Exercise 7.A.16

Proof. Suppose $u\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$. Then, by 7.20,

which implies that $u\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{\text{∗}}$. Hence $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{\text{∗}}$ (because ${(T^{\text{∗}})}^{\text{∗}}=T$ and the same argument can be repeated). Now we have

where the first and last equality follow from items (d) and (b) of 7.7. □