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+Using the GF python bindings
+============================
+
+This is how to use some of the functionalities of the GF shell inside Python.
+
+Loading a pgf file
+------------------
+First you must import the library:
+
+>>> import gf
+
+then load a .pgf file, like this tiny example:
+
+>>> pgf = gf.read_pgf("Query.pgf")
+
+We could ask for the supported languages:
+
+>>> pgf.languages()
+[QueryEng, QuerySpa]
+
+The *start category* of the PGF module is:
+
+>>> pgf.startcat()
+Question
+
+Parsing and linearizing
+-----------------------
+
+Let's us save the languages for later:
+
+>>> eng,spa = pgf.languages()
+
+These are opaque objects, not strings:
+
+>>> type(eng)
+<type 'gf.lang'>
+
+and must be used when parsing:
+
+>>> pgf.parse(eng, "is 42 prime")
+[Prime (Number 42)]
+
+Yes, I know it should have a '?' at the end, but there is not support for other lexers at this time.
+
+Notice that parsing returns a list of gf trees.
+Let's save it and linearize it in Spanish:
+
+>>> t = pgf.parse(eng, "is 42 prime")
+>>> pgf.lin(spa, t[0])
+'42 es primo'
+
+(which is not, but there is a '?' lacking at the end, remember?)
+
+
+Getting parsing completions
+---------------------------
+One of the good things of the GF shell is that it suggests you which tokens can continue the line you are composing.
+
+We got this also in the bindings. 
+Suppose we have no idea on how to start:
+
+>>> pgf.complete(eng, "")
+['is']
+
+so, there is only a sensible thing to put in. Let's continue:
+
+>>> pgf.complete(eng, "is ")
+[]
+
+Is it important to note the blank space at the end, otherwise we get it again:
+
+>>> pgf.complete(eng, "is")
+['is']
+
+But, how come that nothing is suggested at "is "? 
+At the current point, a literal integer is expected so GF would have to present an  infinite list of alternatives. I can not blame it for refusing to do so.
+
+>>> pgf.complete(eng, "is 42 ")
+['even', 'odd', 'prime']
+
+Good. I will go for 'even', just to be in the safe side:
+
+>>> pgf.complete(eng, "is 42 even ")
+[]
+
+Nothin again, but this time the phrase is complete. Let us check it by parsing:
+
+>>> pgf.parse(eng, "is 42 even")
+[Even (Number 42)]
+
+Deconstructing gf trees
+-----------------------
+We store the last result and ask for its type:
+
+>>> t = pgf.parse(eng, "is 42 even")[0]
+>>> type(t)
+<type 'gf.tree'>
+
+What's inside this tree? We use ``unapply`` for that:
+
+>>> t.unapply()
+[Even, Number 42]
+
+This method returns a list with the head of the **fun** judgement and its arguments:
+
+>>> map(type, _)
+[<type 'gf.cid'>, <type 'gf.expr'>]
+
+
+Notice the argument is again a tree (``gf.tree`` or ``gf.expr``, it is all the same here.)
+
+>>> t.unapply()[1]
+Number 42
+
+
+We will repeat the trick with it now:
+
+>>> t.unapply()[1].unapply()
+[Number, 42]
+
+and again, the same structure shows up:
+
+>>> map(type, _)             
+[<type 'gf.cid'>, <type 'gf.expr'>]
+
+One more time, just to get to the bottom of it:
+
+>>> t.unapply()[1].unapply()[1].unapply()
+42
+
+but now it is an actual number:
+
+>>> type(_)
+<type 'int'>
+
+We ended with a full decomposed **fun** judgement.
+
+
+Note
+----
+
+This file can be used to test the bindings: ::
+
+    python -m doctest example.rst
+
+
+