Using natural deduction to prove that $p \implies q \vdash \lnot p \lor q$

Question

Using only rules of natural deduction, I am trying to prove that $$p \implies q \vdash \lnot p \lor q$$ but am having a lot of difficulty. I was able to prove the other direction.

Could anyone show me or point me to the right direction?

Answer 1

I assume that you are trying to prove : $p \to q \vdash \lnot p \lor q$.

1) $p \to q$ --- premise

2) $\lnot (\lnot p \lor q)$ --- assumed [a]

3) $\lnot p$ --- assumed [b]

4) $\lnot p \lor q$ --- from 3) by $\lor$-intro

5) $\bot$ --- from 2) and 4)

6) $p$ --- from 3) and 5) by $\lnot$-elim, discharging [b]

7) $q$ --- from 1) and6) by $\to$-elim

8) $\lnot p \lor q$ --- from 7) by $\lor$-intro

9) $\bot$ --- from 2) and 8)

10) $\lnot p \lor q$ --- from 2) and 9) by $\lnot$-elim, discharging [a].