Consider three statements p : person 'A' passed in mathematics exam q : Person 'A' passed in physics exam r : Person 'A' passed in chemistry exam, Then the statement `~((~(p =>q)=>r)` is equivalent to

Options

A. Person A passed only in mathematics & physics & chemistry
B. Person A failed only in mathematics & physics & chemistry
C. Person A passed in all the three subjects mathematics & physics & chemistry
D. Person A passed in chemistry but failed in mathematics & physics.

Answer

To solve the problem, we need to simplify the statement `~((~(p => q) => r)` step by step.

### Step-by-Step Solution:

1. **Understand the Implication**:
The implication \( p \Rightarrow q \) can be rewritten using logical equivalences:
\[ p \Rightarrow q \equiv \neg p \lor q \]
Therefore, \( \neg(p \Rightarrow q) \) can be rewritten as:
\[ \neg(\neg p \lor q) \equiv p \land \neg q \]

2. **Substituting into the Original Statement**:
Now, we substitute this back into the original statement:
\[ \neg((p \Rightarrow q) \Rightarrow r) \equiv \neg((\neg p \lor q) \Rightarrow r) \]
We can rewrite \( (p \Rightarrow q) \Rightarrow r \) as:
\[ \neg(\neg p \lor q) \lor r \equiv (p \land \neg q) \lor r \]

3. **Negating the Expression**:
Now we need to negate the entire expression:
\[ \neg((p \land \neg q) \lor r) \]
Using De Morgan's laws, this becomes:
\[ \neg(p \land \neg q) \land \neg r \]

4. **Applying De Morgan's Laws**:
We can further simplify \( \neg(p \land \neg q) \):
\[ \neg(p \land \neg q) \equiv \neg p \lor q \]
Therefore, the expression becomes:
\[ (\neg p \lor q) \land \neg r \]

5. **Final Expression**:
So, the final equivalent expression for the original statement `~((~(p => q) => r)` is:
\[ (\neg p \lor q) \land \neg r \]

### Conclusion:
The statement `~((~(p => q) => r)` is equivalent to \( (\neg p \lor q) \land \neg r \).