Direct way to show that 2-out-of-6 holds for weak equivalences in a model category? In a model category, when weak equivalences are inverted, nothing else gets inverted. It follows that weak equivalences satsify 2-out-of-6. But the first sentence takes some work to show. Is there a more direct way to demonstrate 2-out-of-6?
For reference, 2-out-of-6 says that if $v \circ u$ and $w\circ v$ are weak equivalences, then $w\circ v \circ u$ is a weak equivalence (in light of 2-out-of-3, the conclusion is equivalent to saying that any of $u,v,w$ is a weak equivalence). This property apparently features in the definition of a homotopical category. It follows from the property above because isomorphisms satisfy 2-out-of-6, so the preimages of isomorphisms under a functor also satsify 2-out-of-6.
I'm interested in having a direct argument because 2-out-of-6 leads to a simple proof that if $f$ is a homotopy equivalence (i.e. there exists $g$ such that $gf$ and $fg$ are both either left or right homotopic to identity maps) then $f$ is a weak equivalence. I would expect this fact to have a simple proof, but in fact Dwyer-Spalinski and Hovey at least only prove this directly with some fibrancy/cofibrancy restrictions on the objects involved. The proof using 2-out-of-6 goes like this: it's easy to see that if $h$ is left or right homotopic to $k$, then if one is a weak equivalence then so is the other; so if $f$ has homotopy inverse, then $fg$ and $gf$ are both weak equivalences, so 2-out-of-6 implies that $f$ and $g$ are each weak equivalences. 
 A: For some reason, I have the urge to write this out; it's not that hard really.  The proof I sketch is only just what you get if you unwind the proof of saturation in Hovey.  I refer to a sequence $(w,v,u)$ of composable maps, as in the question: I want to show that if $wv$ and $vu$ are weak equivalences, so are the others.


*

*The special case when either $wv$ or $vu$ is an identity map is immediate using the 2-of-3 and retract axioms.  For instance, if $wv=1$, then $u$ is a retract of $vu$.  

*In particular, we can apply point (1) to $(s,r,s)$ when $rs=1$: if $sr$ is a weak equivalence, so are $r$ and $s$. 

*The general problem really happens in the model category of factorizations of the map $wvu$.  Thus, without loss of generality, we may reduce to the following Claim: In any model category, if $T'$ is weakly equivalent to the terminal object $T$, and $I'$ is weakly equivalent to the initial object $I$, then any map $r\colon T'\to I'$ is a weak equivalence (if such a map exists).  [E.g., $T'$ is $(wv,u)$, $I'$ is $(w,vu)$, and $r$ is $v$.]

*By a straightforward argument using factorization and 2-of-3,  you can show that without loss of generality we may assume that $I'$ and $T'$ are cofibrant and that $r$ is a fibration.

*As $I\to I'$ is a trivial cofibration and $r\colon T'\to I'$ is a fibration, by lifting the map $r$ admits a section $s$, so $rs=1_{I'}$.

*2-of-3 implies that any self-map $T'\to T'$ of $T'$ is a weak equivalence, since the unique map $T'\to T$ is a weak equivalence.

*Thus $sr\colon T'\to T'$ is a weak equivalence by point (6), whence $r$ must be a weak equivalence by point (2), as desired.
