Question

Construct derivations to show that the argument below is valid in SD:

N->(O^M) (If N then (O and M))
L<->M (Biconditional L M)
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N->L ((Entail/Solve) If N then L)

Answer 1

Explanation:N -> (O ^ M) (Premise)

L <-> M (Premise)

(N -> O) ^ (N -> M) (Equivalence of 1)

N -> M (Simplification of 3)

M -> L (Equivalence of 2)

N -> L (Transitivity of 4 and 5, i.e., if N implies M and M implies L, then N implies L)

Therefore, N -> L is valid in SD (assuming SD stands for classical propositional logic).