To solve the logical statement `~(p ∨ q) → (p → q)` and find its equivalent expression, we will follow these steps:
### Step 1: Understand the Components
We have two logical statements, p and q. The expression consists of:
- Negation of (p or q): `~(p ∨ q)`
- Implication: `p → q`
### Step 2: Construct the Truth Table
We will create a truth table to evaluate the expression.
| p | q | p ∨ q | ~(p ∨ q) | p → q |
|-------|-------|-------|----------|-------|
| T | T | T | F | T |
| T | F | T | F | F |
| F | T | T | F | T |
| F | F | F | T | T |
### Step 3: Evaluate `~(p ∨ q)`
From the truth table, we can see:
- When both p and q are true (T), `p ∨ q` is true (T), so `~(p ∨ q)` is false (F).
- When p is true and q is false (T, F), `~(p ∨ q)` is still false (F).
- When p is false and q is true (F, T), `~(p ∨ q)` is false (F).
- When both p and q are false (F), `p ∨ q` is false (F), so `~(p ∨ q)` is true (T).
Thus, the column for `~(p ∨ q)` is:
- F, F, F, T
### Step 4: Evaluate `p → q`
From the truth table, we can see:
- When both p and q are true (T), `p → q` is true (T).
- When p is true and q is false (T, F), `p → q` is false (F).
- When p is false and q is true (F, T), `p → q` is true (T).
- When both p and q are false (F), `p → q` is true (T).
Thus, the column for `p → q` is:
- T, F, T, T
### Step 5: Evaluate the Full Expression `~(p ∨ q) → (p → q)`
Now we need to evaluate the implication:
- The implication `A → B` is false only when A is true and B is false.
| ~(p ∨ q) | p → q | ~(p ∨ q) → (p → q) |
|----------|-------|---------------------|
| F | T | T |
| F | F | T |
| F | T | T |
| T | T | T |
Thus, the final column for `~(p ∨ q) → (p → q)` is:
- T, T, T, T
### Conclusion
The expression `~(p ∨ q) → (p → q)` is always true, which means it is a tautology.
### Equivalent Expression
The equivalent expression for `~(p ∨ q) → (p → q)` is `p → (p ∨ q)`.
