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: \(q \land \sim p\).
Consider the following reaction occurring in the blast furnace. \[ {Fe}_3{O}_4(s) + 4{CO}(g) \rightarrow 3{Fe}(l) + 4{CO}_2(g) \] ‘x’ kg of iron is produced when \(2.32 \times 10^3\) kg \(Fe_3O_4\) and \(2.8 \times 10^2 \) kg CO are brought together in the furnace.
The value of ‘x’ is __________ (nearest integer).
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
X g of benzoic acid on reaction with aqueous \(NaHCO_3\) release \(CO_2\) that occupied 11.2 L volume at STP. X is ________ g.
Standard entropies of \(X_2\), \(Y_2\) and \(XY_5\) are 70, 50, and 110 J \(K^{-1}\) mol\(^{-1}\) respectively. The temperature in Kelvin at which the reaction \[ \frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5 \quad \Delta H = -35 \, {kJ mol}^{-1} \] will be at equilibrium is (nearest integer):
37.8 g \( N_2O_5 \) was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K: \[ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) \]
The total pressure at equilibrium was found to be 18.65 bar. Then, \( K_p \) is: Given: \[ R = 0.082 \, \text{bar L mol}^{-1} \, \text{K}^{-1} \]