Re: something's wrong here: Qianosuchus phylogeny

[Mike Taylor <mike@indexdata.com>]

T. Michael Keesey writes:
 > >  > > Really?  What is the reason for knowing a priori that scleromochlids
 > >  > > could not possibly be paraphyletic with respect to pterosaurs?
 > >  >
 > >  > There is only one known scleromochlid, *Scleromochlus* itself, and
 > >  > it has autapomorphies, so the most parsimonious assumption is that
 > >  > it's not an ancestor of anything else we know -- never mind its
 > >  > geological age, which IIRC isn't older than the oldest known
 > >  > pterosaurs.
 > >
 > > OK, but that's an accident of what we currently happen to know about
 > > scleromochlids.  I guess you were making a point about that specific
 > > taxon; I misread your comment as meaning that no taxon ever can be
 > > directly ancestral to any other -- which of course is what I was
 > > disagreeing with.
 > 
 > Well, for that matter, why couldn't it be an accident of what we
 > currently happen to know about pterosaurs? Why couldn't we use
 > pterosaurs as a paraphyletic grade that gave rise to scleromochlids?
 > There's nothing to prevent a taxonomist who uses paraphyletic taxa
 > from including the final common ancestor of Scleromochlus,
 > rhamphorhynchoids, and pterodactyloids in Pterosauria, with
 > Scleromochlus as part of a derived offshoot.

True.

 > (See, this is why higher-level paraphyletic taxa are a bad idea....)

Well, no ... you've not shown that higher-level paraphyletic taxa are
a bad idea, you've just shown one possible consequence (that one taxon
can be located inside another) and _asserted_ that it's bad.  And that
consequence is really just a restatement of the definition of
paraphyly, so your statement seems to reduce to "paraphyly is bad
because of the badness of paraphyly" :-)

I still don't get this -- never have.  Despite being very fond of
formal phylogenetic nomenclature (most of my published and in-press
work is on that very subject) I simply don't see what would be bad
about definitions of the form X = Y-Z where X is a paraphyletic taxon
and Y and Z are other taxa such that Z is a subset of Y.  Every
argument I've ever seen against such definitions boils down to
fundamentalism: That Would Be Bad Because It Is Forbidden.

 _/|_    ___________________________________________________________________
/o ) \/  Mike Taylor    <mike@indexdata.com>    http://www.miketaylor.org.uk
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