How do I show that p → q ⊢ ¬p ∨ q is valid without using LEM (Law of Excluded Middle)?
Edit:
I want to know if this is okay:
\begin{align} &\hspace{5pt}1.\; p\to q \hspace{43pt}\textrm{is given}\\ &\boxed{ \begin{align} &2.\; p &\textrm{Assumption}\\ &3.\; q &\textrm{from 1}\\ &4.\; \lnot p\vee q&\\ &5.\; \lnot p &\textrm{from 4}\\ &6. \perp \hspace{12pt}&\textrm{from 5 and 1} \end{align} }\\ &\hspace{5pt}7. \lnot p\\ &\hspace{5pt}8. \lnot p\vee q\\ \end{align}
$$\begin{array}{ccc} 1& & &p\to q &\text{Premise}\\ 2&|& &\neg(\neg p \lor q)&\text{Assumption}\\ 3&|&|&p&\text{Assumption}\\ 4&|&|&q&\to\text{-elim (1,3)}\\ 5&|&|&\neg p\lor q&\lor\text{-intro (4)}\\ 6&|&|&\bot&\neg\text{-elim}\\ 7&|& & \neg p&\neg\text{-intro (3-6)}\\ 8&|& &\neg p\lor q&\lor\text{-intro (7)}\\ 9&|& &\bot&\neg\text{-elim (2,8)}\\ 10&&&\neg\neg(\neg p\lor q)&\neg\text{-intro (2-9)}\\ 11&&&(\neg p\lor q)&\neg\neg\text{-elim (10)} \end{array} $$
