If I want to do a mathematical double induction on $n$ and $t$ where $n \in \mathbb{N}$ and $1 \leqslant t \leqslant n$ then obviously the base step is $P(n=1,t=1)$ but how do I implement the induction hypothesis. Im very confused as I can’t assume $P(n,t+1)$ to be true nor $P(n+1, t)$ to be true either as I don’t know whether $t+1>n$ or not.
What statements should I then assume and which statement should I prove in the inductive step?
On
There are different solutions possible, depending on your exact problem.
The first one, which probably works, but is not elegant at all is to prove by induction on $n$ the property $\forall 1\leq t\leq n, P(n,t)$. You would prove the base step by setting $n=1$ which forces $t=1$, and for the induction case, you suppose the property for $n$, and prove by induction on $t$ that for all $1\leq t\leq n+1$, you have $P(n+1,t)$. Note that you have to adapt this method to your case, so that you can reduce your problem to a subproblem in the induction step.
The second solution I can think of, which might not be applicable, but is worth trying, is to try and reduce it to a single induction. For that you can try and find a combination of $n$ and $t$ (say $n+t$ for instance) such that $P(n,t)$ with $n+t = k$ implies $P(n',t')$ with $n'+t' = k+1$, and you can then perform the induction on $k$.
These are very generic, and without any more precisions, I can't give you a more specific answer
This is going to depend a lot on exactly what you're trying to prove, but here is a general idea for an inductive approach:
Let $Q(n)$ be the statement "$P(n,t)$ for all $t$ between $1$ and $n$", and prove that $Q$ is satisfied for all natural numbers using induction. In each inductive step, you can again use induction: prove $P(n,t+1)$ from assuming both $Q(n-1)$ and $P(n,t)$.