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.
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)