The statement `(~(phArr q))^^p` is equivalent to

Options

A. `p^^ q`
B. `qhArr p`
C. `p^^~q`
D. `~p^^q`

Answer

To solve the problem of determining the equivalence of the statement `(~(P ↔ Q)) ∧ P`, we will follow a systematic approach using truth tables. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Variables
Let’s define the variables:
- P: A proposition
- Q: Another proposition

### Step 2: Construct the Truth Table
We will construct a truth table that includes all possible truth values for P and Q.

| P | Q | P ↔ Q | ~(P ↔ Q) | ~(P ↔ Q) ∧ P |
|-------|-------|-------|----------|---------------|
| True | True | True | False | False |
| True | False | False | True | True |
| False | True | False | True | False |
| False | False | True | False | False |

### Step 3: Calculate P ↔ Q
The biconditional (↔) is true if both P and Q are either true or false. Thus:
- When P = True and Q = True, P ↔ Q = True
- When P = True and Q = False, P ↔ Q = False
- When P = False and Q = True, P ↔ Q = False
- When P = False and Q = False, P ↔ Q = True

### Step 4: Calculate ~(P ↔ Q)
Negating the results of P ↔ Q:
- For P = True, Q = True: ~(P ↔ Q) = False
- For P = True, Q = False: ~(P ↔ Q) = True
- For P = False, Q = True: ~(P ↔ Q) = True
- For P = False, Q = False: ~(P ↔ Q) = False

### Step 5: Calculate ~(P ↔ Q) ∧ P
Now we will perform the conjunction (∧) with P:
- For P = True, Q = True: False ∧ True = False
- For P = True, Q = False: True ∧ True = True
- For P = False, Q = True: True ∧ False = False
- For P = False, Q = False: False ∧ False = False

### Step 6: Analyze the Results
From the truth table, the final column for the expression `(~(P ↔ Q)) ∧ P` is:
- False, True, False, False

### Step 7: Compare with Given Options
Now we will compare the results with the provided options:
1. P ∧ Q: True, True, False, False (Not equivalent)
2. Q ↔ P: True, True, False, False (Not equivalent)
3. P ∧ ~Q: False, True, False, False (Equivalent)
4. ~P ∧ Q: False, False, True, False (Not equivalent)

### Conclusion
The expression `(~(P ↔ Q)) ∧ P` is equivalent to option 3: P ∧ ~Q.