Multitape Turing Machine Reduces to Turing Machine

Theorem
Let $T$ be a multitape Turing machine.

Then, there exists a Turing machine $T'$ that accepts precisely the same language as $T$, and halts on exactly the inputs that $T$ does.

Proof
Let $k$ be the number of tapes in $T$.

Let:
 * $\Gamma_T$ be the symbols of $T$, with $B$ the blank symbol.
 * $\Sigma \subset \Gamma_T \setminus \set B$ be the input symbols of $T$.
 * $Q_T = \set {q_0, q_1, \dotsc, q_\beta}$ be the states of $T$, with $q_0$ the start state.
 * $\delta_T$ be the transition function of $T$.

Symbols
The symbols of $T'$ will be:
 * $\Gamma_{T'} = \set B \cup \Sigma \cup \paren {\Gamma_T \times 2}^k$

where the unions are assumed to be disjoint.

That is, a symbol is either:
 * $s \in \Sigma$, one of the input symbols from $T$, which are preserved so that the language remains identical
 * $B$, the blank symbol from $T$
 * A tuple $\tuple {s_1, h_1, s_2, h_2, \dotsc, s_k, h_k}$, where each $s_i$ is a tape symbol from $T$, and $h_i$ is either $0$ or $1$

The blank symbol of $T'$ is also $B$, and the input symbols are again $\Sigma$.

States
The states of $T'$ will be:
 * $Q_{T'} = 3 \cup Q_T \times \set R_R \times \Gamma_T^k \cup Q_T \times \set {W_L} \times \paren {\Gamma_T \times \set {L, R, S, M, N} } \cup Q_T \times \set {W_R} \times \paren {\Gamma_T \times \set {L, R, S, M, N} }$

where the unions are again assumed to be disjoint.

In other words, a state is either:
 * One of $\set {q'_0, q'_1, q'_2}$
 * A tuple $\tuple {q, R_R, s_1, s_2, \dotsc, s_k}$ where each $s_i \in \Gamma_T$
 * A tuple $\tuple {q, W_L, s_1, m_1, s_2, m_2, \dotsc, s_k, m_k}$, where each $s_i \in \Gamma_T$ and each $m_i \in \set {L, R, S, M, N}$
 * A tuple $\tuple {q, W_R, s_1, m_1, s_2, m_2, \dotsc, s_k, m_k}$, where each $s_i \in \Gamma_T$ and each $m_i \in \set {L, R, S, M, N}$

Note that the constants $\set {R_R, W_L, W_R}$ are arbitrary, and are just used to differentiate between distinct but similar-looking states.

The start state is $q'_0$, and the accepting states $F'$ are:
 * $\tuple {q_f, R_R, B, B, \dotsc, B}$

for each $q_f \in F$

Initialization
$\map {\delta'} {q'_0, s} = \tuple {q'_1, \tuple {s, 1, B, 1, \dotsc, B, 1}, R}$ for every $s \in \set B \cup \Sigma$

$\map {\delta'} {q'_1, s} = \tuple {q'_1, \tuple {s, 0, B, 0, \dotsc, B, 0}, R}$ for every $s \in \Sigma$

$\map {\delta'} {q'_1, B} = \tuple {q'_2, B, L}$

$\map {\delta'} {q'_2, s} = \tuple {q'_2, s, L}$ for every $s \in \Gamma_{T'} \setminus \set B$

$\map {\delta'} {q'_2, B} = \tuple {\tuple {q_0, R, B, B, \dotsc, B}, B, R}$

Reading
$\map {\delta'} {\tuple {q, R_R, r_1, r_2, \dotsc, r_k}, s} = \tuple {\tuple {q, R_R, r'_1, \dotsc, r'_k}, s, R}$
 * where:
 * $s = \tuple {s_1, h_1, \dotsc, s_k, h_k}$ for any $s_i \in \Gamma_T$ and $h_i \in \set {0, 1}$
 * $r'_i = \begin{cases} r_i & \text {if } h_i = 0 \\ s_i & \text {if } h_i = 1 \end{cases}$

$\map {\delta'} {\tuple {q, R_R, r_1, \dotsc, r_k}, B} = \tuple {\tuple {q', W_L, w_1, m_1, w_2, m_2, \dotsc, w_k, m_k}, B, L}$
 * where:
 * $\tuple {q', w_1, w_2, \dotsc, w_k, m_1, m_2, \dotsc, m_k} = \map \delta {q, r_1, r_2, \dotsc, r_k}$

