Does a reusable barrier solution have a dead end?

I read The Little Book of Semaphores, and on page 41 there is a solution to the reusable barrier problem. The problem is why it will not create a deadlock.

1 # rendezvous
2
3 mutex.wait()
4     count += 1
5     if count == n:
6         turnstile2.wait() # lock the second
7         turnstile.signal() # unlock the first
8 mutex.signal()
9
10 turnstile.wait() # first turnstile
11 turnstile.signal()
12
13 # critical point
14
15 mutex.wait()
16     count -= 1
17     if count == 0:
18         turnstile.wait() # lock the first
19         turnstile2.signal() # unlock the second
20 mutex.signal()
21
22 turnstile2.wait() # second turnstile
23 turnstile2.signal()

In this solution, between lines 15 and 20, is it not a bad habit to call wait () on the semaphore (on line 18) while holding the mutex that causes the deadlock? Please explain. Thank.