1. (10 points) Show (by hand) how Haskell evaluates: myOR [False, True, False] Given the code: foldr:: (a -> b -> b) => b -> [a] => b foldr bf [] = b foldr bf (x:xs) fx (foldr bf xs) myOR:: [Bool] -> Bool myOR foldr myor False myor:: Bool -> Bool -> Bool myor False False False myor True 1. (10 points) Show (by hand) how Haskell evaluates: myOR [False, True, False] Given the code: foldr:: (a -> b -> b) => b -> [a] => b foldr bf [] = b foldr bf (x:xs) fx (foldr bf xs) myOR:: [Bool] -> Bool myOR foldr myor False myor:: Bool -> Bool -> Bool myor False False False myor True