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: \(q \land \sim p\).
Let \(P(S)\) denote the power set of \(S = \{1, 2, 3, \ldots, 10\}\). Define the relations \(R_1\) and \(R_2\) on \(P(S)\) as \(A R_1 B\) if \[(A \cap B^c) \cup (B \cap A^c) = ,\]and \(A R_2 B\) if\[A \cup B^c = B \cup A^c,\]for all \(A, B \in P(S)\). Then:
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 : \(A_2 \rightleftharpoons 2A\)