Writing
$\map {\delta'} {\tuple {q, W_L, w_1, m_1, \dotsc, w_k, m_k}, \tuple {s_1, h_1, \dotsc, s_k, h_k} } = \tuple {\tuple {q, W_d, w_1, m'_1, \dotsc, w_k, m'_k}, \tuple {s'_1, h'_1, \dotsc, s'_k, h'_k}, d}$
 * where:
 * $d = \begin{cases} R & \text {if } \exists i : h_i = 1 \text { and } m_i = R \\ L & \text {otherwise} \end{cases}$
 * $s'_i = \begin{cases} w_i & \text {if } h_i = 1 \text { and } m_i \in \set {S, d} \\ s_i & \text {otherwise} \end{cases}$
 * $h'_i = \begin{cases} 1 & \text {if } m_i = M \\ 0 & \text {if } h_i = 1 \text { and } m_i = d \\ h_i & \text {otherwise} \end{cases}$
 * $m'_i = \begin{cases} M & \text {if } h_i = 1 \text { and } m_i = d \\ N & \text {if } m_i = M \\ N & \text {if } m_i = S \text { and } h_i = 1 \\ m_i & \text {otherwise} \end{cases}$

$\map {\delta'} {\tuple {q, W_L, w_1, m_1, \dotsc, w_k, m_k}, B} = \tuple {q', s', d}$
 * where:
 * $d = \begin{cases} L & \text {if } \exists i : m_i = M \\ R & \text {otherwise} \end{cases}$
 * $q' = \begin{cases} \tuple {q, W_L, w_1, N, \dotsc, w_k, N} & \text {if } d = L \\ \tuple {q, R_R, B, \dotsc, B} & \text {if } d = R \end{cases}$
 * $s' = \begin{cases} \tuple {B, h'_1, \dotsc, B, h'_k} & \text {if } d = L \\ B & \text {if } d = R \end{cases}$
 * $h'_i = \begin{cases} 1 & \text {if } m_i = M \\ 0 & \text {otherwise} \end{cases}$

$\map {\delta'} {\tuple {q, W_R, w_1, m_1, \dotsc, w_k, m_k}, \tuple {s_1, h_1, \dotsc, s_k, h_k} } = \tuple {\tuple {q, W_L, w_1, m'_1, \dotsc, w_k, m'_k}, \tuple {s_1, h'_1, \dotsc, s_k, h'_k}, L}$
 * where:
 * $h'_i = \begin{cases} 1 & \text {if } m_i = M \\ h_i & \text {otherwise} \end{cases}$
 * $m'_i = \begin{cases} N & \text {if } m_i = M \\ m_i & \text {otherwise} \end{cases}$

$\map {\delta'} {\tuple {q, W_R, w_1, m_1, \dotsc, w_k, m_k}, B} = \tuple {\tuple {q, W_L, w_1, m'_1, \dotsc, w_k, m'_k}, \tuple {B, h'_1, \dotsc, B, h'_1}, L}$
 * where:
 * $h'_i = \begin{cases} 1 & \text {if } m_i = M \\ 0 & \text {otherwise} \end{cases}$
 * $m'_i = \begin{cases} N & \text {if } m_i = M \\ m_i & \text {otherwise} \end{cases}$

Proof of Correctness
Let $A$ be an instananeous description of $T$ with state $q$, and let $A_i$ be the $i$-th term of $A$.

Let $A_i$ be written as:
 * $X_{i,m_i} \dotsm X_{i,1} q Y_i Z_{i,1} \dotsm Z_{i,n_i}$

Define the minimal tape encoding $\map {E_t} {A_i}$ of $A_i$ to be:
 * $\tuple {X_{i,m_i}, 0} \dotsm \tuple {X_{i,1}, 0} \tuple {Y_i, 1} \tuple {Z_{i,1}, 0} \dotsm \tuple {Z_{i,n_i}, 0}$

Define a tape encoding $\map {E_t^*} {A_i}$ of $A_i$ to be any finite string of the form:
 * $\tuple {B, 0} \dotsm \tuple {B, 0} \map {E_t} {A_i} \tuple {B, 0} \dotsm \tuple {B, 0}$

Define a multitape encoding $\map {E_m} A$ of $A$ to be:
 * $\tuple {\map {E_t^*} {A_1}_1, \dotsc, \map {E_t^*} {A_k}_1} \dotsm \tuple {\map {E_t^*} {A_1}_p, \dotsc, \map {E_t^*} {A_k}_p}$

where for some tape encodings of the $A_i$, all of the same length $p$.

Define the state encoding $\map {E_q} q$ of $q$ to be:
 * $\tuple {q, R_R, B, \dotsc, B}$

Finally, define a description encoding $\map E A$ of $A$ to be:
 * $\map {E_q} q \map {E_m} A$

Initialization Lemma
Let $T$ have instantaneous description $D$ when started on input $I$.

