From the David Williams book problem e10.5, for $N \in \mathbb{N}$ and for every $n$, there is an equality:
$P(T>kN) = P(T>kN;T>(k-1)N)$
Is that supposed to mean:
$P(T>kN) = P(T>kN\textbf{1}_{T>(k-1)N})$
In the book, $E(X;Y) = E(X\textbf{1}_{Y} )$
Is it the same for probability?
For two events $A$ and $B$, $P(A;B)$ denotes $P(A\cap B)$.
In the particular setting of the question, $A=\{T>kN\}$ and $B=\{T>(k-1)N\}$ hence $A\subset B$ and the equality is then clear.