How can I prove this dFa is minimal?

You can prove your DFA is minimal by proving that every state is both reachable and distinguishable.

To prove a state st is reachable, you must give a word (a possibly empty sequence of symbols) that goes from the starting state (q0 in your diagram) to state st. So for your diagram, you must give six words: one for each of q0, q1, q2, q3, q4, and X. I'll get you started:

To prove two states s1 and s2 are distinguishable, you must give a word that goes from s1 to an accepting state and from s2 to a rejecting state, or vice versa. So for your diagram, you need to provide 6 choose 2 = 15 words: one to distinguish q0 from q1, and one to distinguish q0 from q2, and one to distinguish q1 from q2, and so on. For example, the word a distinguishes q0 from q3, because a goes from q0 to q1 (a rejecting state), but a goes from q3 to q4 (an accepting state).

I'll get you started:

I would suggest two ways:

• Perform minimization (perhaps by a computer) and show the number of states is the same.
• Check that all states are reachable and distinguishable, i.e. For any state s there is a word w∈Σ∗ such that q0−→ws. For each pair (s1,s2) of states present a word w∈Σ∗ such that for s1−→ws′1 and s2−→ws′2 we have s′1 is an accepting state xor s′2 is an accepting state (i.e. s′1∈F⟺s′2∉F).