# Logic Text Chapter 6 Solutions

## Chapter 6: Conditionality

### Basic

Question {6.1}

A & (AB) ⊨ □ B, since if w ⊩ □ A and wAB, then for every v, vA, and for every v, if vA then vB. Therefore, for every v, vB, and hence, w ⊩ □ B, as we wanted.

A & (AB) ⊭ □ 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 wAB (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, vA & □ B, so vA, and since v ⊩ □ B, at any world x, xB. 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 ⊨ ~ □ (~ AB), since if w ⊩ □ A &~ □ B, then there is some world v where v ⊩ ~ B, and vA. Therefore v ⊩ ~(~ AB), and hence, w ⊩ ~ □(~ AB), as desired.

Question {6.2}

w ⊩ ◇ A if and only if for some v, vA.

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, ◇ ( AB) ⊨ ◇ A ∨ ◇ B, since if w ⊩ ◇ ( AB), then there is a world v where vAB, and hence, either vA, or vB. 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 (pr) ∨ (qr) is false, since in world b, p is true but r is false (so pr is false everywhere), and in world c, q is true and r is false (so qr 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 pr 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 qr 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 pr and qr, and this cannot happen if (p & q) ⊃ r is true.