CS 541 Lecture -*- Outline -*-

* Resolution (logical basis for Prolog)

** An abstract interpreter

	keeps a set of goals, called a resolvent
----------------
input: a logic program P, and a goal G
output: a substitution s, if s[[G]] is deducible from P,
	or failure if such a deduction is not possible

algorithm: (which may fail to terminate)
	initialize the resolvent to be G

	while the resolvent is not empty do
	  choose a goal A from the resolvent
		and a (renamed clause) A' :- B1,...,Bn from P
		such that A and A' unify with mgu s
	        (exit if no such goal and clause exist)
	  remove A from the resolvent and add B1,...,Bn
	  apply s to the resolvent and to G
	If the resolvent is empty, output s, else output failure
----------------

	to ensure that variables are local to a clause, rename variables
		each time the clause is chosen to effect a reduction.
		the new names are fresh

** Why is the abstract interpreter correct?

	want to know that if a query is deducible from the program,
		then there is some execution (some choices)
		that will deduce it.  (And conversely, if not deducible,
		will output failure.

	so must tie this interpreter to the deduction rules...

	recall "universal modus ponens":

	(forall X . (A <== B1,...,Bn)), s[[B1]], ..., s[[Bn]]
	_____________________________________________________ s a substitution
				s[[A]]

	rule "conjunction":
		     s[[P1]], s[[P2]]
		________________________	s a substitution, Xi in dom(s)
		exists X1...Xn . P and P2

*** simpler rules, resolution
	Example (for use below)
-------------
%facts
        scientist(sue). %1
        scientist(ron). %2
        spanish(sue).   %3
        american(ron).  %4
%rule
        logician(X) :- scientist(X). %5
%goal
        ?- logician(Y), american(Y).
        % (is there) some Y is both a logician and an american?
---------------

**** to eliminate quantifiers, use fresh variables (skolemize)
	------------------------
	rule 5 becomes
		logician(X) <== scientist(X)

	universal modus ponens becomes:
	(A <== B1,...,Bn), s[[B1]], ..., s[[Bn]]
	_________________________________________ s a substitution
			s[[A]]
	-------------------------

**** to remove implications, use formula: (A <== B) = (A or not B)
	truth table for implication 
			B\A true false
		       true  t    f
		      false  t    t

	----------------------
	rule 5 becomes
		logician(X) or not(scientist(X))

	universal modus ponens becomes

	(A or not(B1) or ... or not(Bn)), s[[B1]], ..., s[[Bn]]
	______________________________________________________ s a substitution
			s[[A]]
	----------------------

**** to ask a query Q, see if not(Q) produces a contradiction
	this converts all existential queries into universal statements.

	------------------------------------------------
	query was exists Y . logician(Y) and american(Y)
	negation is
	 forall Y . not (logician(Y) and american(Y))
	= by deMorgan's laws
	 forall Y . not(logician(Y)) or not(american(Y))

	now skolemize this
	 	not(logician(Y)) or not(american(Y))
	------------------------------------------------

	instead of trying to prove the statement in a query,
		try to disprove, refute, the negation

	so instead of conjuction and modus ponens deduction rules,
		have resolution rule

--------------------------------------
resolution:
not(s[[A1]]) or ...
	or not(s[[Ai]]) or
	... or not(s[[Am]]),
	(Ai or not(B1) or ... or not(Bn))
_________________________________________ s a substitution
not(s[[A1]]) or ...
	or not(B1) or ... or not(Bn) or
	...or not(s[[Am]])

simpler version:
    	not(R) or not(P), (P or not(Q))
        _______________________________
               not(R) or not(Q)

some consequences:
	not(R) or not(P), P
	___________________
	      not(R)

	not(R), R
	_________
	 *poof*
---------------------------------------
	clear connection betwene modus ponens and resolution
		not(R) or not(P), P <== Q
		_________________________
		     not(R) or not(Q)

	look at the negation of this
		R and P, P <== Q
		_________________
		     R and Q

------------------------
suppose clause A contains some term s,
 clause B contains negation of term t,
 s and t are unifiable by some subst sigma,
then resolvent of A and B is generated by
 1. combining the clauses to form A or B
 2. removing s and t (from the resulting term)
 3. applying substitution sigma to result
------------------------

        e.g., resolvent of: not scientist(sue)
                with            scientist(sue)
        formed by combining: not scientist(sue) or scientist(sue),
                then deleting matching parts:   *poof!*

        e.g., resolvent of: not logician(Y) or not american(Y)
                with:           logician(X1) or not scientist(X1)
	unifying substitution s(Y) = U0, s(X1) = U0
                is: not american(U0) or not scientist(U0).

	summary: to ask a question Q, resolve (not Q) against set of clauses,
                if reach a contradiction (empty clause)
                shows that original goal was true

** Why Horn clauses?
        lhs of rules is always positive
                e.g., A if B and C  ==> A or not B or not C
                not every statement in predicate logic can be stated this way.