proof verification for natural deduction

Question

Could someone please let me know if I got the following natural deduction correct for the following formula

(p ∧ q) ⇒ r ├ p ⇒ (q ⇒r)
1       (p ∧ q) ⇒ r    assump 0
2        p ├ (q ⇒r)    
2.1      p              assump 2.1
2.2      q ├ r         
2.2.1    q             assump 2.2
2.2.2    r             ⇒E 1,2.2
2.3      q⇒r          ⇒I 2.2
3        p⇒(q⇒r)      ⇒I 2, 2.3

Answer 1

The notation is a bit weird to me. Does 2. announce what you're going to prove?

And you need an explicit step introducing $p \land q$ in the system I was taught. The rest seems fine.

So in my notation:

1. $(p \land q) \Rightarrow r$ (assumption/axiom)
2. $p$ (assumption)
3. $q$ (assumption)
4. $p \land q$ (from 2 and 3)
5. $r$ (from modus ponens, 4 and 1)
6. $q \Rightarrow r$ and 3 is dropped. (Introduction of $\Rightarrow$)
7. $p \Rightarrow (q \Rightarrow r)$ and 2 is dropped (Introduction of $\Rightarrow$ again).

Proof complete, as the only undropped assumption is the "axiom" on the left hand side of $\vdash$.