(I also asked this in MathOverflow)
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 $I(\phi) \ge 1/2 \Rightarrow I(\psi) \geq I(\phi)$
My teacher has provided a solution:
Take a $[0, 1]_*$-interpretation with $I(\phi \rightarrow \phi * \phi ) = 1$, and say $a:=I(\phi)$.
Define the function \begin{align*} h: [0, 1] & \rightarrow [0, 1]\\ x&\mapsto \begin{cases} 1 &\quad\text{if $x \geq a$}\\ x &\quad\text{if $x < a$} \end{cases} \end{align*}
We can check this function preserves all of the connectives of the logic. So $h \circ I$ preserves connectives and is thus a $[0, 1]_*$-interpretation. And $h \circ I (\phi \rightarrow \phi * \phi) = h \circ I (\phi) = 1 $, so by our supposition $h \circ I (\psi) = 1$ and thus $I(\phi) \leq I(\psi)$.