You have seen regular induction:
$\forall P() : \left( \begin{array} {rrl} & & P(0) \\ \land & \forall x ~:~ & P(x) \implies P(x+1) \\ \implies & \forall y ~:~ & P(y) \end{array}\right)$
Now what about when $P(x)$ is defined as $P(x) := \forall z ~:~ z \le x \implies Q(z)$? In other words, instead of "$x$ has property $P$", we are saying "everything less than $x$ has property $Q$".
$\forall Q() : \left( \begin{array} {rrl} & & \forall z : z < 0 \implies Q(z) \\ \land & \forall x ~:~ & (\forall z : z
Now look at each part:
$\forall z : z < 0 \implies Q(z)$
Since there are no values less than $0$, this is always true, so it can be removed as an assumption.
$(\forall z : z
Since $\forall z : z is the same as $\forall z : z, this can be simplified to
$(\forall z : z
And lastly:
$\forall y ~:~ \forall z ~:~ z < y \implies Q(z)$
This can be simplified to $\forall y ~:~ Q(z)$, since there is no upper bound on the $+1$ function.
So altogether that is:
$\forall Q() : \left( \begin{array} {rrl} & \forall x ~:~ & (\forall z : z
Which is strong induction.