p_∨q, or p⊻q means that p and can not be true simultaneously or false simultaneously. So, p⊻q ⇔ (p ∧ ∼q) ∨ (∼p ∧ q). Note that ∼ is also written ¬.
Let’s verify this in a table. Under p and q, read the four events. Under each logical operator, read its truth-value, for example, if p is T and q is F, under p∧q, you will read F.
| p⊻q | ⇔ | (p | ∧ | ∼ | q) | ∨ | (∼ | p | ∧ | q) |
|---|---|---|---|---|---|---|---|---|---|---|
| F | T | T | F | F | T | F | F | T | F | T |
| T | T | T | T | T | F | T | F | T | F | F |
| T | T | F | F | F | T | T | T | F | T | T |
| F | T | F | F | T | F | F | T | F | F | F |
QED.
Therefore, p⇔q ⇔ ∼(p ⊻ q) ⇔ ∼((p ∧ ∼q) ∨ (∼p ∧ q)) ⇔ (∼p ∨ q) ∧ (p ∨ ∼q) because of de Morgan’s law.
The solution in the book is a bit different, try to see why we are both right.