(sum.1) sum([]) = 0 (sum.2) sum(L) = head(L) + sum(tail(t)) (when L is not empty)Evaluation is as follows.
sum([3,5,7]) = 3 + sum([5,7]) by (sum.2) = 3 + (5 + sum([7])) by (sum.2) = 3 + (5 + (7 + sum([]))) by (sum.2) = 3 + (5 + (7 + 0)) by (sum.1) = 15 by arithmetic