Proving the Logical Statement ?p ∧ q ∨ q ≡ T

Proving the logical statement ?p ∧ q ∨ q ≡ T, or stating that ?p ∧ q ∨ q is logically equivalent to True (T), can be approached through various methods, including the use of logical identities and truth tables. This article will walk you through both methods to establish the equivalence.

Using Logical Identities to Prove Equivalence

Let's start by using logical identities to show the equivalence of the given statement.

Step-by-Step Proof Using Logical Identities

We start with the logical expression:

?p ∧ q ∨ q

According to De Morgan's Law, we can transform the expression ?p ∧ q as follows:

?p ∧ q ?(p ∨ ?q)

Substituting this into our original expression, we get:

?(p ∨ ?q) ∨ q

Now, we can rearrange the expression:

?(p ∨ ?q) ∨ q

According to the Law of Excluded Middle, q ∨ ?q is always true:

q ∨ ?q ≡ T

Substituting this into our expression, we get:

?(p ∨ ?q) ∨ T

Using the Identity Law, where A ∨ T ≡ T for any proposition A, we can further simplify our expression:

?(p ∨ ?q) ∨ T ≡ T

Thus, we have shown that:

?p ∧ q ∨ q ≡ T

Alternative Approach using Truth Tables

Alternatively, we can prove the tautology using a truth table. A tautology is a statement that is always true, regardless of the assignment of truth values to the propositions involved. Let's use a truth table to demonstrate that ?p ∧ q ∨ q is a tautology:

p q ?p ∧ q ?p ∧ q ∨ q T T T T T T T F F F T F T F T T F F F F T

As we can see from the truth table, every possible combination of truth values for p and q results in the expression ?p ∧ q ∨ q being equivalent to True (T).

Proof by Contradiction

For completeness, let's prove the tautology ?p ∧ q ∨ q using a proof by contradiction.

Proof by Contradiction

Assume the negation of the statement, i.e., ?(?p ∧ q ∨ q).

Using De Morgan's Law, we can transform the expression as follows:

?(?p ∧ q ∨ q) (p ∨ ?q) ∧ ?q

Now, we can separate the conjunction into two parts:

(p ∨ ?q) ∧ ?q

Eliminate the double negation in the first part:

(p ∨ ?q) ∧ ?q (p ∨ ?q) ∧ ?q

Combine the two parts:

p ∨ ?q and ?q

Since ?q is true, combining it with ?q results in a contradiction:

q ∧ ?q is a contradiction.

Thus, our assumption that ?(?p ∧ q ∨ q) is false. Therefore, the original statement ?p ∧ q ∨ q must be true:

?p ∧ q ∨ q ≡ T

In conclusion, we have proven that ?p ∧ q ∨ q is logically equivalent to True (T) using both logical identities and a proof by contradiction. This establishes that ?p ∧ q ∨ q is a tautology and a theorem in logic.