7.Combining Turing Machines (1)§
Lemma: If
\((q_1,\ w_1\underline{a_1}u_1) \vdash_M^* (q_2,\ ww_2\underline{a_2}u_2)\)
for string $w$ and
\((q_2,\ w_2\underline{a_2}u_2) \vdash^*_M (q_3,\ w_3\underline{a_3}u_3)\),
then
\((q_1,\ w_1\underline{a_1}u_1) \vdash^*_M (q_3,\ ww_3\underline{a_3}u_3)\).
Insight: Since \((q_2,\ w_2\underline{a_2}u_2) \vdash^*_M (q_3,\ w_3\underline{a_3}u_3)\),
this computation must take place without moving the head left of \(w_2\).
The machine cannot “sense” the left end of the tape.
And if it had moved left, it would have hung.