Induction on degree of ice cream

Let $P$ be a property that satisfies the following condition which we will call $S$ :

For all natural numbers $k$ ; If $P$ holds for all natural numbers smaller than $k$ , then $P$ holds for $k$ .

Now, I claim that $P$ holds for all natural numbers. Assume the opposite, then there exists at least one natural number for which $P$ fails to hold. (Note that this number can not be $0$ since every natural number less than $0$ (namely none) has property $P$ 'vacuously' . Thus by $S$ (case $k=0$ ) $P$ holds for $0$ .) Now let $n$ be the least natural number that does not have property $P$ . By $S$ there exists a natural number smaller than $n$ which does not have property $P$ , which contradicts with the minimality of $n$ . Thus the assumption that $P$ does not hold for all natural numbers leads to a contradiction and we conclude that $P$ holds for all natural numbers.

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Further remarks : To repeat an important point - the so called 'base case', which guarantees that $0$ has property $P$ is vacuously established by $S$ (case $k=0$ ) since every property (in particular $P$ ) holds for all natural numbers less than $0$ (namely all elements of the empty set) vacuously . Thus a property which holds for no natural numbers can not satisfy $S$ .

Note : This principle is also known as complete (strong) induction .