First order logic - Identity. Show $\vDash \forall x\exists y (x = y)$. I want to learn math on my own and I started with logic. I am studying the book "Logic and structure" written by Dirk van Dalen. I really like his style but he doesn't give detailed explanations neither solutions to exercises. There is a small chapter in his book called Identity where he talks about the axioms (first order logic).
The axioms
The exercises don't seem to be extremely hard, but I don't know how to prove them rigorously. (All exercises)

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*Show $\vDash \forall x\exists y (x = y)$

*Show $\vDash \forall x(\varphi(x) \Leftrightarrow \exists y(x = y  \land \varphi(y)))$ and $\vDash \forall x(\varphi(x) \Leftrightarrow \forall y(x = y  \implies \varphi(y)))$ where $y$ does not occur in $\varphi(x)$.

For example, the first one: $\forall x\exists y (x = y)$, seems so obvious, but I don't know if it requires a more complex proof (maybe a proof based on the axioms).
For the second exercise, do I need to use Induction on $\varphi$ ? (like van Dalen used for the second part of the 4 axiom).
Induction
I would really appreciate any help! (at least for the first 2 exercises so I can understand and try to solve the other ones by myself.)
 A: Using the given Identity Axioms:
(E1) $\forall x \exists y (x=y)$
If (E1) is not true, then Axiom I1:
$\forall x (x = x)$
will not be true. By Contradiction, (E1) is true.
(E2) It looks like we can not show this with Only the Identity Axioms; There may be more Axioms listed earlier in the Text Book, which, together with Axioms I1 & I4, must be used in (E2) here.
I think that (E2) is trying to state that variable $x$ can be substituted by variable $y$ in a statement, Provided that $y$ does not occur in the statement.
Eg $x=zx+1$ is equivalent to $y=zy+1$, but $x=yx+1$ is not equivalent to $y=yy+1$
UPDATE:
The first Part of (E2) can be shown via "Contradiction", with DeMorgans Laws:
$\varphi(x) \Leftrightarrow \exists x (x=y \land \varphi(y))$
When LHS is true, RHS must be true. Assume RHS is not true; then the negation is true.
$\lnot (\exists x (x=y \land \varphi(y)))$
$\forall x ( \lnot (x=y) \lor \lnot(\varphi(y)))$
$( \forall x \lnot (x=y) \lor \forall x \lnot(\varphi(y)))$
By (E1), $\forall x \lnot (x=y)$ is not true.
Thus, $\forall x \lnot(\varphi(y))$ must be true.
By given LHS, there is $\varphi(x)$, then $\forall x \lnot(\varphi(y))$ is not true.
It is a "Contradiction" here.
Thus Negation of RHS is not true.
Hence, RHS is true.
Here $\varphi(x)$ must not have $y$, because that might make this argument invalid:
$\varphi(x) : x=z+1$ can have at least one Integer Solution, and $\varphi(x)$ does not have $y$.
$\varphi(y) : y=z+1$ is same.
But:
$\varphi(x) : x=y+1$ can have at least one Integer Solution, but $\varphi(x)$ does have $y$.
$\varphi(y) : y=y+1$ is not the same and has no Integer Solutions.
Alternate Proof of (E2) with I4:
We could take I4 with $n=1$ to get:
$\forall x,y (x=y \rightarrow (\varphi(x) \rightarrow \varphi(y))) $
$\forall x,y ((x=y \land \varphi(x)) \rightarrow (\varphi(y))) $
$\forall x,y ((\varphi(x) \land x=y) \rightarrow (\varphi(y))) $
$\forall x (\varphi(x) \rightarrow \exists y (x=y \land \varphi(y))) $
