By Nicolaas Govert de Bruijn

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**Example text**

1. We will now abbreviate n × m as nm, and use the usual convention that multiplication takes precedence over addition, thus for instance ab + c means (a × b) + c, not a × (b + c). 3 (Positive natural numbers have no zero divisors). Let n, m be natural numbers. Then n × m = 0 if and only if at least one of n, m is equal to zero. In particular, if n and m are both positive, then nm is also positive. Proof. 2. 4 (Distributive law). For any natural numbers a, b, c, we have a(b + c) = ab + ac and (b + c)a = ba + ca.

So suppose ﬁrst that x is an element of (A ∪ B) ∪ C. 4, this means that at least one of x ∈ A ∪ B or x ∈ C is true. We now divide into two cases. 4 again we have x ∈ A ∪ (B ∪ C). 4 again x ∈ A or x ∈ B. 4 we have x ∈ B ∪ C and hence x ∈ A ∪ (B ∪ C). Thus in all cases we see that every element of (A ∪ B) ∪ C lies in A ∪ (B ∪ C). A similar argument shows that every element of A∪(B ∪C) lies in (A∪B)∪C, and so (A∪B)∪C = A∪(B ∪C) as desired. Because of the above lemma, we do not need to use parentheses to denote multiple unions, thus for instance we can write A ∪ B ∪ C instead of (A ∪ B) ∪ C or A ∪ (B ∪ C).

A0 := c for some number c, and then by letting a1 be some function of a0 , a1 := f0 (a0 ), a2 be some function of a1 , a2 := f1 (a1 ), and so forth. In general, we set an++ := fn (an ) for some function fn from N to N. By using all the axioms together we will now conclude that this procedure will give a single value to the sequence element an for each natural number n. 16 (Recursive deﬁnitions). Suppose for each natural number n, we have some function fn : N → N from the natural numbers to the natural numbers.