Using truth table , prove that :`(ptoq)to[(~ptoq)toq]` is a tautology.

Answer

To prove that the expression \((p \to q) \to [(\neg p \to q) \to q]\) is a tautology using a truth table, we will follow these steps:

### Step 1: Set Up the Truth Table
We will create a truth table with columns for \(p\), \(q\), \(\neg p\), \(p \to q\), \(\neg p \to q\), \((\neg p \to q) \to q\), and finally \((p \to q) \to [(\neg p \to q) \to q]\).

| \(p\) | \(q\) | \(\neg p\) | \(p \to q\) | \(\neg p \to q\) | \((\neg p \to q) \to q\) | \((p \to q) \to [(\neg p \to q) \to q]\) |
|-------|-------|------------|--------------|-------------------|---------------------------|------------------------------------------|
| T | T | F | T | T | T | T |
| T | F | F | F | T | F | T |
| F | T | T | T | T | T | T |
| F | F | T | T | F | F | T |

### Step 2: Fill in the Values
1. **Column for \(p\) and \(q\)**: List all combinations of truth values for \(p\) and \(q\) (True (T) and False (F)).
2. **Column for \(\neg p\)**: This is the negation of \(p\). It will be True when \(p\) is False and vice versa.
3. **Column for \(p \to q\)**: This is True unless \(p\) is True and \(q\) is False.
4. **Column for \(\neg p \to q\)**: This is True unless \(\neg p\) is True (which means \(p\) is False) and \(q\) is False.
5. **Column for \((\neg p \to q) \to q\)**: This is True unless \((\neg p \to q)\) is True and \(q\) is False.
6. **Final Column for \((p \to q) \to [(\neg p \to q) \to q]\)**: This is True unless \(p \to q\) is True and \((\neg p \to q) \to q\) is False.

### Step 3: Analyze the Final Column
In the final column, we observe that all entries are True (T). This indicates that the expression \((p \to q) \to [(\neg p \to q) \to q]\) is always true regardless of the truth values of \(p\) and \(q\).

### Conclusion
Since the final column contains only True values, we conclude that the expression is a tautology.