Question {6.1}
□ A & (A → B) ⊨ □ B, since if w ⊩ □ A and w ⊩ A → B, then for every v, v ⊩ A, and for every v, if v ⊩ A then v ⊩ B. Therefore, for every v, v ⊩ B, and hence, w ⊩ □ B, as we wanted.
□ A & (A ⊃ B) ⊭ □ B, for we can have a model with two worlds w and v where A is true at both w and v, B is true at w but not at v. Therefore, w ⊩ □ A, and w ⊩ A ⊃ B (since A and B are both true at w) but w does not give us □ B, as B is not true at v.
~ □ A ⊨ □ ~ □ A, since if w ⊩ ~ □ A, then there is some world v where v ⊩ ~ A. Therefore in any world x, x ⊩ ~ □ A, so w ⊩ □ ~ □ A, as we wanted.
□( A & □ B) ⊨ □( A & B), since if w ⊨ □(A & □ B), it means that at any world v, v ⊩ A & □ B, so v ⊩ A, and since v ⊩ □ B, at any world x, x ⊩ B. But if B is true at any world, and A is true at any world, A & B is true at all worlds too, so w ⊩ □ ( A & B), as desired.
□ A & ~ □ B ⊨ ~ □ (~ A ∨ B), since if w ⊩ □ A &~ □ B, then there is some world v where v ⊩ ~ B, and v ⊩ A. Therefore v ⊩ ~(~ A ∨ B), and hence, w ⊩ ~ □(~ A ∨ B), as desired.
Question {6.2}
w ⊩ ◇ A if and only if for some v, v ⊩ A.
◇ A & ◇ B ⊭ ◇( A & B), since we have a model with two worlds, w and v. At w, A is true but B is false, while at v, B is true and A false. Therefore ◇ A is true at both worlds, as is ◇ B, but ◇ ( A & B) is not true, as there is no world where A and B are both true.
On the other hand, ◇ ( A ∨ B) ⊨ ◇ A ∨ ◇ B, since if w ⊩ ◇ ( A ∨ B), then there is a world v where v ⊩ A ∨ B, and hence, either v ⊩ A, or v ⊩ B. In the first case, w ⊩ ◇ A (and hence w ⊩ ◇ A ∨ ◇ B) and in the second case w ⊩ ◇ B (and hence w ⊩ ◇ A ∨ ◇ B). In either case, w ⊨ ◇ A ∨ ◇ B, as desired.
Questions {6.3}
Have a simple little situation in which there are four worlds, a, b, c and d. In world a, p and q are both true, in world b, p is true and q false, in c, p is false and q true, and in world d, p and q are both false. r is true only in world a. Therefore the premise (p & q) → r is true in every world. The truth of p & q means you’re in world a, and in that world, r is true. The conclusion (p → r) ∨ (q → r) is false, since in world b, p is true but r is false (so p → r is false everywhere), and in world c, q is true and r is false (so q → r is false everywhere). Therefore, we can have the premise true and the conclusion false. This model has four worlds, one for each possible state of the electrical circuitry. ( p & q) → r is true because in each state where p & q is true, so is r. But p → r fails because in some state, p is true but r false (namely, when switch one is flipped, but switch two is not — in world b) and q → r fails because in some other state, q is true but r is false (in world c). If we used the material conditional, you need the same state to fail both p ⊃ r and q ⊃ r, and this cannot happen if (p & q) ⊃ r is true.