Self Doubt

Question

Is this TRUE?

(∃!x)(∃!y)J(x,y)≡(∃!y)(∃!x)J(x,y)

Comments

yes true

$\sim \left ( \forall x \right )\wedge \sim \left ( \forall y \right )J\left ( x,y \right )=\sim \left ( \forall y \right )\wedge \sim \left ( \forall x \right )J\left ( x,y \right )$

$\left ( \forall x \right )\wedge \left ( \forall y \right )J\left ( x,y \right )= \left ( \forall y \right )\wedge \left ( \forall x \right )J\left ( x,y \right )$