Suppose, to the contrary, that there is a function, "f", on the natural numbers with "f"("n"+1) an element of "f"("n") for each "n". Define "S" = {"f"("n"): "n" a natural number}, the range of "f", which can be seen to be a set from the axiom schema of replacement. Applying the axiom of regularity to "S", let "B" be an element of "S" which is disjoint from "S". By the definition of "S", "B" must be "f"("k") for some natural number "k". However, we are given that "f"("k") contains "f"("k"+1) which is also an element of "S". So "f"("k"+1) is in the intersection of "f"("k") and "S". This contradicts the fact that they are disjoint sets. Since our supposition led to a contradiction, there must not be any such function, "f".