Language of all words accepted by a TM by at most $t$ steps is regular Let $M$ be a Turing machine, $\Sigma$ an alphabet, $t \in \mathbb{N}$

$L = \{ w \in \Sigma^* : w$ is accepted by $M$ by at most $t$ steps$\}$

I want to show that $L$ is regular.

My attempt:

I'm trying to do this directly by constructing a DFA for $L$.

Let $M = (Q, \Sigma, \Gamma, \delta, q_0, q_{accept}, q_{reject})$

Define $\Gamma_0 = \Gamma \cup \{\sqcap\}$ where $\sqcap \notin \Gamma$.

Set $Q_0 = Q \times \Gamma_0^{t+1} \times \{i\}_{i=1}^{t+1}$ a finite set.

For $q \in Q$, $(a_1,..,a_{t+1}) \in \Gamma^{t+1}$, $1 \leq i \leq t+1,$ $b \in \Sigma$:

Say that $\delta(q, b) := (q_1, b_0, R $ or $L)$ and that $\delta(q, a_i) := (q_2, a_0, R$ or $L)$. We define $\delta_0: Q_0 \times \Sigma \to Q_0$ by

$\delta_0(q, (a_1,..,a_{t+1}), i) =
  \begin{cases}
    (q_2, (a_1,..,a_0,..,a_{t+1}), i+1)       &amp; \quad \text{if } R \text{ and }a_i \in \Gamma\\
    (q_2, (a_1,..,a_0,..,a_{t+1}), i-1)       &amp; \quad \text{if } L \text{ and }a_i \in \Gamma\\
(q_1, (a_1,..,b_0,..,a_{t+1}), i+1)       &amp; \quad \text{if } R \text{ and }a_i = \sqcap \\
(q_2, (a_1,..,b_0,..,a_{t+1}), i-1)       &amp; \quad \text{if } L \text{ and }a_i = \sqcap\\
  \end{cases}
$

Where the  change of letter in the $t+1$ tuple occured in the $i^{th}$ spot.

**Edit: also make the machine loop upon entering the accept or reject states

Then consider the DFA $A = (Q_0, \Sigma, \delta_0, g_0, F)$ where $g_0 = (q_0, (\sqcap,...,\sqcap), 1)$ and $F = \{(q_{accept}, x, j) : x \in \Gamma_0^{t+1}, 1 \leq j \leq t+1\}$

Now intuitively I think that this machine accepts $L$, but am having a hard time proving it.

So, is it true that $L(A) = L$? If so can you point some directions on how to prove it?
The language $L$ only depends on the first $t$ letters of the input (since the Turing machine never gets to any of the other letters). This implies that it is regular, since
$$
L = (L \cap \Sigma^{&lt;t}) \cup (L \cap \Sigma^t) \Sigma^*,
$$
where $\Sigma^{&lt;t} = \bigcup_{i=0}^{t-1} \Sigma^i$. The right-hand side is regular since $L \cap \Sigma^{&lt;t}$ and $L \cap \Sigma^t$ are both finite, and so, regular.