Question
Using the definition[] ++ ys = ys(x:xs) ++ ys = x : (xs ++ ys)verify the following two properties, by induction on xs:xs ++ [] = xsxs ++ (ys ++ zs) = (xs ++ ys) ++ zs
Step 1
Base case: When xs is an empty list, the left-hand side is [] ++ [] = [] and the right-hand side is also [] since xs is empty. Show more…
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