Let $T'$ have instantaneous description $D'$ under the same conditions.

Then, $D' \vdash^* \map E D$ under $T'$.

Proof
If $I$ is a finite string of length $1$, then $D_1 = q_0 Z$, $D_{i \mathop > 1} = q_0 B$, and $T'$ starts with description $q'_0 Z$.

If $I$ is the null string, it is equivalent to the above case, where $Z = B$.

By the transition function:

which is a description encoding of $D$.

If $I$ has length at least $2$, then $D_1 = q_0 Z_1 Z_2 \dotsm Z_n$, each other $D_i = q_0 B$, and $T'$ starts with description $q'_0 Z_1 Z_2 \dotsm Z_n$.

By the transition function:
 * $q'_0 Z_1 Z_2 \dotsm Z_n \vdash \tuple {Z_1, 1, B, 1, \dotsc, B, 1} q'_1 Z_2 \dotsm Z_n$

After applying the transition function $n - 1$ additional times, each time with the same transition rule since $Z_i \in \Sigma$, the description is:
 * $\tuple {Z_1, 1, B, 1, \dotsc, B, 1} \tuple {Z_2, 0, B, 0, \dotsc, B, 0} \dotsm \tuple {Z_n, 0, B, 0, \dotsc, B, 0} q'_1 B$

The next description is:
 * $\tuple {Z_1, 1, B, 1, \dotsc, B, 1} \tuple {Z_2, 0, B, 0, \dotsc, B, 0} \dotsm \tuple {Z_{n - 1}, 0, B, 0, \dotsc, B, 0} q'_2 \tuple {Z_n, 0, B, 0, \dotsc, B, 0}$

And, after applying the transition function another $n$ times:
 * $q'_2 B \tuple {Z_1, 1, B, 1, \dotsc, B, 1} \tuple {Z_2, 0, B, 0, \dotsc, B, 0} \dotsm \tuple {Z_n, 0, B, 0, \dotsc, B, 0}$

One more transition results in:
 * $\tuple {q_0, R_R, B, \dotsc, B} \tuple {Z_1, 1, B, 1, \dotsc, B, 1} \tuple {Z_2, 0, B, 0, \dotsc, B, 0} \dotsm \tuple {Z_n, 0, B, 0, \dotsc, B, 0}$

which is a description encoding of $D$, where each $\map {E_t^*} {D_{i \mathop > 1} }$ is postfixed with $n - 1$ copies of $\tuple {B, 0}$.

Therefore, in either case, the lemma holds.

Transition Lemma
Let $A$ be an instantaneous description of $T$, and suppose $A \vdash B$.

Then, for every $\map E A$, there is a $\map E B$ such that:
 * $\map E A \vdash^* \map E B$.

Proof
Let $\map E A = \tuple {q, R_R, B, \dotsc, B} \tuple {Z_{1,1}, h_{1,1}, \dotsc, Z_{k,1}, h_{1,1} } \dotsm {Z_{1,n}, h_{1,n}, \dotsc, Z_{k,n}, h_{k,n} }$

After $n$ steps, each of which uses the same formula above, the state will be:
 * $\tuple {Z_{1,1}, h_{1,1}, \dotsc, Z_{k,1}, h_{1,1} } \dotsm \tuple {Z_{1,n}, h_{1,n}, \dotsc, Z_{k,n}, h_{k,n} } \tuple {q, R_R, R_1, \dotsc, R_k} B$

But for each $1 \le i \le k$, there is a unique $j$ such that $h_{i,j} = 1$, by the definition of instantaneous description.

As $R_i$ is left unchanged in the transition, except for in that case:
 * $R_i = Z_{i,j}$

But from the definition of description encoding, it follows that $Z_{i,j}$ is the symbol $Y_i$ immediately to the right of $q$ in $A_i$.

Then, the state of $T'$ at the end of those transitions is:
 * $\tuple {q, R_R, Y_1, \dotsc, Y_k}$

Thus, the next step results in:
 * $\tuple {Z_{1,1}, h_{1,1}, \dotsc, Z_{k,1}, h_{k,1} } \dotsm \tuple {Z_{1,n - 1}, h_{1,n - 1}, \dotsc, Z_{k,n - 1}, h_{k,n - 1} } \tuple {q', W_L, Y'_1, \dotsc, Y'_n} \tuple {Z_{1,n}, h_{1,n}, \dotsc, Z_{k,n}, h_{k,n} }$

Suppose that with some description $A$ of $T'$, $d = R$.

