `^∆ where φ⁰represents φ, C⁰represents C ∨l andCloφ,C∨l(∆) = Cloφ,C∨l(∆⁰). Let C = l₁∨ . . . ∨ l_n. From ∆ |= ¬C we get ∀i, l^⊥_i ∈ ∆ ⊆ Cloφ,C∨l(∆) = Cloφ,C∨l(∆⁰).

We extend π1 into π2by replacing this leaf by the following (incomplete) proof-tree:

∆⁰, l, φ⁰, C⁰ `^∆;l

∆⁰, φ⁰, C⁰ `^∆l^⊥

∆⁰, φ⁰, C⁰ `^∆ [l^⊥]



 ∆⁰, φ⁰, C⁰ `^∆ [l_i^⊥]





l_i∈C

··

··(∧⁺).

∆⁰, φ⁰, C⁰ `^∆[C^0⊥]

∆⁰, φ⁰, C⁰ `^∆

The top-right rules can be applied since l^⊥_i ∈ ∆ and ∆⁰, li |=T and the new leaves are closed.

The top-left leaf is tagged as open.

Noticing that ∆, l ::J∆, lK = (∆, l) :: J∆K, we get that π2 is a ](C⁰) + 3-extension of π₁ that corresponds to ∆, lkφ, C ∨ l (and note that ](C⁰) ≤ 2](φ)).

• Fail: ∆kφ, C ⇒ UNSAT with ∆ |= ¬C and there is no decision literal in ∆.

Let π1 be an incomplete proof-tree corresponding to ∆kφ, C. Since there are no decision literals in ∆, π₁ has exactly one open leaf, and it is labelled by ∆⁰, φ⁰, C⁰ `^∆ where φ⁰ represents φ, C⁰ represents C andCloφ,C(∆) =Cloφ,C(∆⁰). Let C = l₁∨ . . . ∨ ln.

From ∆ |= ¬C we get ∀i, l_i^⊥∈ ∆ ⊆Cloφ,C(∆) =Cloφ,C(∆⁰).

We extend π₁ into π₂by replacing the open leaf by the following (complete) proof-tree:



 ∆⁰, φ⁰, C⁰ `^∆[l^⊥_i ]





l_i∈C

··

··∧⁺.

∆⁰, φ⁰, C⁰ `^∆[C^0⊥]

∆⁰, φ⁰, C⁰ `^∆

The top rules can be applied since l^⊥_i ∈ ∆ and ∆⁰, l_i |=T. All the leaves are closed. π2 is a ](C⁰) + 1-extension of π1 that is complete, and therefore corresponds to theUNSAT state of theDPLL(T ) run (and note that ](C⁰) ≤ 2](φ)).

• Backtrack: ∆1, l^d, ∆₂kφ, C ⇒ ∆₁, l^⊥kφ, C if ∆₁, l, ∆₂|= ¬C and no decision literal is in ∆₂.

Let π1 be an incomplete proof-tree corresponding to ∆1, l^d, ∆2kφ, C. The open leaf corres-ponding to ∆1, l^d, ∆2is of the form ∆⁰, φ⁰, C⁰ `^∆1,l,∆2 where φ⁰ represents φ, C⁰ represents C andCloφ(∆1, l, ∆2) =Cloφ(∆⁰). Let C = l1∨ . . . ∨ ln.

From ∆1, l, ∆2|= ¬C we get ∀i, l_i^⊥∈ ∆1, l, ∆2⊆Cloφ,C(∆1, l, ∆2) =Cloφ,C(∆⁰).

We extend π₁ into π₂by replacing this leaf by the following (complete) proof-tree:



 ∆⁰, φ⁰, C⁰ `^∆1,l,∆2 [l^⊥_i ]





l_i∈C

··

··∧⁺.

∆⁰, φ⁰, C⁰ `^∆1,l,∆2 [C^0⊥]

∆⁰, φ⁰, C⁰ `^∆1,l,∆2

The top rules can be applied since l^⊥_i ∈ ∆₁, l, ∆₂ and ∆⁰, l_i|=_T.

Noticing that J∆¹, l^d, ∆2K = (∆¹, l^⊥) ::J∆¹K, we get that π² is a ](C⁰) + 1-extension of π1

that corresponds to ∆1, l^⊥kφ, C (and note that ](C⁰) ≤ 2](φ)).

• PropagateT: ∆kφ ⇒ ∆, lkφ where l ∈Cloφ(∆) and l 6∈ ∆, l^⊥6∈ ∆.

Let π1 be an incomplete proof-tree corresponding to ∆kφ. The open leaf corresponding to

∆ is of the form ∆⁰, φ⁰ `^∆ where φ⁰ represents φ andCloφ(∆) =Cloφ(∆⁰). We extend π1

into π2 by replacing this leaf by the following (incomplete) proof-tree:

∆⁰, φ⁰ `^∆,l

∆⁰, φ⁰ `^∆

Noticing thatCloφ(∆) =Cloφ(∆, l), π2 is a 1-extension of π1 that corresponds to ∆, lkφ.

• FailT: ∆kφ ⇒UNSAT with ∆ |=_T and there is no decision literal in ∆.

Let π1be an incomplete proof-tree corresponding to ∆kφ. Since there are no decision literals in ∆, π1has exactly one open leaf, and it is labelled by ∆⁰, φ⁰ `^∆ where φ⁰represents φ and Cloφ(∆) =Cloφ(∆⁰).

We extend π₁ into π₂by replacing the open leaf by the following (complete) proof-tree:

∆⁰, φ⁰|=_T

∆⁰, φ⁰ `^∆

Here, π2 is a 1-extension of π1 that is complete and corresponds to UNSAT state of the DPLL(T ) run.

• BacktrackT: ∆₁, l^d, ∆₂kφ ⇒ ∆₁, l^⊥kφ if ∆₁, l, ∆₂|=_T and no decision literal is in ∆₂. Let π1be an incomplete proof-tree corresponding to ∆1, l^d, ∆2kφ. The open leaf correspond-ing to ∆1, l^d, ∆2 is of the form ∆⁰, φ⁰ `^∆1,l,∆2 where φ⁰ represents φ andCloφ(∆1, l, ∆2) = Cloφ(∆⁰).

We extend π₁ into π₂by replacing this leaf by the following (complete) proof-tree:

∆⁰, φ⁰ |=_T

∆⁰, φ⁰ `^∆1,l,∆2

Noticing that J∆1, l^d, ∆₂K = (∆₁, l^⊥) ::J∆1K, we get that π₂ is a 1-extension of π₁ that corresponds to ∆₁, l^⊥kφ state of theDPLL(T ) run.



Corollary 50 If ∅kφ ⇒ⁿUNSAT and φ⁰ represents φ then there is an complete proof in LK^p(T ) of φ⁰ ` , of size smaller than (2](φ) + 3)n.

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