If p,q and r are simple propositions with truth values T,F and T , respectively, then the truth value of `(~pvvq) ^^~r to p` is

Options

A. True
B. False
C. true if r is false
D. true if q is true

Explanation

To solve the problem, we need to evaluate the expression `(~p ∨ q) ∧ ~r → p` using the truth values of the propositions p, q, and r.

### Step-by-step Solution:

1. **Identify the truth values**:
- p = True (T)
- q = False (F)
- r = True (T)

2. **Find the negations**:
- ~p = Not True = False (F)
- ~q = Not False = True (T)
- ~r = Not True = False (F)

3. **Evaluate the expression inside the parentheses**:
- Compute `~p ∨ q`:
- ~p = F
- q = F
- Therefore, `~p ∨ q` = F ∨ F = F (since both are false, the disjunction is false).

4. **Evaluate the intersection with ~r**:
- Now we need to compute `(F) ∧ (~r)`:
- ~r = F
- Therefore, `F ∧ F` = F (since both are false, the conjunction is false).

5. **Evaluate the implication**:
- Now we need to evaluate `F → p`:
- p = T
- The implication `F → T` is True (an implication is only false when the first part is true and the second part is false).

### Final Result:
The truth value of the expression `(~p ∨ q) ∧ ~r → p` is **True (T)**.