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:The set of all natural numbers is [[well-ordered]].
A simplified version is given here. This proof does not use the standard mathematical symbols for '''there exists''' and '''for all''' to make it more accessible to less mathematically motivated readers.▼
▲A simplified version is given here. This proof does not use the standard mathematical symbols for '''there exists''' and '''for all''' to make it more accessible to less mathematically motivated readers. The key technique is [[natural deduction logic]] and [[proof by contradiction]].
Suppose
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which is the principle of mathematical induction.
==Converse==
Conversely, the axiom can be proved by the principle of [[mathematical induction]]. Indeed, the two are equivalent.
Let ''S'' be a set of natural numbers. We want to prove that either ''S'' has a smallest element or else that ''S'' is empty. Let ''P''(''n'') be the statement that no element of ''S'' is smaller than ''n''. ''P''(0) is certainly true, since there is no natural number smaller than 0. Suppose that ''P''(''n'') is true for some ''n''. If ''P''(''n''+1) were false, then ''S'' would have an element smaller than ''n''+1, but it could not be smaller than ''n'', because ''P''(''n'') was true, and so ''S'' would have a minimal element, namely ''n'', and we would be done. So ''P''(''n'') implies ''P''(''n''+1) for all ''n'', or else ''S'' has a minimal element. But if ''P''(''n'') implies ''P''(''n''+1) for all ''n'', then by induction we know that ''P''(''n'') is true for all ''n'', and therefore for all ''n'', no element of ''S'' is smaller than ''n''. But this can only be vacuously true, if ''S'' has no elements at all, since every natural number is smaller than some other natural number. Thus we are done.
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