Step 1: The original expression We are given the statement \( \sim [p \vee (\sim (p \land q))] \). We need to simplify this expression and find the equivalent logical statement.
Step 2: Apply De Morgan's law First, apply De Morgan’s law to the negation of the disjunction \( \sim [p \vee (\sim (p \land q))] \). De Morgan’s law states that \( \sim (A \vee B) = \sim A \land \sim B \), so we get: \[ \sim p \land \sim (\sim (p \land q)) \]
Step 3: Simplify the inner negation Now, simplify the double negation \( \sim (\sim (p \land q)) \), which cancels out the two negations, giving us: \[ \sim p \land (p \land q) \]
Step 4: Conclusion Thus, the expression simplifies to: \[ (p \land q) \land (\sim p) \] which is the correct equivalent form of the original expression.
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: