To prove using sat matrix:
 negate the theorem
 convert to clause form
 convert to sat matrix
 =====================
 prove matrix unsatisfiable

To prove using resolution:
  negate the theorem
  convert to clause form
  derive an empty clause using the resolution  rule
     (a u b u c u d ) and (~a v m u n...)
	gives:
	   b v c v d m v n .....


Existential Graphs:       (negate the theorem. derive a () on the
sheet.
____________________

   Logic      EGs
     P          P
     ~P        (P)
     P and Q    P Q
     P or Q     ((P) (Q))
     P => Q     (P (Q))

Idea:  Circle Goal derive an empty circle on the sheet!

Rules:  ((x)) ---> x.
      x x     ---> x

       x can kill inner copy at any level of nesting:   x   (z(x y)) --> (z(y))

Example:
[A => (B or C)) and ~C  and B => D  therefore A => D.

   in EGs:    ((A (((B) (C)))) (C) (B (D))      ((A (D)))   )

    Negate goal:
       (((A (((B) (C)))) (C) (B (D))      ((A (D)))   ))
     Remove Double Negation
       (A (((B) (C)))) (C) (B (D))      ((A (D)))   

     Remove double negation:
        (A (B) (C))    (C) (B (D))     A (D)

     Kill A:
	 ((B) (C))   (C) (B (D)) A (D)

     KILL (D):    ((B) (C)) (C)   (B) A (D)

     KILL (B):
		  ((C))   (C)  (B) A (D)

     KILL   (C):      () (C)   (B) A (D)

     We are done since there is a () on the sheet!!!