Tautology concept of one- GATE 2025 previous year gate question paper Let P,Q,R be three atomic.
Contents
- 1 Tautology Concept – GATE 2025 (Discrete Mathematics)
- 2 GATE 2025 Previous Year Question on Tautology
- 3 Key Takeaways for GATE 2025
- 4 Tautology concept of one- GATE 2025 previous year gate question paper Let P,Q,R be three atomic.
- 5 What is a Tautology?
- 6 GATE Previous Year Question Example
- 7 Additional Resources
Tautology Concept – GATE 2025 (Discrete Mathematics)
What is a Tautology?
A tautology is a propositional logic statement that is always true regardless of the truth values of its components.
Mathematical Representation:
If PP is a logical proposition, then PP is a tautology if its truth table always evaluates to “True” (T) for all possible input values.
Example of a Tautology
-
Logical OR with a Truth Statement
P∨¬PP \lor \neg P
- This means “P OR NOT P”
- This is always true because either PP is true, or ¬P\neg P is true.
- Example:
- If P=TP = T, then ¬P=F\neg P = F → T∨F=TT \lor F = T
- If P=FP = F, then ¬P=T\neg P = T → F∨T=TF \lor T = T
- Always True → Tautology
-
Implication Rule as a Tautology
(P→Q)∨(¬P)(P \rightarrow Q) \lor (\neg P)
- This means “If P implies Q, OR not P”
- Always evaluates to True in a truth table.
GATE 2025 Previous Year Question on Tautology
Let P,Q,RP, Q, R be three atomic propositions. The statement is:
(P∨Q)→(P∨Q∨R)(P \lor Q) \rightarrow (P \lor Q \lor R)
Solution Using Truth Table
PP | RR | P∨QP \lor Q | P∨Q∨RP \lor Q \lor R | (P∨Q)→(P∨Q∨R)(P \lor Q) \rightarrow (P \lor Q \lor R) | |
---|---|---|---|---|---|
T | T | T | T | T | T |
T | T | F | T | T | T |
T | F | T | T | T | T |
T | F | F | T | T | T |
F | T | T | T | T | T |
F | T | F | T | T | T |
F | F | T | F | T | T |
F | F | F | F | F | T |
The last column is always TRUE, so this statement is a tautology.
Key Takeaways for GATE 2025
- Tautology: A statement that is always true for all values of P, Q, and R.
- Common Tautology Forms:
- P∨¬P=TP \lor \neg P = T
- (P→Q)∨¬P(P \rightarrow Q) \lor \neg P
- P∨Q→P∨Q∨RP \lor Q \rightarrow P \lor Q \lor R
- GATE Questions often test tautology using truth tables, logical identities, and set theory.
Would you like more solved GATE questions on Tautology for practice?
Tautology concept of one- GATE 2025 previous year gate question paper Let P,Q,R be three atomic.
Certainly! Let’s delve into the concept of tautology in propositional logic, particularly in the context of GATE (Graduate Aptitude Test in Engineering) examinations.
What is a Tautology?
In propositional logic, a tautology is a compound proposition that is always true, regardless of the truth values of its constituent variables. This means that no matter how you assign truth values to the individual propositions, the overall expression evaluates to true.
Example:
Consider the expression:
(p∧q)→(p∨q)(p \land q) \rightarrow (p \lor q)
This is a tautology because:
-
If both pp and qq are true, then p∨qp \lor q is also true, making the implication true.
-
If either pp or qq is false, then p∧qp \land q is false, and a false antecedent in an implication makes the entire implication true.
Therefore, in all possible truth assignments, the expression evaluates to true.
GATE Previous Year Question Example
Let’s analyze a question from GATE CS 2017 Set 1:
Question:
Let p,q,p, q, and rr be propositions, and consider the expression:
(p→q)→r(p \rightarrow q) \rightarrow r
Suppose this expression is a contradiction, meaning it is always false. Then, evaluate the nature of the expression:
(r→p)→q(r \rightarrow p) \rightarrow qYouTube
Solution:
-
Understanding the Contradiction:
The expression (p→q)→r(p \rightarrow q) \rightarrow r being a contradiction implies that it is false for all truth assignments. An implication A→BA \rightarrow B is false only when AA is true and BB is false. Therefore, for the entire expression to be always false:
-
p→qp \rightarrow q must be true.
-
rr must be false.
-
-
Analyzing (r→p)→q(r \rightarrow p) \rightarrow q:
Let’s examine the truth value of this expression under the conditions:
-
rr is false.
-
p→qp \rightarrow q is true.
Since rr is false, r→pr \rightarrow p is true (an implication with a false antecedent is always true). Therefore, the expression simplifies to:
true→q\text{true} \rightarrow q
The truth value of this implication depends on qq:
-
If qq is true, then the implication is true.
-
If qq is false, then the implication is false.
However, since p→qp \rightarrow q is true and pp could be true or false, the only way p→qp \rightarrow q is always true is if qq is true. Therefore, under these conditions, qq must be true.
Thus, (r→p)→q(r \rightarrow p) \rightarrow q is always true when qq is true.
-
Answer: The expression (r→p)→q(r \rightarrow p) \rightarrow q is always true when qq is true.
Additional Resources
For a more in-depth understanding and further practice, consider exploring the following resources:
-
GATE Overflow – Mathematical Logic Questions: A comprehensive collection of previous GATE questions on mathematical logic. (gateoverflow.in)
-
GeeksforGeeks – Propositional Logic Notes: Detailed notes on propositional and first-order logic, including tautologies and other logical connectives. (geeksforgeeks.org)
-
YouTube – Propositional Logic GATE PYQs: A video explanation of previous year GATE questions on propositional logic. (youtube.com)
If you have any specific questions or need further clarification on any topic, feel free to ask!