Understanding the concept of $P(u/v)$ in the context of First Order Logic. Working on the book: Angelo Margaris. "First Order Mathematical Logic" (p. 47)

Let $u$ and $v$ be variables and $P$ a formula. $P$ admits $u$ for $v$ if and only if every free occurrence of $v$ in $P$ becomes a free occurrence of $u$ in $P(u/v)$.

The author provides this example:

$P$ is $\exists y(y>x)$. Below we list some terms, and alongside each term the result of substituting this term for $x$ in $P$.

$$
\begin{align*}
2 \quad & \exists y(y>2) & P(2/x)\\
y \quad & \exists y(y>y) & P(y/x)
\end{align*}
$$
In the second case, the first and only free ocurrence of $x$ in $P$ becomes bound in $P(y/x)$.
However, I have two questions:

*

*In the first case, I cannot see why 2 becomes a free occurrence in P as it is a constant.

The definition of "$P$ admits $u$ for $v$" does not mention anything about bound ocurrences of $v$ becoming free.

*

*Let $P$ be $\exists y(y > x)$, then $P(x/y)$ would be $\exists y(x > x)$ ? Is this forbidden ?

 A: The author states some examples of substitution: some correct and some not.
The example $[∃y(y>x)](2/x)$ is a correct one because it substitute the term $2$ in place of the free variable $x$ causing no harm [see the definition of substitution: "Let $P$ a formula, $v$ a variable and $t$ a term. Then $P(t/v)$ is..."]
The second example, instead, is an example of "forbidden" substitution: the formula does not admits $y$ for $x$ because if we replace the free occurrence of $x$ with $y$ into the formula $∃y(y>x)$ what we get is $∃y(y>y)$ and the previous free occurrence of $x$ is now a bound occurrence of $y$, and this modifies the meaning of the formula.

Regarding $[∃y(y>x)](x/y)$, the result is not $∃y(x>x)$; according to the definition: "... $P(t/v)$ is the formula that results when each free occurrence of $v$ in $P$ is replaced by an occurrence of $t$."
In $∃y(y>x)$ there are no free occurrences of $y$; thus $[∃y(y>x)](x/y)$ will be: $∃y(y>x)$.
The substitution operation acts on free occurrences.
