# For multi-tape Turing machines, can we assume that each tape can be in its own state independently of the other tapes?

When we consider multi-tape Turing machines, we can assume that each tape is accessed with a separate head and each head can move independently of the other heads. But can we assume that each tape can be in its own state independently of the other tapes (for example, tape 1 in state A, tape 2 in state F, tape 3 in state X simultaneously)?

Then, for example, the table of instructions for some three-tape two-symbol ("0" and "1") machine can look like this:

    | 000         | 001         | ...
----------------------------------
AAA | 011/BXZ/LRL | 010/XYB/RRR | ...
----------------------------------
AAB | 001/TSH/LLR | 110/AET/LLL | ...
----------------------------------
...


where the first cell of the table can be interpreted as follows: if the first tape is in state A, the first head is over the cell with symbol 0, the second tape is in state A, the second head is over the cell with symbol 0, the third tape is in state A, the third head is over the cell with symbol 0, then the first head writes "0" instead of "0", then the first tape goes into state B, then the first head moves to the left, then the second head writes "1" instead of "0", then the second tape goes into state X, then the second head moves to the right, then the third head writes "1" instead of "0", then the third tape goes into state Z, then the third head moves to the left.

For your example, you can encode state "AAA" into state 1, state "AAB" into state 2, and so on (so there are in total $26^3$ states). Now it becomes a normal multi-tape Turing machine.