Need help to proof that NP⊆NP4

Question

NP4 = { L | There exists a non deterministic polynomial Turing machine M, such that for every x∈L, M accepts x on at least 4 paths in the computational tree of M on x. and for every x∉L,M accepts x on at most 3 paths in the computational tree of M on x. }

Answer 1

Take any language $L \in \mathsf{NP}$ and let $T$ be a non-deterministic polynomial-time Turing machine that decides $L$. Let $\Gamma$ and $q_A$ be the tape alphabet and the accepting state of $T$, respecitvely.

Create a new Turing machine $T'$ that has the same states of $T$ plus four additional states $q_1, q_2, q_3, q_4$. $T'$ is obtained from $T$ by:

• Replacing each transition $(q, a) \to (q_A, b, m)$ of $T$ (where $q$ is a state of $T$, $a,b \in \Gamma$, and $m \in \{\text{left}, \text{right}, \text{stay}\}$ is the movement of the head) with the four transitions $(q, a) \to (q_i, b, m)$ for $i \in \{1,2,3,4\}$.

• Adding the transitions $(q_i, a) \to (q_A, a, \text{stay})$ for $i \in \{ 1, 2, 3, 4\}$ and $a \in \Gamma$.

$T'$ is still a non-deterministic polynomial-time Turing machine that decides $L$. Moreover, if $x \in L$ then $T'$ with input $x$ has at least $4$ accepting paths, therefore $L \in \mathsf{NP4}$, which concludes the proof.