Implicative normal form:
Thus the sentence is now in CNF. In Fact for simplification can take place by removing duplicate literals and dropping any clause that contains both A and ¬A where one will be true because the clause is always true. However in the conjunction of clauses we want everything to be true than we can drop it. Furthermore there is an optional final step that takes it to Kowalski normal form called as implicative normal form (INF) too aS:
Although by reintroduce implication by gathering up all the negative literals in the negated ones and forming them in their conjunction N thus taking the disjunction P of the positive literals so forming the logically equivalent clause N →P.
There is example for understand: just Converting to CNF. Here we will work through a simple propositional example like:
(B ? (A ^ C)) → (B? ¬ A)
So there is this first thing to do is remove the implication sign as:
¬ (B? (A ^ C)) ? (B ? ¬ A)
Now we use De Morgan's laws just to move our negation sign from the outside to the inside of brackets as:
(¬ B ^ ¬ (A^ C)) ? (B ? ¬ A)
So now we can use De Morgan's law again just to move a negation sign inwards as:
(¬ B^ ( ¬ A? ¬ C)) ? (B ?¬ A)
And now next we distribute over as follows:
(¬ B ? (B ? ¬ A)) (( ¬ A ? ¬ C) ? (B ? ¬ A))
There if we flatten our disjunctions so than we get our sentence into CNF form. Notice here the conjunction of disjunctions:
(¬ B? B ? ¬ A)^ ( ¬ A ? ¬ C ? B ? ¬ A)
Now the finally first conjunction has ¬B and B, then the whole conjunction must be true that we can remove the duplicate ¬A in the second conjunction also as:
True ^ (¬ A ? ¬ C ? B)
So however the truth of this sentence is only dependent on the second conjunct. Then if it is false, the complete thing is false hence it is true for the whole thing is true. Thus we can remove the True through giving us a single clause into its final conjunctive normal form as:
¬ A ? ¬ C ?B
There if we want Kowalski normal form we take one more step to get as:
(A ^ C) → B