Question
Using truth table prove that : $p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
A | B | ||||||
P | q | $p \leftrightarrow q$ | $p \wedge q$ | $\sim p$ | $\sim q$ | $\begin{gathered}\sim p \wedge \\ \sim q\end{gathered}$ | $A \vee B$ |
T T F F
| T F T F | T F F T | T F F F | F F T T
| F T F T | F F F T | T F F T |
By column number 3 and 8
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.