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