deducibility from peano axioms I have to solve the following problem:
Using $\exists$ Introduction prove that PA$\vdash x\leq y \wedge y\leq z \longrightarrow x\leq z$: I used that if $x\leq y$ then $\ \exists \ r\  x+r=y$ and in the same way $\exists t \ y
+t=z$, but in logic term I don't know how to use these equalities and the Peano's axioms to get the result.

See Wikipedia Peano Axioms.
The point of this answer is to show that you can define an ordering of the natural numbers before you define addition.
Proposition 1: If $n$ is any number except $0$, there exist an $m$ such that
$n = S(m)$
Moreover, if the successor of both $m$ and $m^{'}$ is equal to $n$, then  $m = m^{'}$.
Proof: Exercise

Definitions:
The Decrement mapping $D$ 
$ n \mapsto D(n)$ 
sends any nonzero number $n$ to $m$, where $S(m) = n$.
We also define $1$ to be $S(0)$. Note that $D(1) = 0$
Note for any nonzero $n$ we have the 'keep decrementing unless you can't' number expressions:
(1) $D(n)$, $D(D(n))$, $D(D(D(n)))$, ...
It is understood that this list might terminate since you can't apply $D$ to $0$.

Proposition 2: If $n$ is any nonzero number, the decrementing terms (1) ends in the number $0$.
Proof: It is true for $n$ = 1.
Assume it is true for $n$, and consider $S(n)$. But $D(S(n)) = n$, so that for $S(n)$, compared to the terms for $n$, you simply get a new starting term. 
QED
These logical expressions,
$m \ne n \;\; iff \;\; D(m) \ne D(n)$
are true provided both $m$ and $n$ are nonzero. By repeatedly applying Proposition 2 you will find that either $m$ or $n$ gets decremented to $0$ first.
Definition: If $m$ and $n$ are two distinct nonzero numbers, we say that $m$ is less than $n$, written $m \lt n$. if $m$ is decremented to zero before $n$.
We define the relation $\le$ in the obvious fashion and agree that for all natural numbers $n$, $\;0 \le n$.
Proposition 3: $x\leq y \wedge y\leq z \longrightarrow x\leq z$
Proof: Assume $x, y, z$ are three distinct nonzero numbers.
$x\leq y \wedge y\leq z \longrightarrow x\leq z$
IF AND ONLY IF
$D(x)\leq D(y) \wedge D(y)\leq D(z) \longrightarrow D(x)\leq D(z)$
As before, you see that $x$ is decremented to zero before both $y$ and $z$, so that $x \le z$.
We leave it to the reader to finish the proof.
