Which of the following is the principal conjunctive normal form for [(pVq) ∧ ~p → ~q]?pV~qpVq~pVq~pV~q

To find the principal conjunctive normal form for [(pVq) ∧ ~p → ~q], we first need to understand the logical operators used in the expression. Here, V stands for OR, ∧ stands for AND, and → stands for IMPLIES. ~ is the NOT operator.

The expression can be broken down as follows:

1. (pVq) ∧ ~p → ~q: This is the given expression.

2. ~[(pVq) ∧ ~p] V ~q: Using the definition of implication, A → B is equivalent to ~A V B.

3. [~(pVq) V ~~p] V ~q: Using De Morgan's laws, ~(A ∧ B) is equivalent to ~A V ~B.

4. [(~p ∧ ~q) V p] V ~q: Again using De Morgan's laws, ~(A V B) is equivalent to ~A ∧ ~B. Also, ~~A is equivalent to A.

5. [(~p V p) ∧ (~q V p)] V ~q: Distributing OR over AND, A ∧ (B V C) is equivalent to (A ∧ B) V (A ∧ C).

6. (True ∧ (~q V p)) V ~q: As ~p V p is always True.

7. (~q V p) V ~q: True ∧ A is equivalent to A.

8. ~q V (p V ~q): Using the associative law, A V (B V C) is equivalent to (A V B) V C.

So, the principal conjunctive normal form for [(pVq) ∧ ~p → ~q] is ~q V (p V ~q).