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Consider computations while a terminal block of 's is removed from register . Until that block has disappeared, the available instruction and next state depend on the current state and on the visible , not on the untouched prefix .
Take a block longer than . Record the state after each successive has been removed. Two records have the same state , by the pigeonhole principle. If the corresponding removed tail lengths are , the same finite instruction sequences work in front of every prefix . Thus there are fixed times such that
for every word .
Solved by gpt-5.6-sol high.

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