Creating a new connective in Propositional Logic In the exercise below, is included a new connective ($\sqsubset$), and I'm stuck in how to deal with it.
We can view the relation $\vDash$ $\varphi$ → $\psi$ as a kind of ordering. Put $\varphi$ $\sqsubset$ $\psi$ := $\vDash \varphi \rightarrow \psi \space and \nvDash \psi \rightarrow \varphi$.
(i) For each $\varphi, \psi$ such that $\varphi \sqsubset \psi$, find $\sigma$ with $\varphi \sqsubset \sigma \sqsubset \psi$.
I suppose, it's need to prove: 
$\varphi \rightarrow \sigma$ and $\sigma \nrightarrow \varphi$,
$\sigma \rightarrow \psi$ and $\psi \nrightarrow\sigma$
Considering $\varphi \sqsubset \psi := \space\vDash\varphi\rightarrow\psi$ and $\nvDash \psi\rightarrow\varphi$, we can write:
$\varphi \sqsubset\psi \leftrightarrow(\varphi\rightarrow\psi)\space\land\space\lnot\space(\psi\rightarrow\varphi)$
$$\begin{array}{rcc|c|cccc} 
(\varphi & \rightarrow & \psi ) &\land & \lnot & (\psi&\rightarrow&\varphi)  \\ \hline
\ 0&1&0&0&0&0&1&0 \\ 
\ 0&1&1&1&1&1&0&0 \\
\ 1&0&0&0&0&0&1&1 \\
\ 1&1&1&0&0&1&1&1 \\ 
\end{array}$$
So, we can define the truth table of $\space\sqsubset$
$$\begin{array}{c|cc} 
\sqsubset & 0 & 1   \\ \hline
\ 0&0&1 \\ 
\ 1&0&0 \\
\end{array}$$
The problem is find $\sigma$ such that $\varphi\sqsubset\sigma\sqsubset\psi$
Truth table:
$$\begin{array}{ccc|c|ccc} 
(\varphi & \sqsubset&\sigma)&\land&(\sigma&\sqsubset&\psi)   \\ \hline
\ 0&0&0&0&0&0&0 \\ 
\ 0&0&0&0&0&1&1 \\
\ 0&1&1&0&1&0&0 \\
\ 0&1&1&0&1&0&1 \\
\ 1&0&0&0&0&0&0 \\ 
\ 1&0&0&0&0&1&1 \\
\ 1&0&1&0&1&0&0 \\
\ 1&0&1&0&1&0&1 \\
\end{array}$$ 
We found a contradiction, so we can conclude $\sigma = \bot$
Is this a right way? I'm a bit confusing about the symbol $\sqsubset.$

If you consider the language with only one fundamental proposition $P$, and take $\varphi = \perp$, $\psi = P$, then $\varphi \sqsubset \psi$ but it is clear that no formula referring to $P$ alone can lie between them.
So in general (i) is true only if you are allowed to introduce a new propositional letter, and then the formula $\sigma$ needs to have value $T$ everywhere in the new truth table that $\varphi$ does, but not everywhere that $\psi$ does.
