Implication Concept Discrete mathematics previous year paper – GATE 2025- The following resolution.
Implication Concept Discrete mathematics previous year paper – GATE 2025- The following resolution.
Contents [hide]
- 1 Implication Concept in Discrete Mathematics – GATE 2025
- 2 Important Properties of Implication
- 3 GATE 2025 Previous Year Question on Implication
- 4 Conclusion
- 5 Implication Concept Discrete mathematics previous year paper – GATE 2025- The following resolution.
- 6 Discrete Mathematics for Computer Science
- 7
Understanding Implication (→)
- 8
Resolution Principle in Propositional Logic
- 9
GATE Previous Year Question Example
- 10
Additional Resources
- 11 Discrete Mathematics and Its Applications, Eighth Edition
Implication Concept in Discrete Mathematics – GATE 2025
What is an Implication?
In propositional logic, an implication is a logical statement of the form:
P→QP \rightarrow Q
This means “If PP is true, then QQ must also be true.”
Truth Table for Implication (P→QP \rightarrow Q)
PP | P→QP \rightarrow Q | |
---|---|---|
T | T | T |
T | F | F |
F | T | T |
F | F | T |
Key Observations:
- If PP is true and QQ is false, then P→QP \rightarrow Q is false.
- In all other cases, P→QP \rightarrow Q is true.
- When PP is false, the implication P→QP \rightarrow Q is always true, regardless of QQ.
Important Properties of Implication
Contrapositive:
P→Q≡¬Q→¬PP \rightarrow Q \equiv \neg Q \rightarrow \neg P
(Logically equivalent, used in proofs)
Inverse:
¬P→¬Q\neg P \rightarrow \neg Q
(Not logically equivalent to P→QP \rightarrow Q)
Converse:
Q→PQ \rightarrow P
(Not logically equivalent to P→QP \rightarrow Q)
Material Implication:
P→Q≡¬P∨QP \rightarrow Q \equiv \neg P \lor Q
(Can be rewritten in terms of OR operation)
GATE 2025 Previous Year Question on Implication
Question:
Given three propositions P,Q,RP, Q, R, which of the following is a valid logical equivalence?
(A) (P→Q)∨R≡(P∨R)→(Q∨R)(P \rightarrow Q) \lor R \equiv (P \lor R) \rightarrow (Q \lor R)
(B) (P→Q)∨R≡(P→(Q∨R))(P \rightarrow Q) \lor R \equiv (P \rightarrow (Q \lor R))
(C) (P→Q)∨R≡((P∨R)→Q)(P \rightarrow Q) \lor R \equiv ((P \lor R) \rightarrow Q)
(D) (P→Q)∨R≡(¬P∨(Q∨R))(P \rightarrow Q) \lor R \equiv (\neg P \lor (Q \lor R))
Solution Approach:
We use Material Implication:
P→Q≡¬P∨QP \rightarrow Q \equiv \neg P \lor Q
Now, rewriting option (D):
(P→Q)∨R=(¬P∨Q)∨R(P \rightarrow Q) \lor R = (\neg P \lor Q) \lor R =¬P∨(Q∨R)= \neg P \lor (Q \lor R)
Correct Answer: Option (D)
Conclusion
Implication (P→QP \rightarrow Q) is False only when PP is True and QQ is False.
Logical Equivalences:
- P→Q≡¬P∨QP \rightarrow Q \equiv \neg P \lor Q
- P→Q≡¬Q→¬PP \rightarrow Q \equiv \neg Q \rightarrow \neg P (Contrapositive)
GATE questions often use truth tables and logical identities.
Want more solved GATE questions on Implication?
Implication Concept Discrete mathematics previous year paper – GATE 2025- The following resolution.
Discrete Mathematics for Computer Science
In Discrete Mathematics, particularly in the context of GATE examinations, understanding implication and its application in logical reasoning is crucial. Let’s delve into the concept of implication and explore how it’s utilized in previous GATE questions.
Understanding Implication (→)
An implication, denoted as p→qp \rightarrow q, reads as “if pp, then qq“. This logical statement is false only when pp is true and qq is false; in all other cases, it is considered true.
Truth Table:
pp | p→qp \rightarrow q | |
---|---|---|
T | T | T |
T | F | F |
F | T | T |
F | F | T |
This table illustrates that the only scenario where the implication fails is when the antecedent pp is true, but the consequent qq is false.GGC Discrete Math
Resolution Principle in Propositional Logic
The resolution principle is a fundamental rule of inference used in propositional logic and automated theorem proving. It allows for deriving a new clause by eliminating a pair of complementary literals.
Resolution Rule:
Given two clauses:
-
A∨BA \lor B
-
¬B∨C\neg B \lor C
We can infer:
-
A∨CA \lor CGeeksforGeeks
This rule is instrumental in simplifying complex logical expressions and proving the validity of arguments.
GATE Previous Year Question Example
Let’s consider a GATE-style question that involves implication and resolution:
Question:
Let a,b,c,da, b, c, d be propositions. Assume the following:
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a↔(b∨¬b)a \leftrightarrow (b \lor \neg b)
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b↔cb \leftrightarrow c
Determine the truth value of the formula:
-
(a∧b)→((a∧c)∨d)(a \land b) \rightarrow ((a \land c) \lor d)
Solution:
-
Analyze a↔(b∨¬b)a \leftrightarrow (b \lor \neg b):
-
The expression b∨¬bb \lor \neg b is a tautology (always true).
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Therefore, a↔Truea \leftrightarrow \text{True} implies aa is true.
-
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Given b↔cb \leftrightarrow c:
-
This means bb and cc have the same truth value.
-
-
Evaluate the antecedent a∧ba \land b:
-
Since aa is true, the truth value of a∧ba \land b depends on bb.
-
-
Evaluate the consequent (a∧c)∨d(a \land c) \lor d:
-
Given b↔cb \leftrightarrow c, cc has the same truth value as bb.
-
Therefore, a∧ca \land c has the same truth value as a∧ba \land b.
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Thus, (a∧c)∨d(a \land c) \lor d is true if either a∧ca \land c is true or dd is true.ExamSIDE
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-
Determine the truth value of the implication:
-
If a∧ba \land b is false, the implication is vacuously true.
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If a∧ba \land b is true, then a∧ca \land c is also true (since b↔cb \leftrightarrow c), making the consequent true.
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Therefore, in all cases, the implication is true.
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Answer: The formula is always true.
Additional Resources
For a more in-depth understanding, you may refer to the following resources:
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GATE Overflow – Mathematical Logic Questions: A comprehensive collection of previous GATE questions on mathematical logic. Gate Overflow
-
GeeksforGeeks – Propositional Logic Notes: Detailed notes on propositional and first-order logic, including implications and other logical connectives. GeeksforGeeks
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YouTube – Propositional Logic GATE PYQs: A video explanation of previous year GATE questions on propositional logic. YouTube
If you have any specific questions or need further clarification on any topic, feel free to ask!