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Proof: Suppose f is an automorphism of (E,<=). Consider a set D, a set of non-fixed points under f. If D is empty, f is an identity mapping. Suppose, toward a contradiction, that D is nonempty. Then D has a least element, say a. Since E is well-ordered, either f(a) < a or a < f(a).

*. So f fixes f(a), hence f(f(a)) = f(a). But then f(a) = a since f is injective, contradicting that a is an element of D. The case a < f(a) follows similarly applying the inverse of f.*

__Since f(a) < a, f(a) is not an element of D__Why does f(a) < a imply that f(a) is not a fixed point?