*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.