Then, every $m_i = R$ such that $h_i = 1$, is replaced by $m'_i = M$.

Therefore, the $d = L$ move one step later returns to a symbol in which no $m_i = R$ has $h_i = 1$.

Thus, the step after is also $d = L$.

From this, it follows that every $d = R$ move is followed by at least two $d = L$ moves.

Therefore, there are at most $3n$ steps until $q \in Q$ is at the left end of the description string, scanning $B$, while in a $W_L$ state.

Fix an $i$ such that $1 \le i \le k$, and consider the tape $A_i$.

As before, there is exactly one $j_0$ such that $h_{i,j_0} = 1$.

Relabel the sequence $\sequence {Z_j }_{1 \le j \le n}$ as follows:
 * $X_m \dotsm X_1 Y Z_1 \dotsm Z_n$

where $Y$ corresponds with $Z_{j_0}$

This sequence corresponds exactly with the format of the instantaneous description.

By the definition of the transition function, $m_i$ will only change from $S$, $R$, or $L$ when $h_i = 1$.

Additionally, unless $m_i = S$, it will only do so when $d = m_i$.

We will analyze the move for tape $i$ case by case:


 * If $m_i = S$, then the first time $Y$ is scanned, regardless of $d$, $h'_i = 1$ and $s'_i = w_i$.
 * That corresponds to the step:
 * $X_m \dotsm X_1 q Y Z_1 \dotsm Z_n \vdash X_m \dotsm X_1 q' Y' Z_1 \dotsm Z_n$, which is correct.
 * $m'_i = N$, which ensures that the transition will not be activated again, even if $Y'$ is scanned a second time.


 * If $m_i$ is $R$ or $L$, the symbol $Y$ will eventually scanned such that $d = m_i$.
 * For, if $m_i = R$, $d = R$ by the transition function.
 * If $m_i = L$ and $d = R$, the next step has $d = L$, re-scanning $Y$, and $d = L$ again, as proven earlier.


 * If there is a symbol in the appropriate direction on $A_i$, its $h_{i,j'}$ will be updated to $1$ as $h_{i,j}$ is updated to $0$.
 * This corresponds with the appropriate transition, using $d = R$ as an example:
 * $X_m \dotsm X_1 q Y Z_1 \dotsm Z_n \vdash X_m \dotsm X_1 Y' q' Z_1 Z_2 \dotsm Z_n$
 * If there is no symbol in the direction, but there are additional $\tuple {B, 0}$ appended in that direction by $\map {E_t^*} {A_i}$, the same transition process introduces a new $B$, as it should.
 * If there is no extension already present either, that is when $Y = B$, the transition extends every tape by one instance of $\tuple {B, 0}$ in that direction, and performs the same steps.
 * If we are writing $B$ at the end of the tape, and moving away from the end, the $\tuple {B, 0}$ is considered as a part of the extensions introduced by $\map {E_t^*} {A_i}$, rather than a part of $A_i$ itself.

When $Y = B$ in a $W_L$ state, every update has been performed, except possibly an extension on the left, indicated by some $m_i = M$.

After performing this extension, if necessary, every tape move has been completed, so every $m_i = N$, and $Y = B$ again, as $T'$ moved left from the left end of the tape.

At this point, the description of $T'$ is:
 * $\tuple {q', W_L, w_1, N, \dotsc, w_k, N} B \tuple {s'_{1,1}, h'_{1,1}, \dotsc, s'_{k,1}, h'_{k,1} } \dotsm \tuple {s'_{1,n}, h'_{1,n}, \dotsc, s'_{k,n}, h'_{k,n} }$

One last step results in:
 * $\tuple {q', R_R, B, \dotsc, B} \tuple {s'_{1,1}, h'_{1,1}, \dotsc, s'_{k,1}, h'_{k,1} } \dotsm \tuple {s'_{1,n}, h'_{1,n}, \dotsc, s'_{k,n}, h'_{k,n} }$

which is a description encoding of the next description of $T$.

The only point in this process in which $\delta'$ could fail to exist is in finding $q'$ from the read symbols and previous state.

But this only happens when those symbols and state would cause $\delta$ to not exist in $T$.

Therefore, $T'$ halts exactly when an undefined $\delta$ appears in the sequence of steps from the input description of $T$, which is when it halts itself.

The accepting states of $T'$ are exactly those in any description encoding of an accepting state of $T$.

Therefore, $T'$ accepts when one of those states appears in the sequence of steps from the input description of $T$, or when it accepts itself.

Thus, $T'$ halts and/or accepts exactly when $T$ does.