Can we prove $\varphi\rightarrow (p \wedge \neg p)\vdash\neg\varphi$ by natural deduction and contradiction this way?

Question

I'm thinking since $(p \wedge \neg p)$ is a contradiction we can do this:

1. $\;\varphi\rightarrow (p \wedge \neg p)$ --- premise
2. $\;\bullet\quad \varphi$ --- assumption
3. $\;\bullet\quad p \wedge \neg p$ --- $\rightarrow$ elim 1,2
4. $\;\neg\varphi$ --- $\neg$ intro 2 - 3

Is there anything wrong with it?

Answer 1

Going by one of your earlier proofs:

Having already proved $A\vdash B$, $B\vdash C$ and $\neg C$, is this a fine proof proving $\vdash\neg A$?

it seems that $\neg Intro$ in your system requires an explicit $\bot$. So, you probably want:

1. $\;\varphi\rightarrow (p \wedge \neg p)$ --- premise
2. $\;\bullet\quad \varphi$ --- assumption
3. $\;\bullet\quad p \wedge \neg p$ --- $\rightarrow$ elim 1,2
4. $\;\bullet\quad p$ --- $\land$ elim 3
5. $\;\bullet\quad \neg p$ --- $\land$ elim 3
6. $\;\bullet\quad \bot$ --- $\bot$ intro 4,5
7. $\;\neg\varphi$ --- $\neg$ intro 2 - 6

Answer 2

$$\dfrac{\varphi\rightarrow (p \land \neg p)}{\dfrac{\dfrac{\dfrac{\dfrac{[\varphi]}{p \land \neg p}{~\small\text{MP}}}{p \qquad \neg p}{~\small\land\text{E}}}{\bot}}{\neg\varphi}{~\small\neg\text{I}}}$$