Page 149
Location: Chapter 14, bonus exercise 1, proposition (d)
It is
(d) If Px(Pyz) sings on all days, so does P(Pxy)(Pyz).
It should disappear
[no sentence (d) to prove at all]
Short explanation
From the first three of Byrd's laws, proposition (d) does not follow. (The rest of the exercise proceeds without problems, though.)
In the terms of Problem 1 of the Chapter, the particular arrangement
x \notin S~~,~~~y \in S~~,~~~z \notin S
acts as a counterexample, making it impossible to derive the fact that P(Pxy)(Pyz) \in S from Px(Pyz)\in S using the first three laws of Problem 1 alone.
In terms of the usual Aristotelian logic (i.e. under the isomorphism presented in "Discussion" on page 145), the following truth table makes the counterexample explicit:
| x | y | z | y\to z | z\to y | (x\to y)\to(y\to z) | x\to(y\to z) | [x\to(y\to z)]\to[(x\to y)\to(y\to z)] |
|---|---|---|---|---|---|---|---|
| T | T | T | T | T | T | T | T |
| T | T | F | F | T | F | F | T |
| T | F | T | T | F | T | T | T |
| T | F | F | T | F | T | T | T |
| F | T | T | T | T | T | T | T |
| F | T | F | F | T | F | T | F |
| F | F | T | T | T | T | T | T |
| F | F | F | T | T | T | T | T |