Consider the following t-norm:

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

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

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

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

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 $I(\phi) \ge 1/2 \Rightarrow I(\psi) \geq I(\phi)$