PROVE: If {$\Delta_{i}$} are all deductively closed set of formulae, so is $\cap \Delta_i$. Show with predicate Calculus.
Definition: {$\Delta_{i}$} a set $\Delta$ of formulae is deductively closed iff $\Delta \vdash \sigma$ implies that $\sigma \in \Delta$ .
My Progress: In Propositional Calculus I think the proof would be: let {${\Delta_{i}}$} each be deductively closed. Let $\Delta = \cap \Delta_i $ and let $\Delta \vdash \sigma$ . There is a derivation D = x1, ... xn of $\sigma$ from $\Delta$. Since all assumptions are from $\Delta$ and $\Delta$ is a subset of {$\Delta_{i}$} for all i, all assumptions are from {$\Delta_{i}$} for any i. thus $\sigma \in \Delta = \cap \Delta_i$
I just don't understand how to demonstrate this through predicate calculus instead. There are a few other similar proofs I am assigned where I understand the propositional calculus approach but don't know how that translates to predicate calculus.
Any help is much appreciated!