III. Logical variables and unification
 A. logical variables
 B. substitutions and instances
  1.     substitution
		----------------
                s: {X, Y} -> {algol, father(me,dad)}
                 such that s[[X]] = algol, s[[Y]] = father(me,dad).
                empty: {} -> {}
                counterexample: bad: {X} -> {X}
		----------------
  2.        instance of a term
		------------------
                father(me,dad) is an instance of father(X,dad).
		------------------
 C. comparing terms
		--------------
                foo(X,3,Y).
                ?- foo(0,X,X).
                X = 3  % this is the X of the question, one in fact bound to 0
		----------------
  1.     common instance of two terms: term which is instance of both
		------------------
                plus(0,3,3) is common instance of plus(0,3,Y) and plus(0,X,X)
		----------------
  2.     generality
  3.     unifier: a subsitution that makes two terms identical.
		--------------------
                star(algol).
                ?- star(X).
                    X = algol
                                % star(X) unifies with star(algol)
                                %       by substituion s(X)=algol

                g(X,3).
                ?- g(4,Y).
                        Y = 3
                        % and (X = 4) not shown

                h(X,X).
                ?- h(Y,Z).
                        Y = _0   % i.e., X's internal name
                        Z = _0
		----------------------
 D. exercises
	-------
	For each of the following pairs, compute the mgu, if any

	member(3,[3])
	member(X,[X])

	member(Y,[3])
	member(X,[X])


	member(3,[4])
	member(X,[X|L])

	length([3],N)
	member(X,[X|L])

	member(X,[X            |L])
	member(Y,[tree(Y,[],[])|M])
	-------
IV. Resolution (logical basis for Prolog)
----------------
imokay :- youreokay, hesokay.	% C1
youreokay :- theyreokay.	% C2
hesokay.			% C3
theyreokay.			% C4
?- imokay.
----------------
----------------
hesnotokay :- imnotokay.	% C5
shesokay :- hesnotokay.		% C6
shesokay :- theyreokay.		% C7
?- shesokay.
----------------
----------------
hesnotokay :- shesokay.		% C8
hesnotokay :- imokay.		% C9
?- shesokay.
----------------
 A. Example
-------------
%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?
---------------
 B. resolution theorem proving
  1.     rule of inference
			--------------------------------------
                        (resolution)  R or P, (not P or Q)
                                      ____________________
                                           R or Q
			---------------------------------------
			------------------			
			A\B true false
		       true  t    f
		      false  t    t
			------------------
  2.     the resolution rule (mechanical)
		------------------------
                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
		------------------------
  3.     to ask a question Q, resolve (not Q) against set of clauses,
  4.     practical questions in using resolution
  5.     SL resolution
 C. Why Horn clauses?
 D. Top-down vs. bottom up control
  1.     top-down control (backward chaining) as in Prolog
  2.     bottom up control (forward chaining) as in OPS5
 E. Depth-first search vs breadth-first search.
		----------------------
                p :- p, q.
                p.
                q.
		----------------------