prenex normal equivalence challenges in math consider !enter image description here these two following formula are prenex normal equivalence with the above formula? i think yes, but didn't have any idea to explain it. !enter image description here

Now I give an answer without using provable equivalences on Enderton, page 121 and page 130. Main formula is equivalent to $$(\
eg\exists x\varphi(x))\vee(\forall x\exists y\psi(x,y))$$ and this is also equivalent to $$(\forall x\
eg\varphi(x))\vee(\forall x\exists y\psi(x,y))$$ but we have: $$\forall x\exists y\psi(x,y))\leftrightarrow\forall z\exists y\psi(z,y))$$ so Main formula is equivalent to $$(\forall x\
eg\varphi(x))\vee(\forall z\exists y\psi(z,y))$$ Since neither x is free in $\psi(z,y)$ nor y and z are free in $\
eg\varphi(x)$. So we can move quantifiers on prefix of formula and thus we have: $$(\forall z\forall x\exists y(\
eg\varphi(x)\vee\psi(z,y)))\leftrightarrow(\forall x\forall z\exists y(\varphi(x)\rightarrow\psi(z,y)))$$ and both of them are prefix normal