VTU 4th Sem Maths | Logical Equivalence Proof | Module 1 – Mathematical Logic | BCS405A

Описание: In this video, we solve an important logical equivalence problem from Module 1 – Mathematical Logic 1 of the VTU 4th Semester Mathematics syllabus (Subject Code: BCS405A).

📌 Question:
Show that the compound proposition
[(p ↔ q) ∧ (q ↔ r) ∧ (r ↔ p)] ⇔ [(p → q) ∧ (q → r) ∧ (r → p)]
is logically equivalent,
where p, q, and r are primitive statements.

We break down both compound propositions and use truth tables to verify their logical equivalence. This problem strengthens your understanding of biconditional (↔), implication (→), and compound propositions — a key area in discrete mathematics and logic.

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In this video from Module 1 – Mathematical Logic 1 (Subject Code: BCS405A), we’ll solve a very interesting logical equivalence problem involving compound propositions.

We’re given two compound expressions:
[(p ↔ q) ∧ (q ↔ r) ∧ (r ↔ p)] and [(p → q) ∧ (q → r) ∧ (r → p)]
and we have to show that they are logically equivalent.

We’ll solve it step-by-step using a truth table, which is a very important concept for exams and understanding logical reasoning.