The statement `~(p ^^ q) vv q`

Options

A. is a tautology
B. is equivalent to `(p ^^ q) vv ~q`
C. is equivalent to `p vv q`
D. is a contradiction

Answer

To solve the statement `~(p ∧ q) ∨ q`, we will use a truth table to evaluate the expression step by step.

### Step 1: Set Up the Truth Table
We will create a truth table with the columns for `p`, `q`, `p ∧ q`, `~(p ∧ q)`, and `~(p ∧ q) ∨ q`.

| p | q | p ∧ q | ~(p ∧ q) | ~(p ∧ q) ∨ q |
|-------|-------|-------|----------|---------------|
| T | T | | | |
| T | F | | | |
| F | T | | | |
| F | F | | | |

### Step 2: Fill in the Values for `p ∧ q`
The conjunction `p ∧ q` is true only when both `p` and `q` are true.

| p | q | p ∧ q | ~(p ∧ q) | ~(p ∧ q) ∨ q |
|-------|-------|-------|----------|---------------|
| T | T | T | | |
| T | F | F | | |
| F | T | F | | |
| F | F | F | | |

### Step 3: Fill in the Values for `~(p ∧ q)`
Negation `~(p ∧ q)` is true when `p ∧ q` is false.

| p | q | p ∧ q | ~(p ∧ q) | ~(p ∧ q) ∨ q |
|-------|-------|-------|----------|---------------|
| T | T | T | F | |
| T | F | F | T | |
| F | T | F | T | |
| F | F | F | T | |

### Step 4: Fill in the Values for `~(p ∧ q) ∨ q`
The disjunction `~(p ∧ q) ∨ q` is true if either `~(p ∧ q)` is true or `q` is true.

| p | q | p ∧ q | ~(p ∧ q) | ~(p ∧ q) ∨ q |
|-------|-------|-------|----------|---------------|
| T | T | T | F | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | T | T |

### Final Truth Table
After filling in all the values, we have:

| p | q | p ∧ q | ~(p ∧ q) | ~(p ∧ q) ∨ q |
|-------|-------|-------|----------|---------------|
| T | T | T | F | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | T | T |

### Conclusion
The last column `~(p ∧ q) ∨ q` is true for all combinations of `p` and `q`, which means the statement is a tautology.