Given $\phi\sqsubset\psi$, find $\sigma$ such that $\phi\sqsubset\sigma\sqsubset\psi$. I'm working through Van Dalen's Logic and Structure (fifth edition, 2013) independently, and have gotten stuck on problem 2.3.14(i), on p.28.
The author defines: $\phi\sqsubset\psi$ if and only if $\vDash\phi\to\psi$ and $\not\vDash\psi\to\phi$. The question, then, is:

For each $\phi$, $\psi$ such that $\phi\sqsubset\psi$, find $\sigma$ with $\phi\sqsubset\sigma\sqsubset\psi$.

As I understand this, we need $\vDash\phi\to\sigma$ and $\vDash\sigma\to\psi$, which means there can be no valuation $v$ where $[[\phi]]_v=1$ and $[[\sigma]]_v=0$ or where $[[\sigma]]_v=1$ and $[[\psi]]_v=0$. Also, we need $\not\vDash\psi\to\sigma$ and $\not\vDash\sigma\to\phi$, which means there must be a valuation $v$ where $[[\psi]]_v=1$ and $[[\sigma]]_v=0$, and a $v$ where $[[\sigma]]_v=1$ and $[[\phi]]_v=0$.
So:

*

*if $[[\phi]]_v=[[\psi]]_v=1$, then $[[\sigma]]_v=1$, because $\vDash\phi\to\sigma$

*if $[[\phi]]_v=[[\psi]]_v=0$, then $[[\sigma]]_v=0$, because $\vDash\sigma\to\psi$

*if $[[\phi]]_v=0$ and $[[\psi]]_v=1$, then $[[\sigma]]_v=0$, because otherwise there would be no valuation $v$ where $[[\psi]]_v=1$ and $[[\sigma]]_v=0$.

But then there is the case where $[[\phi]]_v=1$ and $[[\psi]]_v=0$. If $[[\sigma]]_v=1$, then there is no valuation $v$ such that $[[\sigma]]_v=1$ and $[[\phi]]_v=0$. But if $[[\sigma]]_v=0$, then $\vDash\phi\to\sigma$ does not hold.
Based on my understanding, then, there can be no such $\sigma$, because its value cannot be determined in the case where $[[\phi]]_v=1$ and $[[\psi]]_v=0$. So what am I missing here?

Edit: as @mohottnad pointed out, the case where $[[\phi]]_v=1$ and $[[\psi]]_v=0$ cannot occur because $\phi\sqsubset\psi$ is given. But then $[[\phi]]_v=0$ and $[[\psi]]_v=1$ is a problem: if $[[\sigma]]_v=1$, then we don't have $\not\vDash\psi\to\sigma$, but if $[[\sigma]]_v=0$, then we don't have $\not\vDash\sigma\to\phi$.
 A: The problem is in your third bullet point, where you write:

If $[[\phi]]_v=0$ and $[[\psi]]_v=1$, then $[[\sigma]]_v=0$, because otherwise there would be no valuation $v$ where $[[\psi]]_v=1$ and $[[\sigma]]_v=0$.

This is not right; in these bullet points, you are considering a specific choice of valuation $v$. Even if $[[\sigma]]_v=1$ and $[[\psi]]_v=1$ for this choice of $v$, it is entirely possible that there exists another valuation $w$ such that $[[\sigma]]_w=0$ and $[[\psi]]_w=1$. Perhaps an example might make things clearer; suppose $\phi=\bot$ and $\psi=\top$. Then $[[\phi]]_v=0$ and $[[\psi]]_v=1$ for any valuation $v$. Can you therefore think of a choice of $\sigma$ such that $\phi\sqsubset\sigma\sqsubset\psi$?

 For example, take $\sigma=p$  for any propositional variable $p$. Then $[[\sigma]]_v=0$ whenever $v(p)=0$ and $[[\sigma]]_v=1$ whenever $v(p)=1$.

As a final remark, the situation you describe in the paragraph following your three bullet points – ie of having $[[\phi]]_v=1$ and $[[\psi]]_v=0$ for some valuation $v$ – is impossible, since we have $\phi\sqsubset\psi$ by hypothesis. Do you think you are now equipped to tackle the problem?
A: There seems no such case as [[ϕ]]v=1 and [[ψ]]v=0 you need to worry about since you have premise below. Although your book may use a certain logic system, but intuitively this case violates the usual material conditional rules, you need to use your logic system's rules to rule out such a pathological case formally.

For each $\phi$, $\psi$ such that $\phi\sqsubset\psi$,

