Prove by contrapositive: Φ∪{β} ⊨ α & Φ∪{¬β} ⊨ α iff Φ ⊨ α

Question

We are to prove this by contrapositive (by the way: Φ is a set of formulas of predicate logic and α a formula of predicate logic)

I've managed the Right to Left proof, but I struggle with the Left to Right direction.

Presumably, what I want is that by assuming Φ ⊭ α, then it is not the case that Φ∪{β} ⊨ α & Φ∪{¬β} ⊨ α - i.e. that either Φ∪{β} ⊭ α or Φ∪{¬β} ⊭ α

This is what I've done so far:

Assume Φ ⊭ α.

Then there is at least one interpretations U such that ⊨(U) Φ but ⊭(U) α.

We also know the Law of Excluded Middle holds for all interpretations. So, for all interpretations U', ⊨(U’) (β v ¬β) ⇒ ⊨(U’) β or ⊨_(U’)¬β.

This also holds for Φ∪{β}: for all interpretations U', either ⊨(U’) Φ∪{β} or ⊨(U’) Φ∪{¬β} ⇒ either ⊨(U’) Φ∪{β} or ⊭(U’) Φ∪{β}.

And that's basically it, so far.

I do not know how to proceed with a proof that shows that if Φ∪{β} ⊨ α, then Φ∪{¬β} ⊭ α and if Φ∪{¬β} ⊨ α, then Φ∪{β} ⊭ α. Which I think is what I need.

Though, as I'm writing this I think this may be on the right lines:

1) If Φ∪{¬β} ⊨ α, then for all interpretations U such that ⊨(U) Φ∪{¬β}, ⊨(U) α.

Need to prove there is at least one interpretations U', such that ⊨(U) Φ∪{β} and ⊭(U) α.

Maybe: since Φ∪{¬β} and Φ∪{β} are contradictory, then if ⊨(U) Φ∪{¬β}, ⊨(U) α, then if ⊨(U) Φ∪{β}, then ⊭(U) α. (a lot of ifs and thens here ...). My rational here is roughly that if all interpretations that have ⊨(U) Φ∪{¬β} also have ⊨(U) α, then all the interpretations that have ⊨(U) Φ∪{β} will also have ⊭(U) α

2) If Φ∪{β} ⊨ α, then for all interpretations U such that ⊨(U) Φ∪{β}, ⊨(U) α. Need to prove there is at least one interpretations U', such that ⊨(U) Φ∪{¬β} and ⊭(U) α. Maybe: the opposite of what I said above.

Is this anywhere near correct? If not, how should one proceede with such a proof?

Answer 1

Suppose there is an interpretation $U$ that models $\Phi$ and where $\alpha$ is false and check $\beta$.

• If $\beta$ is true in $U$, then $\Phi\cup\{\beta\}\not\models\alpha$.
• If $\beta$ is false in $U$, then $\Phi\cup\{\lnot\beta\}\not\models\alpha$.

Conversely, suppose that either $\Phi\cup\{\beta\}\not\models\alpha$ or $\Phi\cup\{\lnot\beta\}\not\models\alpha$. In either case, $\Phi\not\models\alpha$.