Give an inductive proof the fact that consecutively mapping two functions over a list is equivalent to mapping their composition over the list.

Give an inductive proof the fact that consecutively mapping two functions over a list is equivalent t Bool) -> [a] -> [a] takew p [] = [] takew p (x:xs) - if (p x) then x: (takew p xs) else ] dropw :: (a -> Bool) -> [a] -> [a] dropw p (x:xs) = if p x then (dropw p xs) else (x:xs) together with the append function and the standard equations for if C] (x:xs) ++ ys- x : (xs ++ ys) A2 Show, using structural induction on lists, that the property if True then p else- = p if False then- else q = q I1 I2 ++ ys holds for all lists xs and all functions p :: a -> Bool. In all proofs indicate the justification (eg, the line of a definition used) for each stejp a) State and prove the base case of the proof of P. b) State the inductive hypotheses of the proof of P. c) State and prove the step case goal of the proof of P