Find to which $( \forall x)$ , each occurrence of x belongs to. (logic) Find to which $( \forall x) $, each occurrence of x belongs to.
$$ (\forall x)((\forall x)(\forall y)\ x < y \lor x > z ) \rightarrow (\forall y)\ y=x $$
Is it right that the third and fourth occurrence of x belongs to the second occurrence of $ \forall x $ and the last occurrence of x belongs to the first occurrence of $\forall x $ ? ( I count the occurrences from left to right)

Is the following formula calculation right ?
$$ y=x,(\forall y\ y=x),x<y,\ x>z,\ x<y\ \lor x>z,\ ((\forall y)\ x<y\ \lor\ x>z), ((\forall x)(\forall y)\ x<y\ \lor\ x>z), ((\forall x)((\forall x)(\forall y)\ x < y \lor x > z ) \rightarrow (\forall y)\ y=x )$$
 A: With this particular notation, there are two conventions (let the example be $(\forall x)\phi \star \psi$)


*

*quantifier binds as far as it can (the example becomes $(\forall x)(\phi \star \psi)$),

*quantifier binds only the closest subexpression (while here it is $\big((\forall x)\phi\big) \star \psi$).


You are correct in the first, but wrong in the second. You should check the notes on which one you need to use. One nice way to match variables with quantifiers it to number them, e.g.
$$(\forall x_1)\big((\forall x_2)(\forall y_3)\ x_2 < y_3 \lor x_2 > z_4 \big) \rightarrow (\forall y_5)\ y_5=x_1, \tag{1}$$
$$(\forall x_1)\big((\forall x_2)(\forall y_3)\ x_2 < y_3 \lor x_{\color{red}{1}} > z_4 \big) \rightarrow (\forall y_5)\ y_5=x_{\color{red}{6}}. \tag{2}$$
As for the second part, I would recommend drawing a tree, perhaps like this:


*

*$(\forall x)\big((\forall x)(\forall y)\ x < y \lor x > z \big) \rightarrow (\forall y)\ y=x$

*

*$(\forall x)(\forall y)\ x < y \lor x > z$

*

*$(\forall y)\ x < y \lor x > z$

*

*$x < y \lor x > z$

*

*$x < y$

*$x > z$




*$(\forall y)\ y=x$

*

*$y = x$




Going back bottom-up you will get the sequence of terms from your post. Also, with this particular notation you need to be careful with parentheses, e.g. you have a term $(\forall y\ y = x)$, which is confusing because now the reader does not know which convention you try to follow.
I hope this helps $\ddot\smile$
Edit: Some clarification after the comments.
