A logical equivalance is derived from Theorem,supply a reason for each step;
(pv~q)^(~pvq)
�ß(~q v p)^(~q v ~p) by_________
�ß ~q v(p ^ ~p) by_________
�ß ~q v c by_________
�ß ~q by_________
Therefore,(p v~q)^ (~p v ~q)�ß ~q.
To derive the logical equivalence (pv~q)^(~pvq) �ß ~q using the given theorem, we need to provide reasons for each step of the derivation. Here is the step-by-step explanation:
1. (pv~q)^(~pvq) (Given)
To simplify the conjunction (pv~q)^(~pvq) using the given theorem, we will apply the distributive law. The distributive law states that a^ (bvc) is logically equivalent to (a^b) v (avc).
2. (~qvp)^(~q v ~p) (Distribution)
Now, let's simplify the conjunction (~qvp)^(~q v ~p).
3. ~qv(p^~p) (Distribution)
By applying the distributive law again, we see that (p^~p) is a contradiction. Any statement along with its negation results in a contradiction.
4. ~qvC (Contradiction Elimination)
Since (p^~p) is a contradiction, we can replace it with any contradiction symbol, denoted as C.
5. ~q (Simplification)
Finally, we can simplify the disjunction ~qvC to ~q. This is a result of the logical law of contradiction, which states that the disjunction of a statement with a contradiction is logically equivalent to the negation of that statement.
Therefore, we have shown that (pv~q)^(~pvq) �ß ~q, providing reasons for each step of the derivation.