First slide

Introduction to mathematical reasoning

Question

Statement $(p \to q) \leftrightarrow (~ q \to\sim p)$

Moderate

A

is contradiction
B

is tautology
C

is neither contradiction not tautology
D

none of these

Solution

$(~ q \to\sim p)$ is contrapositive of $(p \to q)$ .
Therefore, $(p \to q) \leftrightarrow (~ q \to\sim p)$ is tautology.

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