问如何求两个有限自动机的交

EN

E is the input alphabet, a non-empty finite set of symbols
Q is the set of states, non-empty and finite
q0 is the start state, an element of Q
A is the set of accepting or final states, a subset of Q
f is the transition function, taking pairs from Q x E to Q

Take E''' = E'' = E'
Take Q''' = Q' x Q''
Take q0''' = (q0', q0'')
Take A''' = (x, y) where x in A' and y in A'' (for union, x in A' or y in A''; for difference, x in A' but not y in A'').
Take f'''((x, y), e) = (f'(x, e), f''(y, e)).

E' = {a}
Q' = {s0, s1}
q0' = s0
A' = {s0}
f'(s0, a) = s1, f'(s1, a) = s0

E'' = {a}
Q'' = {t0, t1, t2}
q0'' = t0
A'' = {t0}
f''(t0, a) = t1, f''(t1, a) = t2, f''(t2, a) = t0

E''' = {a}
Q''' = {(s0, t0), (s0, t1), (s0, t2), (s1, t0), (s1, t1), (s1, t2)}
q0''' = (s0, t0)
A''' = {(s0, t0)} for intersection, {(s0, t0), (s0, t1), (s0, t2), (s1, t0)} for union, {(s0, t1), (s0, t2)} for difference.
f'''((s0, t0), a) = (s1, t1), f'''((s1, t1), a) = (s0, t2), f'''((s0, t2), a) = (s1, t0), f'''((s1, t0), a) = (s0, t1), f'''((s0, t1), a) = (s1, t2), f'''((s1, t2), a) = (s0, t0).