For any formulas α and β do they entail each other? Show whether the following is true or false:

α |= β or β |= α, for any two formulas α and β

My thought is that I can prove that it is false if I show α doesn't entail β and β doesn't entail α. If I let α be a contingent formula and β be a contradiction formula. The question can then be translated to:


*

*If contingent comes out true, then so does false

*If false comes out true, then so does contingent


It seems that both of the translations are false, which means that the original statement is also false. I am not sure if my logic makes sense. Am I correct in my thinking?
 A: What happens if you simply let $A$ and $B$ be two distinct atomic (non-logical) formulae of the relevant language -- 'non-logical' to rule out the likes of $\top$ and $\bot$?
A: I make the assumption, since you didn't say what logic you are working in, that you are working in a classical two valued logic, which I'll just call PC.
There is a related (so called) paradox of material implication in PC called Disjunction of Conditional and Converse:   
(P->Q)|(Q->P)  Not *exactly* what was asked for, but is often read as if it were.
               It says that at least one of the two implications must be true.
               But it does *not* say that both are provable.

I hope that this is close enough to what you have in mind.
Aside:
While this expression is valid in PC, it is NOT valid in (for example) intuitionist logic, or logics that don't have the law of the excluded middle. (: There is no "one true logic". :)
The link provides several proofs using different techniques.
For a goodly number of students, the proof by truth tables makes the most convincing argument.
Yet another way to prove this is:  
    ( P ->  Q)|( Q -> P)    Disjunction of Conditional and Converse
    (~P  |  Q)|(~Q |  P)    P->Q is equivalent to ~P|Q  
     ~P  |  Q | ~Q |  P     "|" is associative 
 P | ~P  |  Q | ~Q          "|" is commutative [Right circular shift]
(P | ~P) | (Q | ~Q)         "|" is associative
   ⊤     |    ⊤            two applications of the law of the excluded middle
         ⊤                 ⊤ | ⊤   is equivalent to ⊤

The steps above show (P->Q)|(Q->P) -> ⊤,
and the steps can be run "in reverse" to show that ⊤ -> (P->Q)|(Q->P)
and the two together prove that (P->Q)|(Q->P) is equivalent to ⊤.
