I have to prove that the following statement is valid.

If $φ$ and $ψ$ have no propositional letter in common, then $φ ∨ ψ$ is a tautology iff φ is a tautology or ψ is a tautology.

As this includes an if-and-only-if I have to prove two directions.

I started proving this by stating the following:

Let $φ$ be a contradiction and $φ ∨ ψ$ be a tautology. Because $φ$ is a contradiction, $φ ∨ ψ$ is equivalent to $ψ$. Therefore $ψ$ must be a tautology, because otherwise $φ ∨ ψ$ cannot be a tautology.

Am I on the right track? If so, how can I prove the other way? If not, can someone give me some advice on how to prove that?

  • $\begingroup$ What happens in your argument if $\varphi$ is contingent? $\endgroup$ – Daniel Schepler Nov 10 '17 at 23:00
  • $\begingroup$ @DanielSchepler I have not considered that until now. I do not really know how i can handle this case. $\endgroup$ – Alexander Nov 10 '17 at 23:06

Suppose $φ∨ψ$ is a tautology and $φ$ and $ψ$ have no propositional letter in common.

Now if $φ$ and $ψ$ are contradictions, then for every interpretation $φ∨ψ$ would be false. This contradicts our assumption that $φ∨ψ$ is a tautology.

Otherwise if $φ$ and $ψ$ are not both contradictions and w.l.o.g. $φ$ and $ψ$ consist each of one propositional letters. Let's say the the two possible interpretation structures are

$I(φ) = 1$, $I(ψ) = 0$ and

$I(φ) = 0$, $I(ψ) = 1$

The interpretation of $φ∨ψ$ with both interpretation structures would yield a tautology, which would not be a contradiction.

But the interpretations $φ$ and $ψ$ are only possible if $φ$ and $ψ$ have a common propositional letter (e.g. $φ = A$, $ψ = \neg A$). This is a contradiction to the assumption that $φ$ and $ψ$ must not have propositional letters in common. This is analogue for more than one propositional letter per formula.

So the only way is that $φ$ or $ψ$ must be a tautology.

Can someone review this please? ;)


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