Negation of (p⇒q)⇒(q⇒p) is
Step 1: Understand the implication.
The given expression is \((p \to q) \to (q \to p)\). We want to find the negation of this expression.
Step 2: Rewrite the expression using logical equivalences.
Recall that an implication \(A \to B\) is logically equivalent to \(\sim A \lor B\). So, we can rewrite the expression as: \[ (p \to q) \to (q \to p) \equiv \sim(p \to q) \lor (q \to p). \] Now, apply the equivalence \(p \to q \equiv \sim p \lor q\) and \(q \to p \equiv \sim q \lor p\) to obtain: \[ (\sim p \lor q) \to (\sim q \lor p). \] Step 3: Apply negation to the entire expression.
Next, we apply negation to the entire expression. The negation of an implication \(\sim (A \to B)\) is \(A \land \sim B\). So, we have: \[ \sim \left( \sim(p \to q) \lor (q \to p) \right). \] Using De Morgan’s law, this becomes: \[ \sim(\sim p \lor q) \land \sim(\sim q \lor p). \] Step 4: Simplify the expression.
Simplifying the negations inside: \[ \sim(\sim p \lor q) = p \land \sim q, \quad \sim(\sim q \lor p) = q \land \sim p. \] Thus, the negation becomes: \[ (p \land \sim q) \land (q \land \sim p). \] This simplifies to: \[ q \land \sim p. \] Final Answer: \(q \land \sim p\).
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]