.. Copyright (C) 2001-2025 NLTK Project
.. For license information, see LICENSE.TXT

.. see also: gluesemantics.doctest

==============================================================================
 Glue Semantics
==============================================================================

    >>> from nltk.test.gluesemantics_malt_fixt import setup_module
    >>> setup_module()

    >>> from nltk.sem.glue import *
    >>> nltk.sem.logic._counter._value = 0

--------------------------------
Initialize the Dependency Parser
--------------------------------
    >>> from nltk.parse.malt import MaltParser

    >>> tagger = RegexpTagger(
    ...     [('^(John|Mary)$', 'NNP'),
    ...      ('^(sees|chases)$', 'VB'),
    ...      ('^(a)$', 'ex_quant'),
    ...      ('^(every)$', 'univ_quant'),
    ...      ('^(girl|dog)$', 'NN')
    ... ]).tag
    >>> depparser = MaltParser(tagger=tagger)

--------------------
Automated Derivation
--------------------
    >>> glue = Glue(depparser=depparser)
    >>> readings = glue.parse_to_meaning('every girl chases a dog'.split())
    >>> for reading in sorted([r.simplify().normalize() for r in readings], key=str):
    ...     print(reading.normalize())
    all z1.(girl(z1) -> exists z2.(dog(z2) & chases(z1,z2)))
    exists z1.(dog(z1) & all z2.(girl(z2) -> chases(z2,z1)))

    >>> drtglue = DrtGlue(depparser=depparser)
    >>> readings = drtglue.parse_to_meaning('every girl chases a dog'.split())
    >>> for reading in sorted([r.simplify().normalize() for r in readings], key=str):
    ...     print(reading)
    ([],[(([z1],[girl(z1)]) -> ([z2],[dog(z2), chases(z1,z2)]))])
    ([z1],[dog(z1), (([z2],[girl(z2)]) -> ([],[chases(z2,z1)]))])

--------------
With inference
--------------

Checking for equality of two DRSs is very useful when generating readings of a sentence.
For example, the ``glue`` module generates two readings for the sentence
*John sees Mary*:

    >>> from nltk.sem.glue import DrtGlue
    >>> readings = drtglue.parse_to_meaning('John sees Mary'.split())
    >>> for drs in sorted([r.simplify().normalize() for r in readings], key=str):
    ...     print(drs)
    ([z1,z2],[John(z1), Mary(z2), sees(z1,z2)])
    ([z1,z2],[Mary(z1), John(z2), sees(z2,z1)])

However, it is easy to tell that these two readings are logically the
same, and therefore one of them is superfluous.  We can use the theorem prover
to determine this equivalence, and then delete one of them.  A particular
theorem prover may be specified, or the argument may be left off to use the
default.

    >>> readings[0].equiv(readings[1])
    True
