I am trying to figure out a reduction to prove $W[1]$-hardness for this, but I am having significant trouble. Here is the problem:
Bag Automaton: A non deterministic finite state automaton $M=(Q,I,s,F,d)$. $Q$ is the set of states, $I$ is the set of items, $s\in Q$ is the start state, $F\subseteq Q$ is the set of accepting states, $d\subseteq Q\times 2^I\times Q\times 2^I$ is the set of transitions, where $2^I$ is the set of all subsets of $I$. A computation of a Bag Automaton starts in $s$, with given item set $I' \subseteq I$. At each step the bag automaton in state $q$ and associated item set $I^* \subseteq I$ does a state transition $(q,A,q',B)$, $A\subseteq I^*$, which sets the state to $q'$ and the bag automaton's item set to $(I'-A) \cup B $. $M$ accepts if there is a sequence of transitions from $s$ to an $f \in F$.
Bag Automaton Computation Input: A Bag Automaton $M=(Q,I,s,F,d)$, a set $I' \subseteq I$ and a positive integer $k$.
Parameters: $k$
Question: Can $M$ accept on $I'$ by executing at most $k$ transitions?
I am almost certain a reduction from clique where the initial bag contains all vertices will work, but I cannot figure out how to formalize it.
