The proposition \[ \sim (p\nu \sim q)\nu \sim (p\nu q)\] is logically equivalent to
A. \[ \sim p\]
B. \[p\]
C. \[q\]
D. \[ \sim q\]

Hint: Here we will use De Morgan’s Law to solve disjunction in each bracket and then taking negation of each bracket we apply De Morgan’s Law to the final propositions. We will define each symbol and then solve for the logic of the proposition given in the statement.
* De Morgan’s Law gives us
1) \[ \sim (p \wedge q) \equiv \sim p\nu \sim q\]
2) \[ \sim (p\nu q) \equiv \sim p \wedge \sim q\]


Complete step-by-step answer:
We know that the symbols in the statement are the symbols for negation of a proposition and disjunction of a proposition. First we will define the terms used.
Let \[p\] and \[q\] be two propositions. Then the logical symbols are
Negation: The logical operation of making a proposition is negative is called negation. It is denoted by the symbol (\[ \sim \]).
Conjunction: The logical operation of joining two statements with the operator ‘and’ in between is called conjunction. It is denoted by the symbol (\[ \wedge \]).
Disjunction: The logical operation of joining two statements with the operator ‘or’ in between them is called a disjunction. It is denoted by the symbol (\[\nu \]).
Now, we will solve the proposition \[ \sim (p\nu \sim q)\nu \sim (p\nu q)\] step by step by considering the two parts separately.
First we will solve \[ \sim (p\nu \sim q)\].
We know from De Morgan’s Law that \[ \sim (p\nu q) = \sim p \wedge \sim q\]. Substituting the values of \[p = p, q = \sim q\] we get
\[ \Rightarrow \sim (p\nu \sim q) \equiv \sim (p) \wedge \sim ( \sim q)\]
Also, we know that negation of a negation implies the proposition itself, i.e. \[ \sim ( \sim q) \equiv q\]
\[ \Rightarrow \sim (p\nu \sim q) \equiv \sim p \wedge q\] … (1)
Now we solve the second part \[ \sim (p\nu q)\].
From De Morgan’s Law we know that \[ \sim (p\nu q) \equiv \sim p \wedge \sim q\]. Substituting the values of \[p = p, q = q\] we get,
\[ \Rightarrow \sim (p\nu q) \equiv \sim p \wedge \sim q\] … (2)
Now we use the solutions from equation (1) and (2) and substitute in the given propositions to form a simpler one.
\[ \Rightarrow \sim (p\nu \sim q)\nu \sim (p\nu q) \equiv ( \sim p \wedge q)\nu ( \sim p \wedge \sim q)\]
We know distributive property states that \[p \wedge (q\nu r) \equiv (p \wedge r)\nu (q \wedge r)\]
\[ \Rightarrow ( \sim p \wedge q)\nu ( \sim p \wedge \sim q) \equiv \sim p \wedge (q\nu \sim q)\]
\[ \Rightarrow ( \sim p \wedge q)\nu ( \sim p \wedge \sim q) \equiv \sim p\]

So, the correct answer is “Option A”.

Note: Students are likely to make mistakes while writing all the propositions as they write equal to the sign between the propositions which is wrong, we always write the equivalent sign.