Full question:
Show that if $\models\varphi \supset \psi$, and $\varphi$ is not a contradiction nor $\psi$ a tautology, then there is an interpolant for $\varphi$ and $\psi$.
I'm having trouble wrapping my head around this problem. In my professor's notes he gave us a hint, but I'm still not seeing why we are choosing the interpolant to be $\varphi(\top,q) \vee \varphi(\bot,q)$, or why we defined $\top=q\supset q$, and similarly $\bot=\sim(q\supset q)$.
Can anyone highlight why we chose that as our interpolant?
Below is my professor's hint:
First: The definitions of $\perp$ and $\top$ is just so that $\perp$ is a sentence which is always false, $\top$ is a sentence which is always true, and they each only use symbols in both $\varphi$ and $\psi$.
So: What does this mean for the interpolant "$\varphi(\top, q)\vee\varphi(\perp, q)$"? (Call that formula "$\hat\varphi$" for simplicity.) Well, $\top$ and $\perp$ - "true" and "false" - are the two possible values of the variable $p$, which is the "illicit" variable - the one $\varphi$ uses but $\psi$ doesn't, which we need to get rid of. For instance, if $\varphi$ were the sentence "$p\vee q$," then $\hat{\varphi}$ would be equivalent to the sentence "Either $p$ is true and (True or $q$), or $p$ is false and (False or $q$)." Importantly: any truth assignment $\nu$ making $\varphi$ true will make $\hat{\varphi}$ true - if $\nu$ makes $p$ true, then $\nu$ will make $\varphi(\top, q)$ true, and if $\nu$ makes $p$ false, then $\nu$ will make $\varphi(\perp, q)$ true. So $\hat{\varphi}$ is a consequence of $\varphi$ gotten by basically breaking $\varphi$ into all the possible cases for $p$.$^{footnote}$
Conversely, it is not the case that $\hat{\varphi}$ implies $\varphi$ - this is a good exercise - but the following is true:
Just look at which clause of $\varphi$ winds up being true under $\nu$, and to define $\nu'$ set $p$ to $\top$ or $\perp$ accordingly.
This property $(*)$ is in fact where the relationship between $\hat{\varphi}$ and $\psi$, the other half of our desired result, comes from: suppose $\varphi\implies\psi$ is a tautology. We want to show $\hat{\varphi}\implies\psi$ is a tautology. So let $\nu$ be any truth assignment to $p$ and $q$ which makes $\hat{\varphi}$ true. Then we can find a truth assignment $\nu'$, which agrees with $\nu$ about $q$, making $\varphi$ true. That means this assignment can be extended to one making $\psi$ true.
Does this help?
$^{footnote}$ If $\varphi$ had two illicit variables, $p_1$ and $p_2$, then $\hat{\varphi}$ would have four clauses. Etc.