Show that $(p \to q) \lor (q \to p)$ is a tautology I tried to prove that $(p \to q) \lor (q \to p)$ is a tautology.
I used $p$ and $¬q$ as conditions. (Premises 1 and 5)
I managed to get to a solution, but I'm not sure if it's right.
Can you please check it? 
Thank you!!!:)
 A: The answer is pretty simple to derive using truth tables:

A: Regarding step 11 :

$\lnot q \rightarrow (q \rightarrow p)$

it is a tautology; so, it is provable.
You are trying to prove it from assumptions; you can simplify it as follows : 
1) $p$ --- assumed
2) $\lnot q$ --- assumed
3) $\lnot q \lor p$ --- Add.2
4) $q \rightarrow p$ --- Impl.3
5) $\lnot q \rightarrow (q \rightarrow p)$ --- C.P.2-4
Then discharge assumption $p$ to get :

6) $p \rightarrow (\lnot q \rightarrow (q \rightarrow p))$ --- C.P.1-5

but from it, you cannot derive 13) : $\lnot (p \lor \lnot q) \rightarrow (q \rightarrow p)$ ...
You must have :
13') $(p \land \lnot q) \rightarrow (q \rightarrow p)$ --- by Exportation
Then we have : 
14') $\lnot (p \rightarrow q) \rightarrow (q \rightarrow p)$ --- from 13') by the equivalence : $(p \rightarrow q) \leftrightarrow \lnot (p \land \lnot q)$ ... but we have to justify it !
and finally :


$(p \rightarrow q) \lor (q \rightarrow p)$ --- Impl.14'.


A: There's a quicker way than your approach.
By the law of excluded middle, $p \lor \neg p$
If $p$, then ... 
If $\neg p$, then ... 
Remember, anything implies a true statement, and a false statement implies anything.
edit: I'm just suggesting that you start with $p \lor \neg p$, which is either an axiom or derivable from $p \dashv \vdash \neg \neg p$, depending on which book you use.
A: Here you go.
Given the logical fact that a disjunction is true as long as long as at least one statement is true:



*

*Work from all the variables first

*Implications

*Disjunction from implications


The final result is that it is true in all possible worlds, a tautology.
