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.