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.
There are "standard" conventions in math log regarding the omission of parentheses, based on "precedence" between connectives.
Regarding quantifiers, the omission of parentheses can cause big troubles.
In Herbert Enderton, A Mathematical Introduction to Logic (2nd ed Harcourt - 2001), page 78, we have :
$\lnot$, $\forall$, and $\exists$ apply to as little as possible. For example,
$\forall x \alpha \rightarrow \beta$ is $(\forall x \alpha \rightarrow \beta)$, and not $\forall x (\alpha \rightarrow \beta)$.
According to the above convention, the "correct" reading of your formula is 1).
The same convention is used in :
Stephen Cole Kleene, Mathematical Logic (1967), page 80 : "The quantifiers acts as unary operators in building formulas, and with our other unary operator $\lnot$ are ranked last under the convention for omitting parentheses."
in Dirk van Dalen, Logic and Structure (5th ed - 2013), page 58 : "We agree that quantifiers bind more strongly than binary connectives."
and in :
George Tourlakis, Lectures in Logic and Set Theory. Volume 1 : Mathematical Logic (2003), page 17 : "To minimize the use of brackets in the metanotation we adopt standard priorities of connectives: $\forall$, $\exists$, and $\lnot$ have the highest".
But a good advice is : "do not save ink !"
