Comparing reduction orders
For many functions, the reduction order is immaterial to the computation
fun sqr n = n * n; // from D. Watt
sqr (p+q) [say, p = 2, q = 5]
For applicative-order, we compute p+q=7, bind n to 7, then compute 49
For normal-order, we pass in “p+q” and evaluate 2+5 each time sqr uses n
But we get the same answer regardless