# Logic for A.I. - Solutions

 Abstract Axiom 1 of K is the same as Axiom 1 in L, thus we have nothing to prove. Axiom 2 of K is 2(φ → ψ) → (2φ → 2ψ). We give a derivation of this formula in L: (φ → ψ) ∧ φ → ψ 2((φ → ψ) ∧ φ) → 2ψ (the rule from L) 2(φ → ψ) ∧ 2φ → 2ψ (axiom 3 of L and propositional logic) 2(φ → ψ) → (2φ → 2ψ) (propositional logic) Remain the rules of K. Modus ponens is a rule of both so there is nothing to prove. We show that L proves the Necessitation rule. That is, we have to show that if L φ, then L 2φ. The following derivation in L from assumption φ shows this: φ → φ (propositional logic) 2 → 2φ (the rule from L) 2φ (modus ponens using axiom 2 ) This completes the direction of the proof form left to right. Keywords No keywords specified (fix it) Categories Logic and Philosophy of Logic (categorize this paper) Options Save to my reading list Follow the author(s) My bibliography Export citation Edit this record Mark as duplicate Request removal from index Translate to english
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