Prove that $(p \to q) \lor (q \to p)$ is a tautology using natural deduction I have to prove that '  (p ⊃ q) ∨ ( q ⊃ p) ' is a tautology.I have to start by giving assumptions like     a1 ⇒ p ⊃ q and then proceed by eliminating my assumptions and at the end i should have something like ⇒(p ⊃ q) ∨ ( q ⊃ p) but could not figure out how to start.
As a quick overview: You should be able to show $p\vdash q\supset p$ as well as $\neg p\vdash p\supset q$, and then by case distinction $p\lor\neg p\vdash (p\supset q)\lor (q\supset p)$.
$(p\rightarrow q)\lor (q\rightarrow p)\equiv (\lnot p\lor q)\lor (\lnot q\lor p)\equiv \lnot p\lor (q\lor\lnot q)\lor p\equiv \lnot p\lor T\lor p\equiv \lnot p\lor p\equiv T$