proof that p implies q entails not p or q

Question

I could easily prove

$\neg P \lor Q$ entails $P \rightarrow Q$.

It is well known that

$P \rightarrow Q$ entails $\neg P \lor Q$

but I couldn't find a way to prove it.

Although there is the same question; How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?; it's a little bit confusing and I need to see step by step solution.

Could you show me the way by using Natural Deduction?

Answer 1

Assume : $P \rightarrow Q$ --- premise

1) $\lnot (\lnot P \lor Q)$ --- assumed [a]

2) $\lnot P$ --- assumed [b]

3) $\lnot P \lor Q$ --- from 2) by $\lor$I

4) $\bot$ --- from 1) and 3) by $\lnot$E (or $\rightarrow$E)

5) $P$ --- from 2) and 4) by Double Negation, discharging [b]

6) $Q$ --- from premise and 5) by $\rightarrow$E

7) $\lnot P \lor Q$ --- from 6) by $\lor$I

8) $\bot$ --- from 1) and 7) by $\lnot$E (or $\rightarrow$E)

9) $\lnot P \lor Q$ --- from 1) and 8) by Double Negation, discharging [a]

Thus :

$P \rightarrow Q \vdash \lnot P \lor Q$.