Question

Prove the following results; using truth table:
p↔q≡(p→q)∧(q→p)p↔q≡(p→q)∧(q→p)

plz answer fast​

Answer 1

$\textbf{To prove:}$

$\bf\,p{\iff}q\;\equiv\;(p{\implies}q)\wedge(q{\implies}p)$

$\textbf{Solution:}$

$\text{First we form the truth tables for both the statements}$

$\text{After that we compare the last columns whether they are identitical}$

$\begin{array}{|c|c|c|}\cline{1-3}\bf{p}&\bf{q}&\bf{p{\iff}q}\\\cline{1-3}T&T&T\\T&F&F\\F&T&F\\F&F&T\\\cline{1-3}\end{arrary}$

$\begin{array}{|c|c|c|c|c|}\cline{1-5}\bf{p}&\bf{q}&\bf{p{\implies}q}&\bf{q{\implies}p}&\bf(p{\implies}q)\wedge(q{\implies}p)\\\cline{1-5}T&T&T&T&T\\T&F&F&T&F\\F&T&T&F&F\\F&F&T&T&T\\\cline{1-5}\end{array}$

$\text{Comparing last columns of both the tables, they are identical}$

$\therefore\bf\,p{\iff}q\;\equiv\;(p{\implies}q)\wedge(q{\implies}p)$

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P ↔ q ≡ (p ∧ q) ∨ (∼ p ∧ ∼ q)

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