Hint needed for logic problem

[hasty rose]

The first case: $y$ has no primitives in common with $x$, or $\text{Set}(x) \cap \text{Set}(y) = \varnothing$. Then, the completeness theorem gives that $x \models y$, which means that $y$ entails $1$ always (as no shared primitives, can thus choose all possible valuations of $y$). This means $y$ is a tautology, so it is then $\varnothing$, and clearly $\text{Set}(x) \supseteq \text{Set}(y)$.

The second case: $y$ has some primitive in common with $x$, called $k$. Replace all instances of primitives $y$ does not have in common with $x$ with $k$, to get a new formula $y'$. Clearly, $y \vdash y'$.

What I am having trouble with, is showing that $y' \vdash y$. I believe that this is true, but I don't see any way how to prove it.

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However, if that is proven, then I get a contradiction: $y' \sim y$ but $y'$ has fewer primitives. Then it must have been true from the start that $\text{Set}(y) \subseteq \text{Set}(x)$. So I know that $P$ induces a poset on the set of primitives, and as the set of primitives of each formula is finite, there is no uncountable chain.

[deep stratus]

Anthony Gonsalves