Given: The statement is \(((\sim p) \land q) \implies r\). The converse of a conditional statement \((A \implies B)\) is defined as \((B \implies A)\).
Step 1: Identify A and B.
Here:
\(A = ((\sim p) \land q), B = r.\)
Thus, the converse is:
\(r \implies ((\sim p) \land q).\)
Step 2: Express the negation of the implication.
The negation of \(((\sim p) \land q) \implies r\) is:
\[\sim(((\sim p) \land q) \implies r) = (\sim r) \implies ((\sim p) \land q).\]
Step 3: Derive the logical equivalence.
The converse can also be written equivalently as:
\[r \implies ((\sim p) \land q) \implies (\sim((\sim p) \land q)) \implies (\sim r).\]
Simplifying further:
\[(p \lor (\sim q)) \implies (\sim r).\]
Final Answer: The converse is \((p \lor (\sim q)) \implies (\sim r).\)
Equivalent statement to (p\(\to\)q) \(\vee\) (r\(\to\)q) will be
The number of values of $r \in\{p, q, \sim p, \sim q\}$ for which $((p \wedge q) \Rightarrow(r \vee q)) \wedge((p \wedge r) \Rightarrow q)$ is a tautology, is :
Among the statements :
\((S1)\) \((( p \vee q ) \Rightarrow r ) \Leftrightarrow( p \Rightarrow r )\)
\((S2)\)\((( p \vee q ) \Rightarrow r ) \Leftrightarrow(( p \Rightarrow r ) \vee( q \Rightarrow r ))\)
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 : \(A_2 \rightleftharpoons 2A\)