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Extra info for A Compositional Semantics for Multiple Focus
1◦ ) ∗ α and ∗ β. The claim is obvious. (2◦ ) ∗ α and ∗ β. Then, P(α · β) = P(α) · P(β) ∨ P(β). 1 and (CUT), ∗ P(α), hence ∗ P(β) ⇒ P(α) · P(β), by (R ·), and ∗ P(α · β) ⇒ P(α) · P(β), by (L∨). For (3◦ ) ∗ α, ∗ β, and (4◦ ) ∗ α, ∗ β, the arguments are similar. 4) is left to the reader. Let γ = α · β. By the induction hypothesis, N (α) ⇔ α and N (β) ⇔ β are provable in FNL∗S . Then, N (α) · N (β) ⇔ α · β is provable. We have N (α · β) = N (α) · N (β), which yields N (γ) ⇔ γ. By the induction hypothesis P(α) ⇔ α and P(β) ⇔ β are provable, hence P(α) · P(β) ⇔ α · β is provable.
Every sequent from I ( ⇒ γ) is an axiom of the same kind. Structural rules (a), (e), (i) cause no problem: we apply the induction hypothesis to the premise, then apply the same rule. The remaining rules to be considered are all rules for multiplicative connectives and (L∧), (R∨). For the former rules, the arguments are easy. We only consider (L\). The premises are: [β] ⇒ γ and ⇒ α, and the conclusion is: [( , α\β)] ⇒ γ. By the induction hypothesis, there exist sequents [β ] ⇒ γ ∈ I ( [β] ⇒ γ) and ⇒ α ∈ I ( ⇒ α), provable in NL S .
Consequently, the latter sequent belongs to I ( [α1 ∧ α2 ] ⇒ γ). (R∨). The premise is ⇒ αi , where i = 1 or i = 2, and the conclusion is ⇒ α1 ∨ α2 . By the induction hypothesis, there exists a sequent ⇒ αi ∈ I ( ⇒ αi ), provable in NL S . As above, this sequent also belongs to I ( ⇒ α1 ∨ α2 ). 5 does not hold for logics with (c). For instance, p ∧ q ⇒ p · q is provable in FNLc , but neither p ⇒ p · q, nor q ⇒ p · q is provable. The limitation of occurrences of ∧, ∨ is essential: ( p, r ) ⇒ p · r is provable, ( p ∨ q, r ) ⇒ p · r is not provable, and ( p, r ) ⇒ p · r ∈ I (( p ∨ q, r ) ⇒ p · r ), but the occurrence of ∨ is negative.
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