(This is an assignment)
To prove:
$$\frac{\Gamma \vdash_N B}{\Gamma, A \vdash_N B}$$
($\Gamma, A = \Gamma \cup \{A\}$)
I have what I think is a proof for this. I would like and be grateful for some feedback.
Axioms:
A0: $\Gamma \vdash_N B$, whenever $B \in \Gamma$
A1: $\Gamma \vdash_N A → (B → A)$
A2: $\Gamma \vdash_N (A → (B → C)) → ((A → B) → (A → C))$
A3: $\Gamma \vdash_N (\neg B → \neg A) → (A → B)$
Proof rule: Modus Ponens
Proof
By induction on the length of the proof of $\Gamma \vdash_N B$.
Base cases: length 1, axioms:
A0 :: If $ B \in \Gamma$, then $B \in \Gamma \cup \{A\} $
A1 - A3 :: same argument as for A0
MP :: $$\frac{\Gamma \vdash_N C ; \Gamma \vdash_N C → B}{\Gamma \vdash_N B} $$
IH : $\Gamma, D \vdash_N C$
IH : $\Gamma, D \vdash_N C → B$
1: $\Gamma, D \vdash_N C$ (IH)
2: $\Gamma, D \vdash_N C → B$ (IH)
3: $\Gamma, D \vdash_N B$ (MP: 1,2)
Since $D$ is arbitrary: $A = D$, so
$\Gamma, A \vdash_N B$
as we were after.
Your proof seems valid. You probably could have avoided a completely formal proof by noticing the following fact:
The notation $\Gamma\vdash B$ means that $B$ belongs to $Con(\Gamma)$ - the inductive set generated by the basis: $\mathcal{B}_1=\{A_0,A_1,A_2,A_3\}\cup \Gamma$, and the operation: $\{MP\}$. The notation $\Gamma,A\vdash B$ means that $B$ belongs to $Con(\Gamma \cup \{A\})$ the inductive set generated by the basis: $\mathcal{B}_2=\{A_0,A_1,A_2,A_3\}\cup \Gamma \cup \{A\}$, and the operation: $\{MP\}$.
It is a general result is set theory that "if you start with a larger basis, you will end up with a larger set", formally, in this case since $\mathcal{B}_1 \subseteq \mathcal{B}_2$ then $Con(\Gamma)\subseteq Con(\Gamma \cup \{A\})$ which means: $\Gamma\vdash B\Rightarrow\Gamma,A\vdash B$.