for a transition $\delta(q,\sigma,X)$, if I keep state $q$ and input symbol $\sigma$ same but read different stack symbols in place of $X$ will it still be called deterministic PDA because as per my understanding for PDA to be deterministic this whole set should contain at most one element. So, is it allowed if I read different stack symbol but same input and same state? the transition will also occur at same state
as a side note I am trying to create a deterministic PDA for accepting valid brackets of all 3 types; {,[,( and for opening brackets I want to make a transition to similar states no matter what the stack symbol is.