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Sample text
Now, if the language of C involves ∧, and the condition (again a very natural one) (∧) C(α ∧ β) = C(α, β) holds true, then (→) and (∧) imply α ∈ C(β1 , . . , βn ) iff (β1 ∧ . . e. when C is well-determined, just the condition that defines C(X) in terms of → and ∧ for all finite X. 2. The argument presented above is, I hope, a good one to the effect that L should be treated as distinguished among all logics C such that C(∅) = L. We must realize however, that if one defines a logic L as certain deductive set of formulas one need not necessarily view L to be the logical consequence corresponding to L.
It is perhaps worthwhile mentioning that a connective ∨ is said to be disjunction with respect to C iff the language of C involves ∨ and the following holds true (∨) C(X, α ∨ β) = C(X, α) ∩ C(X, β). Let us adopt the convention under which α1 ∧ . . ∧ αn will be treated as an abbreviation for (α1 ∧ (α2 ∧ . . (αn−1 ∧ β)). A similar convention will be adopted for disjunction. ) they involve. 3. Theorem. If C is a well-determined logic then C(∅) defines C uniquely. What are the conditions to be satisfied by C(∅) in order for C to be well-determined?
A partial function : T × L∼ → {0, 1} will be referred to as an epistemic valuation for L∼ , defined relatively to (T, ) iff for all α, β, all t ∈ T , it satisfies the following conditions (0) (t, α) = 0 implies (t , α) = 0 for all t t, (1) (t, α) = 1 implies (t , α) = 1 for all t t, (∧)+ (t, α ∧ β) = 1 iff (t, α) = (t, be) = 1, 49 (∧)− (t, α∧β) = 0 iff either (t, α) = 0, or (t, be) = 0 or (t, α) = (t, β) = 0, (∨)+ (t, α∨β) = 1 iff either (t, α) = 1, or (t, β) = 1, or (t, α) = (t, β) = 1, (∨)− (t, α ∧ β) = 0 iff (t, α) = (t, be) = 0, (→)+ (t, α → β) = 1 iff for all t t, (t , β) = 1 whenever (t , α) = 1 (→)− (t, α → β) = 0 iff (t, α) = 1 and (t, β) = 0, (¬)+ (t, ¬α) = 1 iff for no t t, (t , α) = 1 (¬)− (t, ¬α) = 0 iff (t, α) = 1 (∼)+ (t, ∼ α) = 1 iff (t, α) = 0 (∼)− (t, ∼ α) = 0 iff (t, α) = 1 Again (cf.