If p and q are logical statements, then `(p^^q)rarr(prarrq)` is equivalent to

Options

A. `p^^q`
B. `prarr(pvvq)`
C. `pvv q`
D. `(pvvq)harr (p^^q)`

Answer

To solve the logical statement `(p ∧ q) → (p → q)`, we will use logical equivalences step by step.

### Step 1: Rewrite the Implication
The implication `A → B` can be rewritten using the equivalence `A → B ≡ ¬A ∨ B`. Thus, we can rewrite our expression as:
\[ (p ∧ q) → (p → q) \equiv ¬(p ∧ q) ∨ (p → q) \]

**Hint:** Remember that an implication can be rewritten as a disjunction of the negation of the antecedent and the consequent.

### Step 2: Rewrite the Inner Implication
Now, we need to rewrite the inner implication `p → q` using the same equivalence:
\[ p → q ≡ ¬p ∨ q \]
So, substituting this back into our expression, we get:
\[ ¬(p ∧ q) ∨ (¬p ∨ q) \]

**Hint:** Use the same equivalence for implications to simplify the inner statement.

### Step 3: Apply De Morgan's Law
Next, we apply De Morgan's Law to `¬(p ∧ q)`, which states that `¬(A ∧ B) ≡ ¬A ∨ ¬B`. Thus:
\[ ¬(p ∧ q) ≡ ¬p ∨ ¬q \]
Substituting this back into our expression gives:
\[ (¬p ∨ ¬q) ∨ (¬p ∨ q) \]

**Hint:** De Morgan's Law allows you to transform negations of conjunctions into disjunctions of negations.

### Step 4: Simplify the Expression
Now, we can simplify the expression using the associative and commutative properties of disjunction:
\[ (¬p ∨ ¬q ∨ ¬p ∨ q) \equiv (¬p ∨ ¬q ∨ q) \]
Since `¬q ∨ q` is always true (a tautology), we can simplify further:
\[ ¬p ∨ T \equiv T \]

**Hint:** Remember that a disjunction with a tautology (always true) results in a tautology.

### Conclusion
Thus, the original expression `(p ∧ q) → (p → q)` is equivalent to `T` (true).

### Final Answer
The expression `(p ∧ q) → (p → q)` is equivalent to `True`.