Proof in the sequent calculus I have to prove the following in the sequent calculus:
Given
(a) $\forall x: x \leq x$
(b) $\forall x \forall y \forall z [(x \leq y \wedge y \leq z) \rightarrow x \leq z]$
(c) $\forall x \forall y \exists z [x \leq z \wedge y \leq z]$
$\Gamma := \{(a), (b), (c)\}$
Show that: $\Gamma \vdash \forall x \forall y \forall z \exists w [x \leq w \wedge y \leq w \wedge z \leq w]$
(Common rules like $\forall $-introduction and elimination and $\exists$-introduction and elimination ... etc. can be used)
I know how to prove it intuitively, (with (a) and (c), (b) is not necessary if I am not mistaken), but I have no idea for a formal proof.
Basically it should work if I got to
$\exists w [ x \leq w \wedge y \leq w \wedge z \leq w ]$
(Then I could use 3* $\forall$-introduction)
Can anyone give me a hint?

P.S.: If I eliminate the $\forall$'s:
(a) $ x \leq x $
(b) $(x \leq y \wedge y \leq z) \rightarrow x \leq z$
(c) $\exists z [x \leq z \wedge y \leq z]$
 A: I'll formalize Apostolos' argument with Natural Deduction in "sequent-calculus-style" [see Dirk van Dalen, Logic and Structure, (5th ed - 2013), pages 35, 88 and 91 for the rules].
These are the rules for quantifiers :

$$(\forall I) \, \, {\Gamma \vdash \varphi(x) \over \Gamma \vdash \forall x \varphi(x)}\ x \notin FV(\psi) \, \text{for all} \,\, \psi \in \Gamma$$
$$(\forall E) \, \, {\Gamma \vdash \forall x \varphi(x) \over \Gamma \vdash \varphi(t)}\ t \, \text{free for} \, x \, \text{in} \,\, \varphi$$
$$(\exists I) \, \, {\Gamma \vdash \varphi(t) \over \Gamma \vdash \exists x \varphi(x)}\ t \, \text{free for} \, x \, \text{in} \,\, \varphi$$
$$(\exists E) \, \, {\Gamma, \varphi(x) \vdash \psi \over \Gamma, \exists x \varphi(x) \vdash \psi}\ x \notin FV(\psi) \, \text{and} \, x \notin FV(\gamma) \,\, \text{for all} \,\, \gamma \in \Gamma$$

Proof
From the rule of assumption [i.e. $\Gamma \vdash \varphi$, if $\varphi \in \Gamma$] applied to axiom (c) we derive, following Apostolos' proof, by multiple applications of ($\forall$ E)  :
(1) $\Gamma \vdash ∃w(x≤w∧y≤w)$
(2) $\Gamma \vdash ∃u(y≤u∧z≤u)$
(3) $\Gamma \vdash ∃a(u≤a∧w≤a)$
(4) $\Gamma \vdash x \le w$ --- from (1) by ($\land$ E) and ($\exists$ E)
Note :
By ($\land$ E) we have :

$$\Gamma \vdash x≤w∧y≤w \over \Gamma \vdash x≤w$$

Applying "weakening" [i.e.if $\Gamma \vdash \varphi$, then $\Gamma \cup \Delta \vdash \varphi$], we get :

$$\Gamma \vdash x≤w∧y≤w \over \Gamma, x≤w∧y≤w \vdash x≤w$$

Thus, by ($\exists$ E) we have :

$$\Gamma \vdash x≤w∧y≤w \over \Gamma, ∃w(x≤w∧y≤w) \vdash x≤w$$

But by (1) : $\Gamma \vdash ∃w(x≤w∧y≤w)$, we conclude (4) : $\Gamma \vdash x≤w$.
end note.
In the same way, we have :
(5) $\Gamma \vdash w \le a$ --- from (3)
Thus :
(6) $\Gamma \vdash x \le w \land w \le a$ --- from (4) and (5) by ($\land$ I).
(7) $\Gamma \vdash x \le a$ --- from axiom (b) with multiple applications of ($\forall$ E) and (6) and ($\rightarrow$ E).
Repeating the "procedure" (4)-(7) with $y \le w$ (from (1)) and again with $z \le u$ (from (2)) we get :
(8) $\Gamma \vdash y \le a$
(9) $\Gamma \vdash z \le a$.
Now we apply ($\land$ I) twice with (7), (8) and (9) to get :
(10) $\Gamma \vdash (x \le a \land y \le a) \land z \le a$.
Thus by ($\exists$ I) :
(11) $\Gamma \vdash \exists w(x \le w \land y \le w) \land z \le w$
and finally, by multiple applications of ($\forall$ I) :


(12) $\Gamma \vdash \forall x \forall y \forall z \exists w[(x \le w \land y \le w) \land z \le w]$.


