Let p.q and r be three statements, then `(~prarrq)rarr r` is equivalent to

Options

A. `(~pvvr)^^(qvvr)`
B. `(prarr r)^^(q rarr r)`
C. `(~p ^^ r)vv(q vv r)`
D. `(prarr q)rarr r`

Answer

To solve the problem `(~p → q) → r`, we will break it down step by step using logical equivalences.

### Step 1: Rewrite the Implication
The expression `(~p → q)` can be rewritten using the implication equivalence:
\[ A \to B \equiv \neg A \lor B \]
So, we have:
\[ ~p \to q \equiv \neg(~p) \lor q \equiv p \lor q \]
Thus, we can rewrite the original expression:
\[ (~p \to q) \to r \equiv (p \lor q) \to r \]

### Step 2: Rewrite the Outer Implication
Now, we apply the implication equivalence again to the expression `(p ∨ q) → r`:
\[ (p \lor q) \to r \equiv \neg(p \lor q) \lor r \]

### Step 3: Apply De Morgan's Law
Using De Morgan's Law, we can simplify `¬(p ∨ q)`:
\[ \neg(p \lor q) \equiv \neg p \land \neg q \]
Thus, we can rewrite the expression:
\[ \neg(p \lor q) \lor r \equiv (\neg p \land \neg q) \lor r \]

### Step 4: Final Expression
The final expression we have is:
\[ (\neg p \land \neg q) \lor r \]

### Step 5: Identify the Equivalent Expression
Now, we can express this in terms of implications:
\[ \neg p \land \neg q \lor r \equiv (p \to r) \land (q \to r) \]
This means that the original expression `(~p → q) → r` is equivalent to:
\[ (p \to r) \land (q \to r) \]

### Conclusion
Thus, the expression `(~p → q) → r` is equivalent to:
\[ (p \to r) \land (q \to r) \]