The language of a PDA

There are two ways to define the language of a PDA $P=(Q,\Sigma,\Gamma,\delta,q_0,Z_0,F)$ ( $L(P)\subseteq \Sigma^*$). because there are two notions of acceptance:

Acceptance by final state

$$L(P) = \{ w \mid (q_0,w,Z_0) \vdash_P^* (q,\epsilon,\gamma) \wedge q\in F\}$$

That is the PDA accepts the word $w$ if there is any sequence of IDs starting from $(q_0,w,Z_0)$ and leading to $(q,\epsilon,\gamma)$, where $q\in F$ is one of the final states. Here it doesn't play a role what the contents of the stack are at the end.

In our example the PDA ${P_0}$ would accept $0110$ because $(q_0,0110,\char93 )\vdash_{P_0}^*(q_2,\epsilon,\epsilon)$ and $q_2 \in F$. Hence we conclude $0110\in L({P_0})$.

On the other hand since there is no successful sequence of IDs starting with $(q_0,0011,\char93 )$ we know that $0011\notin L({P_0})$.

Acceptance by empty stack

$$L(P) = \{ w \mid (q_0,w,Z_0) \vdash_P^* (q,\epsilon,\epsilon) \}$$

That is the PDA accepts the word $w$ if there is any sequence of IDs starting from $(q_0,w,Z_0)$ and leading to $(q,\epsilon,\epsilon)$, in this case the final state plays no role.

If we specify a PDA for acceptance by empty stack we will leave out the set of final states $F$ and just use $P=(Q,\Sigma,\Gamma,\delta,q_0,Z_0)$.

Our example automaton ${P_0}$ also works if we leave out $F$ and use acceptance by empty stack.

We can always turn a PDA which use one acceptance method into one which uses the other. Hence, both acceptance criteria specify the same class of languages.