Originally Posted by Klaas van Aarsen
Perhaps the context of the problem defines its Ordinary Turing Machine such that every state - that is not a final state - must have a state transition for all inputs...
After all, we already now that what the problem considers an 'ordinary Turing Machine' is not the definition that wiki gives.

Or perhaps those steps are indeed redundant...
We have the following definition for the transformation:

Do we have those steps maybe because of the last part of that definition?

2.

3. Originally Posted by mathmari
We have the following definition for the transformation:

Do we have those steps maybe because of the last part of that definition?
That transformation converts the Modified Turing Machine into and Ordinary Turing Machine.

First off, it shows that we will indeed have states with only 1 transition in the Ordinary Turing Machine.
That is, the state transition from $s_1'$ to $s$ in the 2nd rule is only defined for the symbol $x'$ and not for any other symbol.

The last part with $\delta(s,x)=\varnothing$ shows that if there are no state transitions at all in state $s$ of the Modified Turing Machine, that we transform it into a state that halts the Ordinary Turing Machine. It means that it is a final state.

Originally Posted by Klaas van Aarsen
That transformation converts the Modified Turing Machine into and Ordinary Turing Machine.

First off, it shows that we will indeed have states with only 1 transition in the Ordinary Turing Machine.
That is, the state transition from $s_1'$ to $s$ in the 2nd rule is only defined for the symbol $x'$ and not for any other symbol.

The last part with $\delta(s,x)=\varnothing$ shows that if there are no state transitions at all in state $s$ of the Modified Turing Machine, that we transform it into a state that halts the Ordinary Turing Machine. It means that it is a final state.
Is the third line at each step of the proposed solution at #16 justified by the above?

5. Originally Posted by mathmari
Is the third line at each step of the proposed solution at #16 justified by the above?
No. It is not.
The above specifies the single transition I suggested and leaves out that third line at each step.

Originally Posted by Klaas van Aarsen
No. It is not.
The above specifies the single transition I suggested and leaves out that third line at each step.
At the given table of the modified TM of #1 do we not have $\delta (s_4 ,\sqcup )=\emptyset$ ?

So according to the definition do we not get the following ordinary TM ?

$(s_0,\sqcup,s_1,|,\ell)$ gets $(s_0,\sqcup,s_1',|), (s_1',\mid,s_1,\ell)$

$(s_0,\mid,s_0,|,\ell)$ gets $(s_0,\mid,s_0',|), (s_0',\mid,s_0,\ell)$

$(s_1,\sqcup,s_2,|,r)$ gets $(s_1,\sqcup,s_2',|), (s_2',\mid,s_2,r)$

$(s_1,\mid,s_1,|,r)$ gets $(s_1,\mid,s_1'',|), (s_1'',\mid,s_1,r)$

$(s_2,\sqcup,s_0,|,\ell)$ gets $(s_2,\sqcup,s_0'',|), (s_0'',\mid,s_0,\ell)$

$(s_2,\mid,s_3,|,r)$ gets $(s_2,\mid,s_3',|), (s_3',\mid,s_3,r)$

$(s_3,\sqcup,s_0,|,\ell)$ gets $(s_3,\sqcup,s_0''',|), (s_0''',\mid,s_0,\ell)$

$(s_3,\mid,s_4,|,r)$ gets $(s_3,\mid,s_4',|), (s_4',\mid,s_4,r)$

$(s_4,\mid,s_2,\sqcup,r)$ gets $(s_4,\mid,s_2'',\sqcup), (s_2'',\sqcup,s_2,r)$

Now we have that \begin{align*}\delta (s_1', \sqcup)&=\delta (s_0', \sqcup)=\delta (s_2', \sqcup)=\delta (s_1'', \sqcup)=\delta (s_0'', \sqcup)\\ & =\delta (s_3', \sqcup)=\delta (s_0''', \sqcup)=\delta (s_4', \sqcup)=\delta (s_4, \sqcup)\\ & =\delta (s_2'', \mid)=\emptyset \end{align*} or not?

So according to the last line of the definition we have to add for each of these ones an additional transition, which are the following:

\begin{align*}&(s_1', \sqcup, s_1',h), \ (s_0', \sqcup, s_0', h), \ (s_2', \sqcup, s_2', h), \ (s_1'', \sqcup, s_1'', h), \ (s_0'', \sqcup, s_0'', h)\\ & (s_3', \sqcup, s_3', h), \ (s_0''', \sqcup, s_0''', h), \ (s_4', \sqcup, s_4', h), \ (s_4, \sqcup, s_4, h), \ (s_2'', \mid, s_2'', h) \end{align*}

Or have I understood that wrong?