Time shift -- two interpretations
Let 

$$h_1(t) = x(t-1) ~~ \text{ for all }~ t, \tag{1}$$
that is, $h_1(t)$ is just
$x(t)$ delayed by one unit. Let $h(t)$ be the result of "flipping" $h_1(t)$, that is, 
$$h(t) = h_1(-t) ~~ \text{ for all }~ t, \tag{2}.$$
But, $(1)$ says that $h_1(-t)$ is the same as $x(-t-1)$ for all $t$,
and so we conclude that
$h(t)$, the result of delaying $x(t)$ by one time unit and then flipping the
result gives us 
$$h(t) = x(-t-1)~~ \text{ for all }~ t, \tag{3}$$
which of course is not the desired $x(-t+1)$.

So let's try O&amp;W's prescription. Let $g_1(t) = x(t+1)$ for all $t$
be the result of
advancing $x(t)$ by one time unit. Now flip $g_1(t)$ to get
$$g(t) = g_1(-t) = x(-t+1)$$
which is what is desired, no?