Prove :(P → Q) ∨ (Q → P) using natural deduction Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E
I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?
Edit:
My proof attempt
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
$3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)$
$4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)$
$5.\qquad  ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)$
$6.\qquad  Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum$
$7.\qquad  (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)$
$8.\qquad  (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)$
$9.\qquad  ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)$
$10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)$
Stuck at this line, can't spot any contradiction to line 1

The formula is valid only in classical logic.
Thus, to prove it we have to use Excluded Middle or Double Negation.
With LEM :
1) $\vdash P \lor \lnot P$
2) $P$ --- assumed [a] from 1) for $\lor$-E
3) $Q \to P$ --- from 2) by $\to$-I

4) $(P \to Q) \lor (Q \to P)$ --- from 3) by $\lor$-I

5) $\lnot P$ --- assumed [b] from 1) for $\lor$-E
6) $P$ --- assumed [c]
7) $\bot$ --- from 5) and 6)
8) $Q$ --- from 7)
9) $P \to Q$ --- from 6) and 8) by $\to$-I, discharging [c]

10) $(P \to Q) \lor (Q \to P)$ --- from 9) by $\lor$-I.

Having derived $(P \to Q) \lor (Q \to P)$ both from 2) and 5) we may conclude with :


$\vdash (P \to Q) \lor (Q \to P)$


by $\lor$-I from 1), discharging [a] and [b].
