吳聰敏老師的討論區

[tmwu]

我的作法可能是

\usepackage{amsmath}
...

\begin{align}
f_{1}(W1_{n}W2_{n}W3_{n}) &=W1_{n}' W2_{n}' W3_{n}'
\stackrel{S_{1}}{\longrightarrow} f_{1}(W1_{n+1}W2_{n+1}W3_{n+1}) \notag \\
& = (W1_{n+1}' W2_{n+1}' W3_{n+1}') \\
f_{2}(W1_{n}W2_{n}W3_{n}) &= W1_{n}'' W2_{n}'' W3_{n}''
\stackrel{S_{2}}{\longrightarrow} f_{1} (W1_{n+1} W2_{n+1} W3_{n+1}) \notag\\
&= (W1_{n+1}'' W2_{n+1}'' W3_{n+1}'') \\
f_{3}(W1_{n}W2_{n}W3_{n}) &= W1_{n}''' W2_{n}''' W3_{n}'''
\stackrel{S_{3}} {\longrightarrow} f_{1} (W1_{n+1} W2_{n+1} W3_{n+1}) \notag\\
&= (W1_{n+1}''' W2_{n+1}''' W3_{n+1}''')
\end{align}

[tmwu]

How about:

Define $\phi(i,j) \equiv f_i(W1_{j}W2_{j}W3_{j}$,
\begin{align}
\phi(1,n) =W1_{n}' W2_{n}' W3_{n}' & \stackrel{S_{1}}{\longrightarrow} \phi(1,n+1)
= (W1_{n+1}' W2_{n+1}' W3_{n+1}') \\
\phi(2,n) =W1_{n}' W2_{n}' W3_{n}' & \stackrel{S_{2}}{\longrightarrow} \phi(1,n+1)
= (W1_{n+1}'' W2_{n+1}'' W3_{n+1}'') \\
\phi(3,n) =W1_{n}' W2_{n}' W3_{n}' & \stackrel{S_{3}}{\longrightarrow} \phi(1,n+1)
= (W1_{n+1}''' W2_{n+1}''' W3_{n+1}''')
\end{align}