When is a subproof provable in natural deduction 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.)
 A: 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

($\Gamma_2 \vdash B_2$ is provable) $\Rightarrow$ ($\Gamma_1, \Gamma_2 \models B_2$ is true)

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

($\Gamma_1, \Gamma_2 \models B_2$ is true) $\Rightarrow$ ($\Gamma_2 \vdash B_2$ is provable)

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.
