Question

Which one of the following is a tautology ?

• A. \((P ∧ (P \to Q)) \to Q\)
• B. \(P ∨ (P ∧ Q)\)
• C. \(Q \to (P ∧ (P \to Q))\)
• D. \(P ∧ (P ∨ Q)\)

Answer 1

The Correct Option is A

 To determine which of the given logical propositions is a tautology, we need to evaluate each option. A tautology is a statement that is always true, regardless of the truth values of its individual components. Let's analyze each option:

1. \((P ∧ (P \to Q)) \to Q\)
   • The expression \((P ∧ (P \to Q)) \to Q\) reads as "if both P and P implies Q are true, then Q is true".
   • Evaluate \(P \to Q\): This is false only if P is true and Q is false.
   • So, for \((P ∧ (P \to Q))\) to be true, both P must be true, and \(P \to Q\) must also be true.
   • If \((P ∧ (P \to Q))\) is true, then Q must be true, making the entire statement \((P ∧ (P \to Q)) \to Q\) always true. Thus, it is a tautology.
2. \(P ∨ (P ∧ Q)\)
   • The expression \(P ∨ (P ∧ Q)\) reads as "P or (P and Q)".
   • This is true if either P is true or both P and Q are true.
   • There could be a situation when both P and Q are false, making the expression false.
   • Thus, it is not a tautology as it's not always true.
3. \(Q \to (P ∧ (P \to Q))\)
   • This expression reads as "Q implies (P and (P implies Q))".
   • If Q is false, the statement is true regardless of the other parts because false implies anything is true.
   • However, if Q is true, we need both P and \(P \to Q\) to be true, which is not guaranteed.
   • Thus, it is not always true, so this is not a tautology.
4. \(P ∧ (P ∨ Q)\)
   • This reads as "P and (P or Q)".
   • For this expression to be true, P must be true (regardless of Q).
   • There could be a scenario where P is false, making the entire expression false.
   • Therefore, it is not a tautology, as it is not always true.

After evaluating all options, the expression \((P ∧ (P \to Q)) \to Q\) is the only one that is always true, making it the tautology among the given options.