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Step 6 is a goal-assertion the input, another algorithm might result. Thus one could resolution that functions similarly to the goal-goal resolution break a into a[1],..., a [length(a)/2] and a [length(a)/ above. The final synthesized program is: 2 1],..., a[length(a)] and find an algorithm that recursively calls f on both the first and second halves of its f(x) if x NIL then 0 else car(x) f(cdr(x)).