Are the two LTL properties $GF(\psi_1 \land F\psi_2 )$ and $GF(\psi_2 \land F\psi_1 )$ equivalent?
You are right, but you should prove it.  Here's a proof sketch:

Proof setup:


  Suppose that $GF(\psi_1 \land F\psi_2)$ is true and assume towards a contradiction that $\neg GF(F\psi_1 \land \psi_2)$.


Observations:


  Then there exists a state $s_1$ in which $\neg F(F\psi_1 \land \psi_2)$, which means that every state $s_2 &gt; s_1$ has $\neg(F\psi_1 \land \psi_2)$, which means that for every state $s_2$, either $\neg F\psi_1$ or $\neg \psi_2$.


Contradiction:


  But since $GF(\psi_1 \land F\psi_2)$, there exists a state $s_3 &gt; s_2$ such that $s_3 \vDash \psi_1$ and a state $s_3 &gt; s_2'$ such that $s_3' \vDash \psi_2$, which is a contradiction.


The other direction is identical.