A refined calculus for Intuitionistic Propositional Logic - DISCo

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The previous lemma is the main step to prove the soundness of the calculusT Int . Indeed, let us assume that there exists a closed proof table τ of T Int for{FA}. If {FA} is realizable, by the the previous lemma there exists a leaf S ofτ such that S is realizable, a contradiction (recall that S is a contradictory set).It follows that {FA} is not realizable, and this means that A is valid in all theKripke models, that is A ∈ Int. Thus:Theorem 1 (Soundness). If there exists a closed proof table for {FA}, thenA is intuitionistically valid.5 CompletenessTo prove the completeness we need to introduce the following complexity measuredeg on formulas:– if p is a propositional variable, then deg(p) = 0;– deg(A ∧ B) = deg(A) + deg(B) + 2;– deg(A ∨ B) = deg(A) + deg(B) + 3;– deg(A → B) = deg(A) + deg(B)+ 1;– deg(¬A) = deg(A) + 1.We extend the function deg to signed formulas as follows:– For a signed formula SA (S ∈ {T, F, F c }), deg(SA) = deg(A).– For a finite set S of signed formulas, deg(S) = ∑ H∈S deg(H).It is easy to check that, if S ′ is a set in the consequence of a rule of T Int appliedto a finite set of signed formulas S, then deg(S ′ ) < deg(S).To describe our proof search strategy, we introduce the notion of rule relatedto S, H, where S is a set of signed formulas and H a signed formula.- If H has not the form T(A → B), the rule related to S, H is the only rule ofTable 1 having H as major premise and S \ {H} as set of minor premises.- If H = T(A → B) and TA ∈ S, the rule related to S, H is the rule MP ofTable 2 having H as major premise and S \ {H} as set of minor premises.- If H = T(A → B), TA ∉ S and S = S c , the rule related to S, H is the ruleT → certain of Table 2 having H as major premise and S \ {H} as set ofminor premises.- If H = T(A → B), TA ∉ S and S ≠ S c , the rule related to S, H is one ofthe rules of Table 2 and 3 having H as major premise and S \ {H} as set ofminor premises (there exists only one applicable rule).We notice that given S and H there exists at most a rule R of T Int related toS, H. If R is a splitting rule, we denote with R 1 S,H and R2 S,H the leftmost set andthe rightmost set in the consequence of R respectively; for non-splitting rules wedenote with R 1 S,H the only set in the consequence of R. A set of signed formulasS is consistent if there exists no closed proof table for S.10
Lemma 3. Let S be a finite set of signed formulas. If S is consistent, then Sis realizable.Proof. We prove the assertion by complete induction on deg(S). Let us assumethe assertion holds for all S ′ such that deg(S ′ ) < deg(S); we prove it for S.Let S ⊆ S be the set of signed formulas H of S satisfying one of the followingconditions:(i). H = SA, where S ∈ {T, F, F c }, and A = B ∧ C or A = B ∨ C.(ii). H = T(¬A) or H = T(A ∧ B → C) or H = T(A ∨ B → C).(iii). H = T(A → B) and TA ∈ S.(iv). H = T(A → B) and S = S c .(v). H = T(¬A → C) and (S \ {H}) ∪ {TC} is consistent.(vi). H = T((A → p) → C), p is a propositional variable, and (S \{H})∪{TC}is consistent.(vii). H = T((A → ¬B) → C) and (S \ {H}) ∪ {TC} is consistent.(viii). H = T((A → (X ∧ Y )) → C) and (S \ {H}) ∪ {TC} is consistent.(ix). H = T((A → (X ∨ Y )) → C) and (S \ {H}) ∪ {TC} is consistent.(x). H = T((A → (X → Y )) → C) and (S \ {H}) ∪ {TC} is consistent.Let us assume that S ≠ ∅ and let H be any formula of S. Since S is consistent,there exists k ∈ {1, 2} such that the set S ′ = R k S,H is consistent; moreover, if His one of the signed formulas in cases (v), (vi), (vii), (viii), (ix) and (x), we takeS ′ = RS,H 2 (i.e. S′ = (S \ {H}) ∪ {TC}). Since deg(S ′ ) < deg(S), by inductionhypothesis there exists a Kripke model K = 〈P, ≤, ρ, ⊩〉 such that K, ρ ✄ S ′ . Itis easy to check that K, ρ ✄ S, and this proves the assertion.Now, let us assume that S = ∅. Let S 1 ⊆ S be the set of formulas H ∈ Ssatisfying one of the following conditions:1. H = Tp or H = F c p or H = Fp, with p a propositional variable.2. H = T(p → B), with p a propositional variable and Tp ∉ S.Let S 2 ⊆ S be the set of formulas H ∈ S satisfying one of the following conditions:3. H = F(A → B), or H = F c (A → B), or H = F(¬A) or H = F c (¬A).4. H = T(¬A → C) and R 1 S,H is consistent.5. H = T((A → B) → C) and R 1 S,H is consistent.Since S is consistent and S is empty, S 1 ∪ S 2 = S. If S 2 = ∅, then S = S 1 canbe realized in the Kripke model K = 〈P, ≤, ρ, ⊩〉 where P = {ρ} and, for everypropositional variable p, ρ ⊩ p iff Tp ∈ S.Now, let us suppose that S 2 = {H 1 , . . . , H n } and let j ∈ {1, . . . , n}. Bydefinition of S 2 , the set T j = R 1 S,H is consistent. Since deg(T j) < deg(S), byinduction hypothesis there exists a Kripke model K j = 〈P j , ≤ j , ρ j , ⊩ j 〉 suchthat K j , ρ j ✄ T j (we assume that the P j ’s are pairwise disjoint). We build the11
Page 1 and 2: A refined calculus forIntuitionisti
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A refined calculus for Intuitionistic Propositional Logic - DISCo

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