Consider the following t-norm:

$   
a * b = \begin{cases}
       \text{$2ab,$} &\quad\text{if $a, b$}\le1/2\\
       \text{$min\{a, b\}$} &\quad\text{otherwise}\\
      \end{cases}
$

We build from it the $\langle [0, 1]_*, \{1\}\rangle$ matrix logic as usual. Thus, the resulting R-implication and associated negation are:

$   
a \rightarrow b = \begin{cases}
       \text{1} &\quad\text{if $a \le b$}\\
       \text{$b/2a$} &\quad\text{if 1/2 $\ge a>b$}\\
        \text{$b$} &\quad\text{if a > b and a > 1/2}\\
      \end{cases}
$

$   
¬ a = \begin{cases}
       \text{1} &\quad\text{if $a = 0$}\\
       \text{0} &\quad\text{otherwise}\\
      \end{cases}
$

As usual, $\wedge$ is the minimum and $\lor$ is the maximum.

For this logic, I need to prove the following implication, for any formulas $\phi$ and $\psi$:

$ \phi \rightarrow \phi * \phi, \phi \models \psi$ implies $ \phi \rightarrow \phi * \phi \models \phi \rightarrow \psi$

This is equivalent to showing the following:

if for any $[0, 1]_*$-interpretation I, we have $I(\phi) = 1 \Rightarrow I(\psi) = 1$

then for any $[0, 1]_*$-interpretation I, we have $I(\phi) \ge 1/2 \Rightarrow I(\psi) \geq I(\phi)$