For a natural deduction proof of $\Gamma_1 \vdash B_1$, I want to know whether a subproof $\Gamma_2 \vdash B_2$ in the proof is provable without uisng natural deduction (this subproof could be anywhere in the proof).
My feeling is that ($\Gamma_2 \vdash B_2$ is provable) $\implies$ ($\Gamma_1, \Gamma_2 \vDash B_2$ is true). Is this correct?
But I'm not sure whether it works both ways i.e.
($\Gamma_2 \vdash B_2$ is provable) $\iff$ ($\Gamma_1, \Gamma_2 \vDash B_2$ is true).
Can someone help?
(A formula $\Gamma \vDash B$ could be checked using a truth table etc.)
Since you were mentioning truth tables, I asssume you're talking about classical propositional logic, and one of the standard natural deduction calculi for it.
Then, your first conjecture that
is correct. By soundness of natural deduction, we have
($\Gamma_2 \vdash B_2$ is provable) $\Rightarrow$ ($\Gamma_2 \models B_2$ is true)
By monotonicity of classical logic, we have
($\Gamma_2 \models B_2$ is true) $\Rightarrow$ ($\Gamma_1, \Gamma_2 \models B_2$ is true)
Application of transitivity yields the desired result.
Your second conjecture that
is not correct, however. Here is a counterexample: Set $\Gamma_1 = \{P\}$, $\Gamma_2 = \{P \rightarrow Q\}$ and $B_2 = Q$. Then $\Gamma_1, \Gamma_2 \models B_2$ is true (check with a truth table, if you like). But $\Gamma_2 \models B_2$ is not true, so by soundness again $\Gamma_2 \vdash B_2$ is also not provable.
Nonetheless (if that should be what you've meant),
($\Gamma_2 \models B_2$ is true) $\Rightarrow$ ($\Gamma_2 \vdash B_2$ is provable)
is correct, because natural deduction is also complete.