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) $\exists z [x \leq z \wedge y \leq z]$