Which of the following is a tautology ?

Options

A. `(~p wedge q) vv (p vv ~p)`
B. `(p to q) vv q`
C. `(p harr q) vv q`
D. `p vv (p harr q)`

Answer

To determine which of the given options is a tautology, we will analyze each option step by step. A tautology is a statement that is always true, regardless of the truth values of its components.

### Given Options:
1. **Option A**: ¬(p ∧ q) ∨ (p ∨ ¬p)
2. **Option B**: (p → q) ∨ q
3. **Option C**: p → (q ∨ q)
4. **Option D**: p ∨ (p → q)

### Step-by-Step Solution:

#### Step 1: Analyze Option A
**Option A**: ¬(p ∧ q) ∨ (p ∨ ¬p)

1. **Evaluate p ∨ ¬p**: This is a tautology because either p is true or p is false.
- Truth Table:
- p = True → ¬p = False → p ∨ ¬p = True
- p = False → ¬p = True → p ∨ ¬p = True
- Conclusion: p ∨ ¬p = True

2. **Evaluate ¬(p ∧ q)**: This expression is true unless both p and q are true.
- Truth Table:
- p = True, q = True → ¬(True ∧ True) = False
- p = True, q = False → ¬(True ∧ False) = True
- p = False, q = True → ¬(False ∧ True) = True
- p = False, q = False → ¬(False ∧ False) = True

3. **Combine**: Since p ∨ ¬p is always true, ¬(p ∧ q) ∨ (p ∨ ¬p) will also always be true regardless of the values of p and q.
- Conclusion: Option A is a tautology.

#### Step 2: Analyze Option B
**Option B**: (p → q) ∨ q

1. **Evaluate p → q**: This is false only when p is true and q is false.
- Truth Table:
- p = True, q = True → p → q = True
- p = True, q = False → p → q = False
- p = False, q = True → p → q = True
- p = False, q = False → p → q = True

2. **Combine**: The expression (p → q) ∨ q is not always true since it can be false when p is true and q is false.
- Conclusion: Option B is not a tautology.

#### Step 3: Analyze Option C
**Option C**: p → (q ∨ q)

1. **Evaluate q ∨ q**: This is simply q, so we have p → q.
2. **Evaluate p → q**: This is not always true, as discussed in Option B.
- Conclusion: Option C is not a tautology.

#### Step 4: Analyze Option D
**Option D**: p ∨ (p → q)

1. **Evaluate p → q**: As before, this is false only when p is true and q is false.
2. **Combine**: The expression p ∨ (p → q) can be false if p is false and q is false.
- Conclusion: Option D is not a tautology.

### Final Conclusion
From our analysis, only **Option A** is a tautology.