A free variable $x$ in $\exists x x<(y\cdot y) \wedge\forall y \neg x\doteq (y+y)$ Is $x$ a free variable in $\exists x x<(y\cdot y) \wedge\forall y \neg x\doteq (y+y)$?
I'm asking because you can read $\exists x x<(y\cdot y) \wedge\forall y \neg x\doteq (y+y)$ as:


*

*$(\exists x x<(y\cdot y) )\wedge (\forall y \neg x\doteq (y+y))$? Or

*$\exists x (x<(y\cdot y) \wedge\forall y \neg x\doteq (y+y))$?
In case 1 $x$ is free since $x$ is free in $\forall y \neg x\doteq (y+y)$ and in case 2 $x$ is not free.
How to read this? What is the rule?

In logic, there is no generally accepted precedence rule for determining how missing parentheses should be inserted.   The "formula" that you started with is simply not a well-formed formula.   
