Exercises

Exercise 1

Exercise 2

until :: (a -> a) -> (a -> Bool) -> a -> a
until f p x =
    if (p x)
        then x
        else until f p (f x)

Exercise 3

Prove that $\text{double}(x)=2\ast x$: Base Case: $\text{double}(0)=2\ast 0=0\equiv 0+0$ Inductive Case: Suppose $\text{double}(k)=k+k$. We want to show $\text{double}(k+1)=(k+1)+(k+1)$.

$$\begin{array}{rl}\text{double}(k+1)& =2+(k+1)\\ & =2+k+2\\ & =\text{double}(k)+2\\ & =k+k+2\\ & =k+1+k+1\\ & =(k+1)+(k+1)\end{array}$$

[2, 3]

Exercise 4