Answer
$f (s,xy) = f (f (s,x), y)$ for all states s ∈ S and all strings $x ∈ I ^∗$ and $y ∈ I^∗$.
Work Step by Step
Using a deterministic finite-state automaton M = $(S, I, f, s_0, F)$, using structural induction and the recursive
definition of the extended transition function f for proving that $f (s,xy) = f (f (s,x), y)$ for all states s ∈ S and all
strings $x ∈ I ^∗$ and $y ∈ I^∗$.
We use structural induction on the input string y.
-The basis step considers $y = λ,$
and for the inductive step we write $ y = wa,$ where $w ∈ I^*$
and $a ∈ I$ .
- For the basis step, we have xy = x, so we must show that $f (s,x) = f (f (s,x), λ).$
-But part (i) of the definition of the extended transition function
says that this is true.
- We then assume the inductive hypothesis that the equation holds for w and
-prove that
$f (s, xwa) = f (f (s, x), wa)$. By part (ii) of the definition,
-the left-hand side of this equation equals $f (f (s, xw), a).$
-By the inductive hypothesis, $f (s, xw) = f (f (s, x), w),$
so
- f (f (s, xw), a) = f (f (f (s, x), w), a).
-The righthand side of our desired equality is, by part (ii) of the
definition, also equal to f (f (f (s, x), w), a), as desired.
